Remarks on the nonlocal Dirichlet problem

We study translation-invariant integrodifferential operators that generate L\'{e}vy processes. First, we investigate different notions of what a solution to a nonlocal Dirichlet problem is and we provide the classical representation formula for distributional solutions. Second, we study the question under which assumptions distributional solutions are twice differentiable in the classical sense. Sufficient conditions and counterexamples are provided.


Introduction
The aim of this article is to provide two results on translation-invariant integrodifferential operators, which are not surprising but have not been systematically covered in the literature. Let us briefly explain these results in case of the classical Laplace operator.
The classical result of Weyl says the following. Assume D ⊂ R d is an open set, f ∈ C ∞ (D), and u ∈ D ′ (D) is a Schwartz distribution satisfying ∆u = f in the distributional sense, i.e. u, ∆ψ = ψ, f for every ψ ∈ C ∞ c (D). Then u ∈ C ∞ (D) and ∆u = f in D. This is the starting point for the study of distributional solutions to boundary value problems. Our first aim is to study distributional solutions to nonlocal boundary value problems of the form where L is an integrodifferential operator generating a unimodal Lévy process. Our second aim is to provide sufficient conditions such that distributional solutions u to the nonlocal Dirichlet problem are twice differentiable in the classical sense. In case of the Laplace operator, it is well known that Dini continuity of f : D → R, i.e. finiteness of the integral 1 0 ω f (r)/r dr for the modulus of continuity ω f , implies that the distributional solution u to the classical Dirichlet problem satisfies u ∈ C 2 loc (D). On the other hand, one can construct a continuous function f : B 1 → R and a distribution u ∈ D ′ (B 1 ) such that ∆u = f in the distributional sense, but u / ∈ C 2 loc (B 1 ). These observation have been made long time ago [24]. They have been extended to non-translation-invariant operators by several authors [11,30] and to nonlinear problems [28,14]. Note that there are many more related contributions including treatments of partial differential equations on non-smooth domains. In the present work we treat the simple linear case for a general class of nonlocal operators generating unimodal Lévy processes.
Let us introduce the objects of our study and formulate our main results. Let ν : The function ν induces a measure ν(dh) = ν(h) dh, which is called the Lévy measure. Note that we use the same symbol for the measure as well as for the density. We study operators of the form This expression is well defined if u is sufficiently regular in the neighbourhood of x ∈ R d and satisfies some integrability condition at infinity. We recall that for α ∈ (0, 2) and ν(dh) = c α |h| −d−α dh with some appropriate constant c α , the operator L equals the fractional Laplace operator −(−∆) α/2 on C 2 b (R d ). The regularity theory of such operators has been intensively studied recently. For instance, it is well known [3,33,18,34,32] that the solution of −(−∆) α/2 u = f with f ∈ C β belongs to C α+β provided that neither β nor α + β is an integer. The same result in more general setting is derived in [2].
Our standing assumption is that h → ν(h) is a non-increasing radial function and that there exists a Lévy measure ν * resp. a density ν * such that ν ν * and ν * (r) Cν * (r + 1), r r 0 The theorem above says that the distributional solution of (1.4) is unique up to a harmonic function. If, additionally, D is a Lipschitz domain and we impose some regularity, then the solution is unique. Boundedness of u, f , g would suffice, of course. It is obvious that one has to impose some regularity condition on f in order to prove uniqueness of solutions. Note that, in the case where L equals the fractional Laplace operator, similar results like Theorem 1.1 are proved in [6]. A result similar to Theorem 1.1 has recently been proved in [26]. The authors consider a smaller class of operators and concentrate on viscosity solutions instead of distributional solutions.
Variational solutions to nonlocal operators have been studied by several authors, e.g., in [17,35]. The problem to determine appropriate function spaces for the data g leads to the notion of nonlocal traces spaces introduced in [15]. It is interesting that the study of Dirichlet problems for nonlocal operators leads to new questions regarding the theory of function spaces.
The formulation of our second main result requires some further preparation. They are rather technical because we cover a large class of translation-invariant operators. The similar condition to the following appears in [7].
(A) and (1.2) are essential for proving that functions with the mean-value property are twice continuously differentiable, see Lemma 2.3. We emphasize that in general this is not the case and usually harmonic functions lack sufficient regularity if no additional assumptions are imposed. The reader is referred to [29,Example 7.5], where a function f with the mean-value property is constructed for which f ′ (0) does not exist.
Let G be a fundamental solution of L on R d (see (2.2) for definition). Note that in the case of the fractional Laplace operator G(x) = c d,α |x| α−d for d = α and some constant c d,α . In what follows we will assume the kernel G to satisfy the following growth condition: ) and there exists a non-increasing function S : (0, ∞) → [0, ∞) and r 0 > 0 such that  2) and the fundamental solution G satisfies (G). Let g ∈ L 1 (D c ) and f : D → R.
belongs to C 2 loc (D) and is unique up to a harmonic function (with respect to L ). The result uses quite involved conditions because the measure ν interacts with the Dinitype assumptions for the right-hand side function f . Looking at examples, we see that the two cases described in the theorem appear naturally. In the fractional Laplacian case (G(x) = c d,α |x| α−d ), finiteness of the expression 1/2 0 |G ′ (t)|t d−1 dt depends on the value of α ∈ (0, 2). We show in Section 6 that the conditions hold true when L is the generator of a rotationally symmetric α-stable process, i.e., when L equals the fractional Laplace operator. Note that Theorem 1.2 is a new result even in this case. We also study the more general class, e.g. operators of the form −ϕ(−∆), where ϕ is a Bernstein function. Note that in the theorem above we do not assume that g is bounded.
Remark 1.4. We emphasize that in the case of L being the fractional Laplace operator of order α ∈ (0, 2) and f ∈ C 2−α loc (D), it is not true that every solution of L u = f belongs to C 2 loc (D) as is stated in [1,Theorem 3.7]. A similar phenomenon has been mentioned in [3] and is visible here as well. Observe that in such case the integrals (1.5) and (1.6) are clearly divergent and consequently, Theorem 1.2 cannot be applied. We devote Section 5 to the construction of counterexamples for any α ∈ (0, 2).
The article is organized as follows: in Section 2 we provide the main definitions and some preliminary results. The proof of Theorem 1.1 is provided in Section 3. Section 4 contains several rather technical computations and the proof of Theorem 1.2. We discuss the necessity of the assumptions of Theorem 1.2 through examples in Section 5. Finally, in Section 6 we provide examples that show that the assumptions of Theorem 1.2 are natural.

Preliminaries
In this section we explain our use of notation, define several objects and collect some basic facts. We write f ≍ g when f and g are comparable, that is the quotient f /g stays between two positive constants. To simplify the notation, for a radial function f we use the same symbol to denote its radial profile. In the whole paper c and C denote constants which may vary from line to line. We write c(a) when the constant c depends only on a. By B(x, r) we denote the ball of radius r centered at x, that is B(x, r) = {y ∈ R d : |y − x| < r}. For convenience we set B r = B(0, r). For an open set D and x ∈ D we define δ D (x) = dist(x, ∂D) and diam(D) = sup x,y∈D |x − y|. The modulus of continuity of a continuous function f : D → R is defined by We say that a Borel measure is isotropic unimodal if it is absolutely continuous on R d \{0} with respect to the Lebesgue measure and has a radial, non-increasing density. Given an isotropic unimodal Lévy measure ν(dx) = ν(|x|) dx, we define a Lévy-Khinchine exponent ψ is usually called the characteristic exponent. It is well known (e.g. [29,Lemma 2.5 The family {p t } t>0 induces a strongly continuous contraction semigroup on C 0 (R d ) and whose generator A has the Fourier symbol −ψ. Using the Kolmogorov theorem one can construct a stochastic process X t with transition densities p t (x, y) Here P x is the probability corresponding to a process X t starting from x, that is P x (X 0 = x) = 1. By E x we denote the corresponding expectation. In fact, X t is a pure-jump isotropic unimodal Lévy process in R d , that is a stochastic process with stationary and independent increments and càdlàg paths (see for instance [36]).
One of the objects of significant importance in this paper is the potential kernel defined as follows: Clearly U (x, y) = U (y − x). The potential kernel can be defined in our setting if In particular, for d 3 the potential kernel always exists (see [36,Theorem 37.8]). If this is not the case, one can consider the compensated potential kernel for some fixed x 0 ∈ R d . If d = 1 and B 1 dξ ψ(ξ) < ∞, we can set x 0 = 0. In other cases the compensation must be taken with x 0 ∈ R d \ {0}. For details we refer the reader to [21] and to the Appendix A.
Slightly abusing the notation, we let W 1 be (2.1) for x 0 = (0, ..., 0, 1) ∈ R d . Thus, we have arrived with three potential kernels: U , W 0 and W 1 . Each one corresponds to a different type of process X t and an operator associated with it. In order to merge these cases in one object, we let For instance, in the case of L = ∆ we have The basic object in the theory of stochastic processes is the first exit time of X from D, Using τ D we define an analogue of the generator of X t , namely, the characteristic operator or Dynkin operator. We say a Borel function f is in a domain D U of Dynkin operator U if there exists a limit Here B → {x} is understood as a limit over all sequences of open sets B n whose intersection is {x} and whose diameters tend to 0 as n → ∞. The characteristic operator is an extension of A, that is D A ⊂ D U and U | D A = A. For a wide description of characteristic operator and its relation with the generator of X t we refer the reader to [16,Chapter V].
Instead of the whole R d , one can consider a process X killed after exiting D. By p D t (x, y) we denote its transition density (or, in other words, the fundamental solution of ∂ t − L in D). We have . We call P D (x, dz) a harmonic measure and its density P D (x, z) on R d \ D with respect to the Lebesgue measure -a Poisson kernel.
We define a Green function for the set D and the Green operator We note that G D (x, y) can be interpreted as the occupation time density up to the exit time τ D , G D [f ] -as a mean value of f (X t ). Using that we obtain G D [1] = E x τ D . For bounded sets D we have sup x∈R d E x τ D < ∞ ( [31], [8]). By the strong Markov property for any open Ω ⊂ D we have s., the well-known Hunt formula holds: In case of compensated potential kernels, a similar formula is valid, namely,

Definition 2.2.
We say that a function g : Here we assume that the integral is absolutely convergent. If g has the mean-value property in every bounded open set whose closure is contained in D then u is said to have the mean-value property inside D.
Clearly if f has the mean-value property inside D, then Uf = 0 in D.
In general, functions with the mean-value property lack sufficient regularity if no additional assumptions are imposed. In our setting, however, we can show that they are, in fact, twice continuously differentiable in D.  [23]). We say that a Borel function f belongs to the Kato class K if it satisfies the following condition This is one of three conditions discussed by Zhao in [38]. A detailed description of different notions of the Kato class and related conditions can be found in [23].
Let r = 2 sup x∈D |x|. Then D ⊂ B r and by [20,

Moreover, by (2.3)
Observe that A straightforward application of the proof of [12,Theorem 4.3] to the last term gives the claim.

By integration by parts
Observe that G ′ is of constant sign. Hence, both lim t→0 + G ′ (t)t d−1 and the integral are finite. In particular, integration by parts once again yields Both lim t→0 + G(t)t d−2 and the integral are positive. Hence, both must be finite. By [19, which is a contradiction. Now let d = 2. By the same argument and we conclude that the integral is finite. Hence, lim t→0 + G(t) < ∞. By [36, Theorems 41.5 and 41.9] we get the contradiction. Finally, for d = 1 we get that lim t→0 + G ′ (t) < ∞. It follows that lim sup t→0 + G(t)/t < ∞. Due to [4,Theorem 16] and [21, Lemma 2.14] we obtain that lim inf which is a contradiction, since lim sup x→∞ ψ(x)/x 2 = 0.

Lemma 2.7. Let D be bounded open and
. Since x 0 was arbitrary, the claim follows by induction.
A consequence of Lemma 2.7 is the following corollary.

Corollary 2.8. Let D be open and bounded and
The following lemma is crucial in one of the proofs.

Weak solutions
The aim of this section is to prove Theorem 1.1. For the fractional Laplacian related results are known, cf. [6,Section 3]. A similar result has recently been obtained in [10] using purely analytic methods instead of probabilistic ones exploited in [6]. When the generalization of these results to more general nonlocal operators is immediate, we omit the proof.
We consider Dynkin characteristic operator U. Since it is an extension of L and is translation-invariant, we obtain The following lemma is a generalization of [6, Theorem 3.9 and Corollary 3.10], where the fractional Laplace operator is considered.
Then u has the mean-value property inside D.
By the strong Markov property we may assume that D 1 is a Lipschitz domain. We claim that u has the mean-value property in The first integral is clearly absolutely convergent. We claim that it is also continuous as a function of x in D 1 . Indeed, by Lemma 2.9 it is continuous in Since the second term goes to 0 as x → x 0 (see [8, Lemmas 2.1 and 2.9]), by arbitrary choice of ǫ we get the claim.
Furthermore, from monotonicity of 1 ∧ ν * (h) we obtain Since u ∈ L 1 (R d ), (1.3) implies the absolute convergence of the second integral. Since by [8, Lemma 2.9 and Remark 2] E x τ D 1 ∈ C 0 (D 1 ), it is continuous as well. Hence u is continuous and has the mean-value property in Let h = u − u. We now verify that h ≡ 0 so that u = u has the mean-value property in D 1 . Since L u = 0 in D 1 , from Lemma 3.1 we have L h(x) = 0 for x ∈ D 1 . Observe h is continuous and compactly supported . Suppose it has a positive maximum at x 0 ∈ D 1 , then If not we can use the chain rule to get for any n ∈ N that h is constant on nsupp(ν) + x 0 and consequently h 0. Similarly, h must be non-negative.
Then u has the mean-value property inside D.
Proof. Let Ω ⊂⊂ D be a bounded Lipschitz domain. By [37] and the Ikeda-Watanabe formula we have that the harmonic measure P Ω (x, dz) is absolutely continuous with respect to the Lebesgue measure. Define ρ = (1 ∧ dist(Ω, D c ))/2 and let V = Ω + B ρ .
Moreover, since φ ǫ * u has the mean-value property in V ρ/2 , by Lemma 2.9 Let c = 2 sup x∈V |x|. Then from boundedness of P r and local integrability of u we get Furthermore, for |y| > c we have |z − y − s| > r, hence P r (z − y − s) P Br (0, z − y − s). From (1.2) and monotonicity of the Lévy measure we get It follows that φ ǫ * u are uniformly integrable with respect to the measure P Ω (x, z) dz in V ρ/2 . By the Vitali convergence theorem Using the fact that Thus u(x) = P Ω [u](x) for a. e. x ∈ Ω.
Combining Lemma 3.1 and Lemma 3.3 we obtain a following result. Proof. First assume f is continuous. Then by [16, Chapter V] we have Let φ ǫ , ǫ > 0, be a standard mollifier. Since U is an extension of L and is translationinvariant we get Passing ǫ → 0 we obtain Using mollification of f we get the claim.
is a solution of (1.4), which is bounded near to the boundary. Let U n ր D be a sequence of Lipschitz domains approaching D. We have . Note that by our additional assumptions on g and ν we have that P D [g] is well-defined. Furthermore, since f ∈ K(D \ V ), there exists n 0 ∈ N such that for n n 0 we have V ⊂ U n . From boundedness of u and Lemma 2.5 we get that h is bounded in D \ U n for n > n 0 and By [37, Theorem 1] we have

The sufficient condition for twice differentiability
In this section, we provide auxiliary technical results and the proof of Theorem 1.   Suppose f is a uniformly continuous function on D and H(x, y) is a continuous function for x, y ∈ D, x = y satisfying for some non-increasing function F : (0, ∞) → [0, ∞). If the following holds Indeed, clearly we have Since 1 [h,∞) (t)h/t 1, the claim follows by the dominated convergence theorem.
Proof. First note that by integration in polar coordinates one can check that the integral defining g actually exists. Set ǫ > 0. Let 0 < h < δ(D) and x i z be arbitrary fixed points in D such that |x − z| = h. Denote j(x, y) := H(x, y) (f (y) − f (x)). Observe that |g(x) − g(z)| is bounded by the sum of two integrals I 1 and I 2 of j(x, ·) − j(z, ·) over the sets D ∩ B(x, 2h) and D \ B(x, 2h) respectively. On D ∩ B(x, 2h) we have H(x, y) dy .

By the mean value theorem
for some x = θx + (1 − θ)z, θ ∈ (0, 1). Note that for y ∈ D \ B(x, 2h) we have |x − y| 2|x − z| = 2h > 0. It follows that | x − y| h and consequently |z − y| |z − x| + | x − y| 2| x − y|. Thus, Thus, by Remark 4.2 we see that I 3 < ǫ/3 for sufficiently small h. Finally, (4.1) implies In particular, I 4 ǫ/3 for sufficiently small h. It follows that |g(x) − g(z)| < ǫ, if h is sufficiently small. Thus, g is uniformly continuous. D and H(x, y) is a continuous function for x, y ∈ D, x = y such that D H(x, y) dy is continuously differentiable with respect to x. Assume there exists a non-increasing function F :

Lemma 4.3. Suppose f is a uniformly continuous function on
If the following holds B(x, 2r)) ω f (h, D).
is continuous on V s . Let x ∈ B(x, r). Integrating (4.6) with respect to x i from x i to x i we obtain a continuously differentiable function Ψ ǫ (x) with respect to x i with (4.6) being its derivative. Denote x = (x, x d ) and x = (x, x d ), wherex = (x 1 , ..., x d−1 ) and x d is fixed. the Fubini theorem and interchanging the order of integration yields Thus, for x ∈ B(x, r) the partial derivative ∂uǫ(x) ∂x d exists and is equal to (4.6). The same argument applies to any i = 1, ..., d. It remains to prove that (4.6) converges uniformly to (4.4), as ǫ → 0. Since f ǫ → f uniformly, as ǫ → 0, it is enough to prove the convergence of first integral in (4.6). Fix δ > 0. Since On the complement of B(x, γ) the function ∂H(x,y) which combined with (4.7) and arbitrary choice of δ ends the proof.
Now we are ready to prove Theorem 1.2. Observe I 3 has the mean-value property in D, thus, by Remark 2.1 and Lemma 2.3 it belongs to C 2 loc (D). Moreover, for x ∈ D from symmetry of G and (G) we obtain that both G and its first and second derivative are bounded either by S(δ D (x)) or S(δ D (x))/δ D (x), depending on the finiteness of 1/2 0 |G ′ (t)|t d−1 dt, and we are allowed to differentiate under the integral sign. Hence, it is enough to prove that g(x) := D G(x, y)f (y) dy is in C 2 loc (D). Fix i, j ∈ {1, ..., d}. Consider two cases.

Proof of Theorem 1.2. Let u be of the form
where the localization functions χ 1 and χ 2 are chosen in dependence of x. Note that in the integral defining w 2 , due to the function χ 2 and (G), integration w.r.t.
y takes place in a region where G and its derivative are bounded. Hence, from (G) we see that differentiating under the integral sign is justified. We obtain If we split w 1 into two integrals where D 1 ⊂ D is such that χ 1 D 1 ≡ 1 then the same argument can be applied to Next, observe that shows that the assumptions of Lemma 4.3 are satisfied with F = S. Note that here we use the additional assumption on G ′′′ . Thus,

Moreover, by Corollary 2.8 the function x → D G(x, y) dy is continuously differentiable and from (G) we see that (4.2) of Lemma 4.3 is satisfied for H(x, y)
We have proved that u ∈ C 2 loc (D). Then by [7,Lemma 4.7] the Dynkin characteristic operator U coincides with L . Hence u indeed is a solution of the problem (1.7). Now suppose u is another solution of (1.7). By Theorem 1.1 we find that it is of the form where h(x) = u + G D [f ](x) and U is any Lipschitz domain such that U ⊂⊂ D. Fix x 0 ∈ D. Then U 0 = B(x 0 , r) ⊂⊂ D for any r < dist(x 0 , D c ) and obviously U 0 is also Lipschitz. Hence, , thus u is twice continuously differentiable in the neighbourhood x 0 . Since x 0 was arbitrary, it follows that every solution of (1.7) is C 2 loc (D).

Counterexamples for the case "α + β = 2"
In this section we provide several counterexamples for Theorem 1.2. These examples are of the nature "α + β = 2", i.e., for α ∈ (0, 2) we give a function f ∈ C 2−α (D) for which the solution of the Dirichlet problem (5.1) is not twice continuously differentiable inside of D. In Section 6 we explain how the counterexamples can be modified in order to match the assumptions of Theorem 1.2.
where α ∈ (0, 2). It is known (see [6] or Theorem 1. y) is Green function for the operator ∆ α/2 and domain D solves (5.1). By the Hunt formula where G is the (compensated) potential for process X t whose generator is ∆ α/2 . Note that since E x G(X τ D , y) is C ∞ , the regularity problem is reduced to the regularity of the function x → g(x) = B(0,1) G(x, y)f (y) dy = G * f (x).

Case α ∈ (0, 1)
We follow closely the idea from the proof of Theorem 1.2 apart from the fact that at the end we will show that the last function w 3 is not continuously differentiable. From Lemma 2.7 we get . χ 1 and χ 2 in (5.2) are chosen for x 0 = 0. Put f (y) = ((y d ) + ) 2−α and calculate ∂ 2 ∂x 2 d g(x) w x = 0. Since in w 2 we are separated from the origin, it follows that If we split w 1 into then the same argument applies for w 4 . Therefore, it remains to calculate the derivative of w 3 . Observe that on B 1/4 we have f χ 1 ≡ f . To simplify the notation we accept a mild ambiguity and by h we denote, depending on the context, either a real number or a vector in R d of the form (0, ..., 0, h).

Case α = 1
Let d = 1. The compensated kernel is of the form G(x, y) = 1 π ln 1 |x−y| . Note that we cannot apply [11, Lemma 2.3] because (ii) does not hold. Instead write Let f be a Lipschitz function. By the mean value theorem It follows that Hence, Put f (y) = y + ln −β 1 + y −1 + , β ∈ (0, 1). It is easy to check that f is a Lipschitz function. Let h < 0. Since f (y) = 0 for y 0, from (5.5) we obtain We calculate the left-sided second partial derivative ∂ 2 ∂x d 2 g(x) in x = 0. Note that some of terms vanish and the remaining limit is f ((0, ..., 0, s)). We have where H s = {y : y d > s}. Denoteỹ = (y 1 , ..., y d−1 ). Then The Dominated Convergence Theorem implies Note that the function H under the integral sign is bounded on B 1 . It follows that By the Fatou lemma We have Thus,

Examples
In the last section we present some examples of operators L resp. corresponding Dirichlet problems that allow for an application of Theorem 1.2. In Example 6.1 we modify the considerations from Section 5 in order to match the assumptions of Theorem 1.2. In Example 6.2 we generalize to subordinated Brownian motion. Finally, in Example 6.4 we extend the above class and discuss the process which is assumed only to have the lower scaling property on the characteristic exponent.
Example 6.1 (fractional Laplace operator). Let X t be strictly stable process whose generator is the fractional Laplace operator −(−∆) α/2 . Let D be a bounded open set.
Example 6.2 (Subordinate Brownian motion). Let (B t , t 0) be a Brownian motion in R d and (S t , t 0) -a subordinator independent from B t , i.e. a Lévy process in R which stars from 0 and has non-negative trajectories. Process (X t , t 0) defined by X t = B St is called a subordinated Brownian motion. Denote by φ the Laplace exponent of S t : It is well known that φ is of the form The corresponding operator is of the form L = −φ(−∆) and we have ψ(ξ) = φ(|ξ| 2 ). An example of subordinated Brownian motion is the process from Example 6.1 with φ(λ) = λ α/2 , α ∈ (0, 2). Another example is geometric stable process with φ(λ) = ln 1 + λ α/2 , α ∈ (0, 2). Denote by G d (r) the potential of d-dimensional subordinated Brownian motion X t . From [13,Theorem 5.17] we have if d 3 and there exist β ∈ [0, d/2 + 1) and α > 0 such that φ −2 φ ′ satisfies weak lower and upper scaling condition at infinity with exponents −β and −α, respectively (see [13] It follows that That and (6.4) imply

By induction
Thus, the necessary conditions involving G and its derivatives hold true for S(r) = G d (r)/r 2 . Note that the density of Lévy measure of X t belongs to C ∞ . By [7,Lemma 7.4] the assumptions of Theorem 1.2 are satisfied with ν * ≡ ν if φ is a complete Bernstein function.
Proof. Observe that for d 3 the potential U always exists. By [19,Theorem 3] there exists c > 0 such that Our aim is to prove (G). By definition and isotropy of p t where byr = (0, ..., 0, r) ∈ R d . Since p t is radially decreasing, by the Tonelli theorem By [22,Theorem 5.6 and Corollary 6.8] Let us estimate |U ′ (r)|. We have The scaling property of h for |x| > h −1 (1/t) yields It follows that For α > 1/2 the integral is finite and we get

A Potential theory for recurrent unimodal Lévy process
In this appendix we establish a formula for the Green function for a bounded open set D in case of recurrent unimodal Lévy process X t . Contrary to the transient case, here the potential kernel U (x) = ∞ 0 p t (x) dt is infinite, so the classical Hunt formula has no application. Instead, one can define the λ-potential kernel U λ by setting Note that both U λ and G λ D exist. An analogue of the Hunt formula for G λ D holds, namely, for x, y ∈ D Proof. In the following part we introduce a mild ambiguity by denoting by 1, depending on the context, either a real number or the vector (0, ..., 0, 1) ∈ R d . Set x 0 = 1. Let f λ (r) = |x|<r dx ∞ 0 e −λu p u (x) du. We have Note that the integrand has a positive sign. Indeed, (p t (y)f (y) − p t (1)f (y)) dy > 0, since 4ǫ < 1. Furthermore, p t (1 + 4ǫ − y)f (y) dy = p t * f (1 + 4ǫ).
Hence, by the Fourier inversion theorem By the monotone convergence theorem and the fact that f (ξ) decays faster than any polynomial Hence, Since W 1 is radially decreasing and positive for |x| < 1, (A.1) implies that it may be infinite only for x = 0. It follows that W 1 is well defined for 0 < |x| 1. Similarly 0 W x 0 < ∞ for 0 < |x| |x 0 |.
It remains to notice that for |x| > |x 0 | we have 0 |W x 0 (x)| = −W x 0 (x) = W x (x 0 ) < ∞ by the first part of the proof.
Lemma A.2 allows us to introduce, following [5], [25], [9], a compensated potential kernel by setting for x ∈ R d \ {0} where x 0 ∈ R d \ {0} is an arbitrary but fixed point. From the proof of Lemma A.2 we immediately obtain the following corollary.
Proof. Let x, y ∈ D. Fix x 0 ∈ D c and observe that We want to pass with λ to 0. The limit of left-hand side is well defined and is equal to G D (x, y). From Lemma A.1 we get Moreover, from Lemma A.2 we obtain that It remains to show the convergence of the middle term of (A.4). Since U λ is radially decreasing, U λ (y − X τ D ) − U λ (x 0 ) is positive on the set {y ∈ R d : |y − X τ D | |x 0 |} and non-positive on its complement. By Lemma A.2 and the Monotone Convergence Theorem Observe that the left-hand side of (A.4) converges to G D so it is finite. The remaining integral on the right-hand side converges as well by the monotone convergence theorem, but since all the other terms are finite, it follows that the integral is also finite and we obtain lim which ends the proof.