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Potential Theory of Truncated Stable Processes

For any 0<alpha<2, a truncated symmetric alpha-stable process is a symmetric Levy process in R^d with a Levy density given by c|x|^{-d-alpha} 1_{|x|<1} for some constant c. In this paper we study the potential theory of truncated symmetric stable processes in detail. We prove a Harnack inequality for nonnegative harmonic nonnegative functions these processes. We also establish a boundary Harnack principle for nonnegative functions which are harmonic with respect to these processes in bounded convex domains. We give an example of a non-convex domain for which the boundary Harnack principle fails.

However in a lot of applications one needs to use discontinuous Markov processes which are not stable processes. Therefore we need to extend the known results on stable processes to other discontinuous Markov processes.
In [25] and [16], sharp estimates on the Green functions of killed relativistic stable processes in bounded C 1,1 domains were established. These estimates can be used to establish various properties of relativistic stable processes.
Another discontinuous Markov process, the censored stable process, was introduced and studied in [6]. Roughly speaking, for α ∈ (0, 2), a censored α-stable process in an open set D ⊂ R d is a process obtained from a symmetric α-stable Lévy process by restricting its Lévy measure to D. The censored process is repelled from the complement of the open set D because it is prohibited to make jumps outside D. Some potential theoretic properties of censored stable processes, such as Green function estimates, Martin boundary, and Fatou type theorem, were established recently (see [11], [12] and [21]).
In this paper we study yet another type of discontinuous Markov processes which we call truncated symmetric stable processes. For α ∈ (0, 2), a truncated symmetric α-stable process is a symmetric Lévy process in R d whose Lévy density l(x) coincides with the Lévy density of a symmetric α-stable process for |x| small (say, |x| < 1) and is equal to zero for |x| large (say, |x| ≥ 1). In other words, a truncated symmetric α-stable process is a symmetric Lévy process in R d with a Lévy density given by c|x| −d−α 1 {|x|<1} for some constant c. Truncated stable processes are very natural and important in applications where only jumps up to a certain size are allowed. One expects that many properties of the truncated stable processes should be similar to those of the symmetric stable processes, but some properties are very different. For instance, the boundary Harnack principle for symmetric stable processes is valid on any κ-fat set, while we will show that on non-convex domains the boundary Harnack principle for truncated stable processes might fail.
In some aspects, truncated stable processes have nicer behaviors and are more preferable than symmetric stable processes, for instance, by Theorem 25.17 of [26] we know that truncated stable processes have finite exponential moments. However, as we shall see later, in some other respects, truncated stable processes are much more difficult and more delicate to study than symmetric stable processes.
The starting point of our research on truncated stable processes was our attempt to establish a Harnack inequality for nonnegative harmonic functions of truncated stable processes. The recent developments in Harnack inequalities for discontinuous Markov processes were initiated in [1]. The method of [1] was extended in [28] to cover a large class of Markov processes. Two other methods for proving the Harnack inequality for discontinuous Markov processes were contained in [8] and [10]. However, none of the methods above apply to truncated stable processes. This gives another indication that truncated stable processes are pretty delicate to deal with.
Our strategy for studying truncated stable processes is as follows. First, we consider killed truncated stable processes on small sets and show that its Green functions are comparable to the Green functions of the corresponding killed symmetric stable processes. Then we study Poisson kernels for truncated stable processes on small sets in detail. Finally we prove the Harnack inequality and boundary Harnack principle for nonnegative harmonic functions of truncated stable processes by using properties of its Poisson kernels and some ideas in [4], [7] and [29].
In this paper we will always assume that d ≥ 2. The case of d = 1 can also be considered, but some arguments need to be modified. We leave this case to the interested reader.
In this paper, we use ":=" as a way of definition, which is read as "is defined to be". The letter c, with or without subscripts, signifies a constant whose value is unimportant and which may change from location to location, even within a line.

Stable Processes and Truncated Stable Processes
Throughout this paper we assume α ∈ (0, 2) and d ≥ 2. Recall that a symmetric α-stable process X = (X t , P x ) in R d is a Lévy process such that for every x ∈ R d and ξ ∈ R d .
The Dirichlet form (E, D(E)) associated with X is given by whereû(ξ) := (2π) −d/2 R d e iξ·y u(y)dy is the Fourier transform of u. As usual, we define E 1 (u, v) := E(u, v) + (u, v) L 2 (R d ) for u, v ∈ D(E). Then we have E 1 (u, u) = R d |û(ξ)| 2 (1 + |ξ| α )dξ, u ∈ D(E). (2.1) Another expression for E is as follows: Here Γ is the Gamma function defined by Γ(λ) := ∞ 0 t λ−1 e −t dt for every λ > 0. By a truncated symmetric α-stable process in R d we mean a symmetric Lévy process The Dirichlet form (Q, D(Q)) of Y is given by The Dirichlet form (Q, D(Q)) of Y can also be written as follows Similar to E 1 , we can also define Q 1 . Then we have By the change of variable y = x/|ξ|, we have from (2.2) Since 1 − cos( ξ |ξ| · x) behaves like |x| 2 for small |x|, it is easy to check that ψ(ξ) behaves like |ξ| 2 near the origin. Also we see that as |ξ| goes to infinity, the integral in the above equation goes to a positive constant. So ψ(ξ) behaves like |ξ| α near infinity. Therefore by the definition of D(E) and D(Q) we see that D(E) = D(Q). From now on we will use F to stand for D(E). Using (2.1), (2.3) and the fact above, we see that there exist positive constants c 1 and c 2 such that Therefore the capacities corresponding to the Dirichlet forms (E, F) and (Q, F) are comparable, hence we get that a set A has zero capacity with respect to (E, F) if and only if it has zero capacity with respect to (Q, F) and that a function u is quasi continuous with respect to the capacity of (E, F) if and only if it is quasi continuous with respect to the capacity of (Q, F). So when we speak of quasi continuous functions or sets of zero capacity, we do not need to specify the Dirichlet forms. For concepts and results related to Dirichlet forms, we refer our readers to [18].
It is well known that any function u ∈ F admits a quasi continuous version. From now on, whenever we talk about a function u ∈ F, we always use the quasi continuous version.
Using the asymptotic behavior of ψ and Proposition 28.1 in [26] we know that the process Y has a smooth density p Y (t, x, y). Since ψ(ξ) behaves like |ξ| 2 near the origin, it follows from Corollary 37.6 of [26] that Y is recurrent when d = 2 and transient when d ≥ 3. By using the smoothness of the density, one can easily check that, when d ≥ 3, the Green function of Y For any open set D, we use τ X D to denote the first exit time from D by the process X and use X D to denote the process obtained by killing the symmetric α-stable process upon leaving D. The process X D is usually called the killed symmetric α-stable process in D. The Dirichlet form of X D is (E, F D ), where F D = {u ∈ F : u = 0 on D c except for a set of zero capacity }.
For any u, v ∈ F D , Similarly, for any open set D, we use τ Y D to denote the first exit time from D by the process Y and use Y D to denote the process obtained by killing the process Y upon exiting D.
Using the continuity of p Y , it is routine (see, for instance, the proof of Theorem 2.4 [17] ) to show that Y D has a continuous and symmetric density p Y D (t, x, y). From this one can easily show that the Green function

Comparability between Green Functions of Stable and Truncated Stable Processes in small sets
In this section we take an open set D with diam(D) ≤ 1 2 , where diam(D) stands for the diameter of D. In this case, we have from (2.4) that J(x, y) = J Y (x, y) for all x, y ∈ D. So we can regard Y D as a Feynman-Kac transform of the process X D with the potential q D (x) : To get an upper bound for G Y D , we need to assume that D is κ-fat. We first recall the definition of κ-fat set from [29]. Note that all Lipschitz domain and all non-tangentially accessible domain (see [20] for the definition) are κ-fat. Moreover, every John domain is κ-fat (see Lemma 6.3 in [24]). The boundary of a κ-fat open set can be highly nonrectifiable and, in general, no regularity of its boundary can be inferred. Bounded κ-fat open set may be disconnected.
Suppose further that D is a κ-fat set. Since q D is bounded, one can use the 3G-type estimates for symmetric stable processes in [29] to check that q D belongs to the class S ∞ (X D ) there, thus it follows from [15] or [16] that G Y D (x, y) is continuous and there exists a positive constant C such that But the constant C above might depend on D. For later applications, we will need the constant C to be invariant under scaling and translation. First we consider the case of balls.
Proposition 3.2 There exists a positive constant r 0 ≤ 1 4 such that for all r ∈ (0, r 0 ] and a ∈ R d , we have Proof. Let B r := B(0, r) with r ≤ 1 4 . For any z ∈ B r , let (P z x , X Br t ) be the G Br (·, z)-transform of (P x , X Br t ), that is, for any nonnegative Borel functions f in B r , It is well known that there exists a positive constant C independent of r such that So there exists a positive constant r 0 such that for any r ∈ (0, r 0 ] and all x, z ∈ B r , where B(d, α) is the constant in (2.5). Hence by (2.5) and Khasminskii's lemma (see, for instance, Lemma 3.7 in [17]) we get that for r ∈ (0, r 0 ∧ 1 4 ] q(X Br s )ds) , x, z ∈ B(0, r), using the translation invariance property of our Green functions, we arrive at our desired result. ✷ The 3G-type estimate for symmetric stable processes on κ-fat open sets was proved in [29]. It is easy to see that the constant appearing in the 3G estimate depends only on the characteristics of the κ-fat open set and the diameter of the set. Moreover, by the scaling and translation invariant property of X, the constant is invariant under scaling and translation of D (3.4) By using (3.4) instead of (3.2) in the proof of Proposition 3.2, we immediately get the following result. Then there exists constant r 1 = r 1 (κ, R, α, d, diam(D)) ≤ 1 2 diam(D) such that for all r ∈ (0, r 1 ] and a ∈ R d , we have Proof. We omit the details. ✷

Harnack Inequality for Truncated Stable Processes
In this section we will prove a Harnack inequality for truncated stable processes. It is well-known (see Lemma 6 of [4]) that for any bounded Lipschitz domain D in R d (see Section 5 for the definition), The process X has a Lévy system (N, H) with N (x, dy) = A(d, −α)|x − y| −(d+α) dy and H t = t (see [18]). Using this and (4.1) we know that for every bounded Lipschitz domain D and f ≥ 0, we have Recall that for any ball B(x, r), we use τ X B(x,r) and τ Y B(x,r) to denote the first exit time from the ball B(x, r) by the processes X and Y respectively. Using Proposition 3.2, we can easily see that, for r ≤ r 0 , . Thus it follows from Theorem 1 of [30] that for any bounded Lipschitz domain D in R d we have (4.4) The process Y has a Lévy system ( and H Y t = t (see [18]). Using this and (4.4) we have the following result.
In the above integral, we will consider the smallest possible open set to integrate on. Let z r := (0, · · · , 0, 1 + r 2 ). The above integral is larger than or equal to and By a direct computation, one can show that (4.14) Putting (4.12) and (4.14) together, we get . We have proved the lower bound.
Lemma 4.6 Suppose that r < r 0 . Then there exists a constant c = c(d, α) > 0 such that for any z ∈ A(x 0 , r, 1 + r 2 ) and x 1 , for every open set B whose closure is a compact subset of D; (2) regular harmonic in D with respect to Y if it is harmonic in D with respect to Y and for each We define (regular) harmonic function with respect to X similarly. The next lemma is a preliminary version of the Harnack inequality for Y and it is an immediate consequence of Lemma 4.3.
for any nonnegative function u which is regular harmonic in B(x, r) and zero in B(x, 2r) c .
Now we are ready to prove a (scale-invariant) Harnack inequality for Y .
Then there exists a constant J > 0 depending only on d and α, such that for every nonnegative function u which is regular harmonic with respect to Y in B(x 1 , r) ∪ B(x 2 , r).
We split into two cases. First we deal with the case |x 1 − x 2 | < 1 4 r. In this case we have ∅ = B 1 ∩ B 2 ⊃ {x 1 , x 2 }. By Theorem 4.1 we have for any y ∈ B 1 ∩ B 2 and i = 1, 2, By Lemma 4.6, for some constant c 1 = c 1 (d, α). Note that by Theorem 4.1 Thus by the strong Markov property, we have .
for z ∈ A(x i , r 2 , r), and zero otherwise. Then we have from the above argument that for some constant c 2 = c 2 (d, α). Combining the two parts together, we get that . Now we consider the case when 1 4 . Thus, by Proposition 3.2 and Theorem 4.1, we have for w ∈ B(x 1 , r 8 ), we can apply the conclusion of the first case with x 2 = y and r replaced by 7r 8 to get that for some constant c 6 = c 6 (d, α). We have thus proved the right hand side inequality in the conclusion of the theorem. The inequality on the left hand side can be proved similarly. ✷ The Harnack inequality above is similar to the Harnack inequality (Lemma 2) for symmetric stable processes in [4], the difference is that we have to require that the two balls are not too far apart. Because our process can only make jumps of size at most 1, one can easily see that, without the assumption above, the Harnack inequality fails.
As a consequence of the theorem above we immediately get the following for any nonnegative function u which is harmonic in B(x, r).
Using this and a standard chain argument, we can get the following for any nonnegative function u which is harmonic in D

Boundary Harnack Principle for Truncated Stable Processes
In this section we will prove two versions of the boundary Harnack principle for truncated stable processes. Throughout this section, r 0 is the constant in Proposition 3.2.
We will use A α to denote the L 2 -generator of Y , and C ∞ c (R d ) to denote the space of continuous function with compact support. It is well-known that C ∞ c (R d ) is in the domain of A α and, for we have, by the symmetry of A α , for any x ∈ D and any nonnegative φ ∈ C ∞ c (R d ), In particular, if φ(x) = 0 for x ∈ D, we have Since G Y,λ D (x, y) increases to G Y D (x, y) as λ ↓ 0 and (A α − λ)φ is bounded for small λ, by letting λ ↓ 0 in the equation above, the dominated convergence theorem gives for any x ∈ D satisfying φ(x) = 0. Take a sequence of radial functions We claim that there exists a constant C > 0 such that for all r ∈ (0, 1), In fact, we have for some constant C 1 = C 1 (d, α) > 0. When D ⊂ B(0, r) for some r ∈ (0, 1), we get, by combining (5.3) and (5.4), that for any x ∈ D ∩ B(0, r 2 ), We have proved the following. Then for some constant C = C(d, α) > 0.
Recall that r 0 is the constant from Proposition 3.2.
for some constant c 11 = c 11 (d, α). This implies that for some constant c 12 = c 12 (d, α). It follows immediately that If r < r 0 , then for some constant C = C(d, α) > 0.
Proof. Without loss of generality, we may assume Q = 0 and x ∈ D ∩ B(0, 3 2 r). The left hand side inequality in the conclusion of the lemma is obvious, so we only need to prove the right hand side inequality. Since u is regular harmonic in D ∩ B(0, 2r) with respect to Y and u = 0 on B(0, 1 − r) c , we know from Lemma 5.4 that there exists σ ∈ ( 10r 6 , 11r 6 ) such that c 1 (d, α). On the other hand, by (4.7), we have that for some constant c 2 = c 2 (d, α). Note that |y − A| ≤ 2|y| on A(0, 3r 2 , 1 − r). Hence The following result is a boundary Harnack principle for nonnegative functions which are harmonic with respect to Y and vanish outside a small ball. The proof is similar to the proof in [29] but we spell out the details for the reader's convenience.
Theorem 5.6 Suppose that D is an open set, Q ∈ ∂D, r > 0 and that B(A, κr) is a ball in D ∩ B(Q, r). If 2r < r 0 , then for any nonnegative functions u, v in R d which are regular harmonic in D ∩ B(Q, 2r) with respect to Y and vanish in (D c ∩ B(Q, 2r)) ∪ B(Q, 1 − r) c , we have for some constant C = C(d, α) > 1.
Proof. Without loss of generality we may assume that Q = 0 and u(A) = v(A). Define u 1 and u 2 to be regular harmonic functions in D ∩ B(0, r) with respect to Y such that and r ≤ |x| < 3r 2 , and note that u = u 1 + u 2 . If D ∩ {r ≤ |y| < 3r 2 } is empty, then u 1 = 0 and the inequality (5.11) below holds trivially. So we assume D ∩ {r ≤ |y| < 3r 2 } is not empty. Then by Lemma 5.5, for some constant c 1 = c 1 (d, α). For x ∈ D ∩ B(0, r 2 ), we have Now using Lemma 5.3 we get for some constant c 2 = c 2 (d, α). Since 2r < r 0 , Corollary 4.10 implies that for some constant c 3 = c 3 (d, α). Therefore for x ∈ D ∩ B(0, r 2 ) From (5.9), the analogue of (5.10) for v and the assumption that u(A) = v(A), it follows that for x ∈ D ∩ B(0, r 2 ), for some constant c 4 = c 4 (d, α). Since u = 0 on B(0, 1 − r) c , we have that for x ∈ D ∩ B(0, r), for some constant c 5 = c 5 (d, α). Applying (5.12) to u and v and Lemma 5.5 to v and v 2 , we obtain for x ∈ D ∩ B(0, r 2 ), for some constant c 6 = c 6 (d, α). Combining (5.11) and (5.13), we have for some constant c 7 = c 7 (d, α). ✷ The theorem above applies to any κ-fat D, but the harmonic functions there are assumed to vanish outside a small ball. Thus the theorem above is very useful in studying properties of positive functions which are harmonic with respect to Y in κ-fat sets with diameters less than 1, and not very useful in the case when the diameters of the κ-fat sets are large.
Comparing the boundary Harnack principle above with the boundary Harnack principle for symmetric stable processes in [29], we notice that in the boundary Harnack principle above we assumed an extra condition that the functions vanish in B(Q, 1 − r) c . This extra condition is not purely technical. In the next section, we will give an example of a bounded non-convex domain showing that, without this extra condition, the boundary Harnack principle for Y fails.
In the remainder of this section, we will prove a boundary Harnack principle for nonnegative functions which are harmonic with respect to Y in bounded convex domains without assuming that they vanish outside small balls.
It is well-known that every convex domain is Lipschitz. Recall that a bounded domain D is said to be Lipschitz if there is a localization radius R 0 > 0 and a constant Λ > 0 such that for every Q ∈ ∂D, there is a Lipschitz function φ Q : and an orthonormal coordinate system y = (y 1 , · · · , y d−1 , y d ) : The pair (R 0 , Λ) is called the characteristics of the Lipschitz domain D. It is easy to see that D is κ-fat with characteristics (R 0 , κ 0 ) with some κ 0 = κ 0 (D).
In the remainder of this section we assume D is a bounded convex domain with the Lipschitz characteristics (R 0 , Λ) and the κ-fat characteristics (R 0 , κ 0 ).
For every Q ∈ ∂D and x ∈ B(Q, R 0 ) ∩ {y : The next result is well-known (see Lemma 6.7 in [17] for the Brownian motion case). Proof. The proof of this result is standard and we omit the details. ✷ Lemma 5.8 Let Q ∈ ∂D. Assume that B(A, κr) ⊂ D ∩ B(Q, r) for some 0 < r < 1 4 (r 0 ∧ R 0 ) and κ ∈ (0, 1 2 ]. Suppose that u ≥ 0 is regular harmonic in D ∩ B(Q, 4r) with respect to Y and u = 0 in D c , then Proof. Without loss of generality, we may assume Q = 0. Let φ(x) := φ 0 (x) and δ(x) := δ 0 (x). Since u is regular harmonic in D ∩ B(0, 4r) with respect to Y and D ∩ B(0, r) is bounded Lipschitz, by Theorem 4.1 we have which is equal to for some constant c 1 = c 1 (d, α) by (4.9). Note that |z − A| ≤ 2|z| for z ∈ A(0, r, 1 − r). Hence for some constant c 2 = c 2 (d, α). Now we will establish a different lower bound for u(A). Since u is regular harmonic in D ∩ B(0, 4r) with respect to Y and is zero outside of D, we have Let Ω κr := y ∈ D ∩ B(0, (4 − 1 2 κ)r) : δ(y) > 1 2 κr .
Proof. Without loss of generality, we may assume Q = 0. Let x ∈ D ∩ B(0, 3 2 r). Since u is regular harmonic in D ∩ B(0, 4r) with respect to Y , we have where σ is the constant from Lemma 5.4. By Lemma 5.4, We have for some constant c 1 = c 1 (d, α). Now we consider u 2 (x). We have We know from Lemma 4.4 that K Y B(0,σ) (x, z) ≤ c 2 r α for some constant c 2 = c 2 (d, α). Therefore for some constant c 3 = c 3 (d, α) since σ ∈ ( 10r 6 , 11r 6 ). Combining (5.17), (5.18) and Lemma 5.8, we have proved the lemma. ✷ We shall follow the "box method" of [7], originally developed by Bass and Burdzy ( [2] and [3]). Since we are going to use results of [7], we will closely follow their notations for the reader's convenience. Recall that for every Q ∈ ∂D and x ∈ B(Q, R 0 ) ∩ {y : We start with the following simple lemma. Proof. For y ∈ ∆ Q (Q, ar, br)) ∪ ∇ Q (Q, ar, br),
We observe that, for any Lipschitz function ϕ : R d−1 → R with Lipschitz constant Λ, its dilation ϕ r (x) := rϕ(x/r) is also Lipschitz with the same Lipschitz constant Λ. For any Q ∈ ∂D, let φ Q be the function in the definition of a Lipschitz domain. For any r > 0, put η = (2(4 + 3Λ)r)/R 0 and ψ = (φ Q ) η . Then it is easy to see that for any Q ∈ ∂D and r > 0, We can show that ∆ ψ ⊂ B(0, 1 2 R 0 ) by the same argument in the proof of Lemma 5.10. On the other hand, it is easy to see that ∆ ψ is a κ 1 -fat open set with κ 1 = κ 1 (Λ, R 0 ) for every Lipschitz function ψ : R d−1 → R with Lipschitz constant Λ. Therefore by Proposition 3.4, there exists positive constant r 2 such that for every Q ∈ ∂D and r ∈ (0, r 2 ], we have The next theorem is a boundary Harnack principle for bounded convex domains and it is the main result of this section. Maybe a word of caution is in order here. The boundary Harnack principle here is a little different from the ones proved in [4] and [29] in the sense that in the boundary Harnack principle below we require our harmonic functions to vanish on the whole complement of the bounded convex domain. However, this will not affect our application later since we are mainly interested in the case when the harmonic functions are given by the Green functions of the convex domain. Recall that D is a bounded convex domain with Lipschitz characteristics (R 0 , Λ) and κ-fat characteristics (R 0 , κ 0 ). Theorem 5.12 There exist constants c > 1 and r 3 > 0, depending on d, α and D such that for any Q ∈ ∂D, r < r 3 and any nonnegative functions u, v which are regular harmonic with respect to Y in D ∩ B(Q, 6(3 + 2Λ)r) and vanish in D c , we have for any x, y ∈ D ∩ B(Q, r 1 + Λ ).

✷
In the remainder of this section, we fix r 3 > 0 from Theorem 5.12. The following result is analogous to Lemma 5 of [4]. We recall from Definition 3.1 that for each z ∈ ∂D and r ∈ (0, R 0 ), A r (z) is a point in D ∩ B(z, r) satisfying B(A r (z), κ 0 r) ⊂ D ∩ B(z, r).
Proof. Without loss of generality, we may assume Q = 0. Fix r < r 3 and let Note that the B k 's are disjoint. So by the harmonicity of u, we have Since r < r 3 , (4.7) and Corollary 4.10 imply that For z ∈ A(0, r, 1 − r k ), by (4.8) and (4.9) we get for some constant c 1 = c 1 (d, α). For z ∈ A(0, 1 − r k , 1 + r k ), we use Lemma 4.4 and get for some constant c 8 = c 8 (D, d, α). ✷ Now the next theorem follows from Lemma 5.13, Theorem 5.12 and Lemma 5.14 (instead of using Lemma 5, Lemma 13 and Lemma 14 in [4] respectively) in very much the same way as in the case of symmetric stable process proved in Lemma 16 of [4]. We omit the details.
Theorem 5.15 There exist positive constants r 4 , M 1 , C and ν depending on D and α such that for any Q ∈ ∂D, r < r 4 and nonnegative functions u, v which are regular harmonic with respect to Y in D ∩ B(Q, M 1 r), vanish in R d \ D, and satisfy u(A r (Q)) = v(A r (Q)) > 0, we have x, y ∈ D ∩ B(Q, r).
In particular, the limit lim D∋x→w u(x)/v(x) exists for every w ∈ ∂D ∩ B(Q, r).
Using the results above and repeating the arguments in the proof of Theorem 4.1 of [29] we can get the following result identifying the Martin boundary of any bounded convex domain. For the definition and basic results on Martin boundary, one can see [23] and [29].
Theorem 5.16 Suppose that D is a bounded convex domain in R d . Then both the Martin boundary and the minimal Martin boundary of D with respect to Y coincide with the Euclidean boundary of D.
Proof. We omit the details. ✷

Counterexample
In this section, we present an example of a bounded non-convex domain on which the boundary Harnack principle for Y fails. Of course there are nothing special about the numbers 100 and 50 above, they are just two big numbers. Suppose the boundary Harnack principle (not necessarily scale invariant) is true for D at the origin. i.e., there exist constants R 1 > 0 and M 1 > 1 such that for any r < R 1 and any nonnegative functions u, v which are regular harmonic with respect to Y in D ∩ B(0, M 1 r) and vanish in D c , we have u(x) v(x) ≤ c u(y) v(y) for any x, y ∈ D ∩ B(0, r), where c = c(D, r) > 0 is independent of harmonic functions u and v. Choose an r 1 < R 1 with M 1 r 1 < 1/2 and let A := (0, 1 2 r 1 ). We define a function v by v(x) := P x Y τ D∩B(0,M 1 r 1 ) ∈ {y ∈ D; y d > 0} .
By definition v is regular harmonic in D ∩ B(0, M 1 r 1 ) with respect to Y and vanishes in D c .
Applying v above to (6.1), we have a Carleson type estimate at 0, i.e., there exists constant c 1 = c 1 (D, r 1 ) > 0 such that for any nonnegative function u which is regular harmonic with respect to Y in D ∩ B(0, M 1 r 1 ) and vanishes in D c we have u(A) ≥ c 1 u(x), x ∈ D ∩ B(0, r 1 ). (6.2) We will construct a bounded positive function u which is regular in D ∩ B(0, M 1 r) with respect to Y and vanishes in D c for which (6.2) fails.
For any n, let T Dn be the first hitting time of D n by the process Y . Note that since ∩ ∞ 1 D n = ∅, P A (τ D∩B(0,M 1 r 1 ) > T Dn ) → 0, as n → ∞.
u is a nonnegative bounded function which is regular harmonic in D ∩ B(0, M 1 r 1 ) with respect to Y and vanishes in D c . It also vanishes continuously on ∂D ∩ B(0, M 1 r 1 ). Note that by Theorem 4.1, P A Y τ D∩B(0,M 1 r 1 ) ∈ C n , τ D∩B(0,M 1 r 1 ) ≤ T Dn = P A Y τ D∩B(0,M 1 r 1 )\Dn ∈ C n = 0.
In the last inequality above, we have used (6.3)-(6.4). But by (6.2), u(A) ≥ c 1 sup x∈D∩B(0,r 1 ) u(x), which gives a contradiction. Thus the boundary Harnack principle is not true for D at the origin.
By smoothing off the corners of D, we can easily construct a smooth bounded non-convex domain on which the boundary Harnack principle fails for the truncated stable process Y .