On Harnack inequality for α-stable Ornstein-Uhlenbeck processes.

On Harnack inequality for α-stable Ornstein-Uhlenbeck processes.

Tomasz Jakubowski

Institute of Mathematics and Computer Science Wroc law University of Technology

Wybrze˙ze Wyspia´ nskiego 27, 50-370 Wroc law, Poland e-mail: Tomasz.Jakubowski@pwr.wroc.pl

Abstract

We consider the α-stable Ornstein-Uhlenbeck process in R^d with the generator L = ∆^α/2− λx · ∇_x. We show that if 2 > α ≥ 1 or α < 1 = d the Harnack inequality holds. For α < 1 < d we construct a counterexample that shows that the Harnack inequality doesn’t hold.

keywords: α-stable Ornstein-Uhlenbeck process, harmonic func- tions, Harnack inequality.

AMS Subject Classification: 60J75 60J45

1 Introduction

We consider the α-stable Ornstein-Uhlenbeck process in R^d with 0 < α < 2 given by the stochastic integral

X_t= e^−λtX₀+ Z t

0

e^−λ(t−s)d bX_s,

∗partially supported by KBN and MEN

where bX_t is the isotropic α-stable process (for more details see section Pre- liminaries). The process X_twas considered by Janicki and Weron in [12] and Samorodnitsky and Taqqu [13]. In the previous paper [11] the author gives the estimates of the mean first exit time of X_tfrom the ball. In this paper we are interested in the behaviour of harmonic functions related to the process Xt, which will be called L-harmonic (for definition see section Preliminar- ies). One of the fundamental questions concerning harmonic functions of the process X_t is the existence of the Harnack inequality which says

Harnack inequality. Let M > 0. There exists a constant C₁ depend- ing only on d, α, λ, M such that for any x₀ ∈ B(0, M ), r ∈ (0, 1/2) and any bounded function h nonnegative on R^d and L-harmonic in B(x0, 2r) it holds h(y) ≤ C₁h(x), x, y ∈ B(x₀, r/2) . (1) The answer to this question is positive for many Markov processes. It is well known that for Brownian motion Harnack inequality holds (see for example [8]). The first result on Harnack inequality for processes with jumps was obtained for the process bXt (for the proof see for example the paper of Bogdan, Byczkowski [3]). The proof bases on exact formula of density of harmonic measure of the ball. In paper [1] Bass and Levin prove Har- nack inequality for processes with symmetric L´evy measure comparable with L´evy measure of bX_t. Completely different methods based on so-called reg- ularization of Poisson kernel were used by Bogdan, St´os and Sztonyk in [4]

for α-stable processes on d-sets with some conditions on α. In [6] Chen and Kumagai generalize results from [4] on α ∈ (0, 2). In recent paper [14] Song and Vondracek applying methods used in [1] and [6] prove Harnack inequal- ity for a class of jump Markov process satisfying certain conditions. We note that in all papers above the constant C₁ doesn’t depend on M . Since the process X_t is not space invariant we cannot avoid this dependence which will be justified in section 5. The main results of this paper are

Theorem 1. For α < 1 < d Harnack inequality (1) doesn’t hold. It means that there exist a sequence of nonnegative functions h_n, L-harmonic in B(x₀, 1/n) and a sequence of points yn such that hn(yn) → 0 while n → ∞ and hn(x0) >

c > 0 for some constant c.

Theorem 2. For α ≥ 1 or α < 1 = d Harnack inequality (1) holds.

In [5] and [7] the authors consider the pure jump Girsanov transform and drift transform to show that the Green functions of the initial process and transformed process are comparable. From Theorem 1 we deduce that we cannot obtain the α-stable Ornstein-Uhlenbeck process from the process bX_t by pure jump Girsanov transform and drift transform at least in the case d > 1 > α and the mentioned papers do not apply to the process Xt.

The work is organized as follows. In section 2 and 3 we introduce basic definitions and facts. In section 4 we deal with the case α < 1. First, for d > 1, we construct a counterexample and show Theorem 1. Next, adapting some methods from [4], we show Theorem 2 for d = 1 > α. The section 5 is devoted to the case α ≥ 1. We prove Theorem 2 for α ≥ 1 by showing that Xt satisfies Song-Vondracek conditions from [14].

2 Preliminaries

By R^d we denote the d-dimensional Euclidean space with norm | · |. We denote a ∧ b = min(a, b). All constants are positive. In this paper we use a convention that constants denoted by small letters may differ in each lemma, while constants denoted by capital letters don’t change.

We denote by ( bXt, P^x) the isotropic α-stable L´evy process in R^d with index of stability α ∈ (0, 2) and characteristic function

E⁰e^{i bXt·ξ} = e^−t|ξ|α, ξ ∈ R^d, t ≥ 0 . (2) As usual, E^x denotes the expectation with respect to the distribution P^x of the process starting from x ∈ R^d. We always assume that sample paths of bX_t are right continuous and have left limits almost surely. ( bX_t, P^x) is a Markov process with transition densities p(t, x, y) satisfying Pb ^x( bXt ∈ D) = R

Dp(t, x, y)dy, and is strong Markov with respect to the so-called ”standardb filtration” (F_t, F ) (see [2]).

We consider the α-stable Ornstein-Uhlenbeck process in R^d with 0 < α <

2 given by the stochastic integral X_t= e^−λtX₀+

Z t 0

e^−λ(t−s)d bX_s, Xb₀ = 0 ,

where the integral is in Stieltjes sense. Hence we may rewrite this equation

in the form (see Corollary 5 in [11]) X_t= e^−λtX₀+ bX_t− λ

Z t 0

e^−λ(t−s)Xb_sds , Xb₀ = 0 , (3)

X_t is a Markov process (see [12], [13]) with transition density (see [11]) p(t, x, y) = bp

1−e^−αλt

αλ , e^−λtx, y

, t > 0, x, y ∈ R^d. (4) We have the following estimates for the function p(t, x, y) from [4]b

C₂⁻¹

 t

|x − y|^d+α ∧ t^−d/α



≤p(t, x, y) ≤ Cb ₂

 t

|x − y|^d+α ∧ t^−d/α



, (5) for t > 0, x, y ∈ R^d and some constant C₂ depending only on d and α.

Proposition 3. X_t is a Feller and strong Feller process. In particular the transition semigroup P_t generated by the process X_t is a strongly continuous contraction semigroup on C₀(R^d).

Proof. Let f ∈ L^∞(R^d) and x_n → x. Then

|P_tf (x_n) − P_tf (x)| =     Z

R^d

(p(t, x_n, y) − p(t, x, y))f (y) dy    

≤ ||f ||_∞ Z

B(xe^−λt,r)

|p(t, x_n, y) − p(t, x, y)| dy + ||f ||∞

Z

B(xe^−λt,r)^c

|p(t, x_n, y) − p(t, x, y)| dy

By (4) and (5) the second integral in the last inequality is arbitrary small for large r while the first integral tends to 0 by bounded convergence. Hence for f ∈ L^∞(R^d) the function P_tf (x) is continuous in x. Obviously P_tf (x) is also bounded which gives the strong Feller property. To show the Feller property it suffices to show that for any f ∈ C0(R^d) it holds

lim

|x|→∞P_tf (x) = 0 (6)

limt→0||P_tf − f ||∞ = 0 (7)

For any ε > 0 there exists r > 0 such that |f (y)| < ε if |y| > r. Hence

|P_tf (x)| ≤ ε + ||f ||∞

Z

B(0,r)

p(t, x, y) dy.

The last integral tends to 0 while |x| → ∞, proving (6). Now we have

||P_tf − f ||∞ ≤ sup

x∈R^d

Z

R^d

p(t, x, y)|f (y) − f (x)| dy .

Since f ∈ C₀(R^d), it is uniformly continous on R^d. For any ε > 0 there exists δ > 0 such that |f (x) − f (y)| < ε if |x − y| < δ. Moreover there is r > 0 such that |f (x)| < ε for |x| ≥ r. We note that for |x| < 2r and t < δ/(4rλ) we have |x − y| < δ if |xe^−λt− y| < δ/2. Therefore

sup

x∈B(0,2r)

Z

R^d

p(t, x, y)|f (y) − f (x)| dy

≤ ε + 2||f ||∞ sup

x∈B(0,2r)

Z

|xe^−λt−y|≥δ/2

t

|xe^−λt− y|^d+αdy

The second term in the last line tends to 0 when t → 0. For |x| > 2r and t ≤ ln(4/3)/λ we have e^−λt|x| > 3r/2. Hence

sup

x∈B(0,2r)^c

Z

R^d

p(t, x, y)|f (y) − f (x)| dy

≤ 2ε + (ε + ||f ||∞) sup

x∈B(0,2r)^c

Z

B(0,r)

t

|xe^−λt− y|^d+αdy .

Like in the previous case the last term tends to 0 when t → 0, which gives (6).

For an open set D ⊂ R^d, we define a Markov time τ_D = inf{t ≥ 0 : X_t∈ D^c}, the first exit time from D. If |D| < ∞ then P^x{τ_D < ∞} = 1, x ∈ R^d. In this case the P^x distribution of XτD is a probability measure on D^c. We define the function p_D(t, x, y) by

p_D(t, x, y) = p(t, x, y) − E^x(p(t − τ_D, X_τD, y)1{τD<t}) .

The function p_D(t, x, y) is a density of the semigroup P_t^D of the process X_t killed on exiting the set D given by

P_t^Df (x) = E^x(f (X_t)1t<τD) .

By standard argument (see [8]) we obtain that for any open set A it holds P^x(τD > t , Xt∈ A) =

Z

A

pD(t, x, y) dy , t > 0, x ∈ R^d (8) and the function p_D satisfies the Chapman-Kolmogorov equation

p_D(t, x, y) = Z

D

p_D(t − s, x, z)p_D(s, z, y) dz , t > s > 0, x, y ∈ R^d. (9) Using the method given in [4] we also get that p_D(t, x, y) > 0 for all t > 0 and x, y ∈ R^d.

Since for any r > 0 and x, z ∈ R^d we have E^x(τ_B(z,r)) < ∞ (see [11]), we get

E^x(τ_B) = Z ∞

0

P^x(τ_B > t) dt = Z ∞

0

Z

B

p_B(t, z, y) dy dt

= Z

B

Z ∞ 0

p_B(t, z, y) dt

 dy .

Hence we may define the Green function of the set B by G_B(x, y) =

Z ∞ 0

p_B(t, z, y) dt , and we have the equality

E^x(τ_B) = Z

B

G_B(x, y) dy . (10)

We will also use the estimates for E^x(τ_B(x,r)) established in [11]:

Lemma 4. Let 0 < r < M . There exists a constant C₃ depending on d, α, λ, M such that for all x ∈ R^d it holds

1 C₃



r^α∧ r

|x|



≤ E^x(τ_B(x,r)) ≤ C₃



r^α∧ r

|x|



Since X_t is a Markov process, it may be identified by its infinitesimal generator denoted in this paper by L. We will show that L = ∆^α/2− λx · ∇_x, where

∆^α/2ψ(x) = Ad,α P.V.

Z

R^d

ψ(y) − ψ(x)

|x − y|^d+α dy

is the infinitesimal generator of the process bX_t and A_d,α is some constant.

Lemma 5. For any bounded function f ∈ C_c^∞(R^d) we have limt→0

1 t

Z

R^d

p(t, x, y)f (y) dy − f (x)



= ∆^α/2f (x) − λx · ∇xf (x) , x ∈ R^d. Proof. Denote β = αλ and (t)_β = ^1−e_β^−βt. By (2) and (4) we have

Z

R^d

e^iy·ξp(t, x, y) dy = Z

R^d

e^iy·ξp((t)b _β, 0, y − e^−λtx) dy = e^{−(t)β|ξ|α+ie−λtx·ξ}. Hence

p(t, x, y) = 1 (2π)^d

Z

R^d

exp −(t)_β|ξ|^α+ ie^−λtx · ξ
e^−iξ·ydξ . (11) If we denote the Fourier transform of f by F (f )(ξ) = R

R^de^−ixξf (x) dx we obtain

Z

R^d

p(t, x, y)f (y) dy − f (x)

= 1

(2π)^d Z

R^d

Z

R^d

exp −(t)_β|ξ|^α+ ie^−λtx · ξ
e^−iξ·yf (y) dξ dy − f (x)

= 1

(2π)^d Z

R^d

e^{−(t)β|ξ|α+ie−λtx·ξ}F (f )(ξ) dξ − 1 (2π)^d

Z

R^d

e^iξ·xF (f )(ξ) dξ

Since f ∈ C_c^∞(R^d) then |F (f )(ξ)| ≤ c₁(1 + |ξ|)^−N for all N ∈ N and some constant c₁ depending on N . We have estimates

e^{−(t)β|ξ|α+i(1−e−λt)x·ξ}− 1 t

≤ |(t)_β|ξ|^α/t + i(1 − e^−λt)x · ξ/t| ≤ |ξ|^α+ λ|x||ξ| . Hence by dominated convergence theorem

limt→0

1 t

Z

R^d

p(t, x, y)f (y) dy − f (x)



= 1

(2π)^dlim

t→0

1 t

Z

R^d



e^{−(t)β|ξ|α+ie−λtx·ξ} − e^iξ·x

F (f )(ξ) dξ

= 1

(2π)^d Z

R^d

limt→0

1 t



e^{−(t)β|ξ|α+ie−λtx·ξ} − e^iξ·x

F (f )(ξ) dξ

= 1

(2π)^d Z

R^d

(|ξ|^α− iλx · ξ)e^ix·ξF (f )(ξ) dξ

= ∆^α/2f (x) − λx · ∇_xf (x) .

Therefore the infinitesimal generator of X_tis equal to L = ∆^α/2− λx · ∇_x. Definition 6. Let u ∈ B(R^d). We say that a function u is harmonic with respect to X_t in an open set D ⊂ R^d if

u(x) = E^xu(X(τU)), x ∈ U, for every bounded open set U satisfying U ⊂ D.

Since L is the infinitesimal generator of the process X_t, harmonic func- tions with respect to X_t will be called L-harmonic.

3 L´ evy system and Ikeda-Watanabe formula

One of the main tools in this paper is the following Ikeda-Watanabe formula expressing the probability that the process on exiting set B hits set A Proposition 7 (Ikeda-Watanabe formula). Let B be an open nonempty bounded set in R^d. For any Borel set A ⊂ ¯B^c it holds

P^x(X_τB ∈ A) = C_d,α Z

A

Z

B

G_B(x, z)

|y − z|^d+αdz dy , (12) where C_d,α is some constant.

To prove it we need only to check the following assumptions from [10].

(A1) The semigroup P_tf (x) =R

R^dp(t, x, y)f (y) dy maps C₀(R^d) into C₀(R^d) and is strongly continuous in t ≥ 0.

(A2) There exists a positive kernel n(x, E), x ∈ R^d, E ⊂ R^d a Borel subset, such that

(a) If dist(x, E) > 0 then n(x, E) < ∞.

(b) For f ∈ C0(R^d) and a bounded open set D with dist(D, supp f ) >

0,

sup

x∈D,t>0

t⁻¹P_tf (x) < ∞ and

lim

t→0⁺t⁻¹P_tf (x) = Z

R^d

f (y)n(x, dy) for every x ∈ D.

A family of measures n(x, E) is called a L´evy system of X_t. For any x, y ∈ R^d define N (x, y) = lim_t→0⁺t⁻¹p(t, x, y).

Lemma 8. We have N (x, y) = C_d,α|x − y|^−d−α, for some constant C_d,α. Proof. It suffices to show that (p(t, x, y) − p(t, x, y))/t → 0 when t → 0.b Denote β = αλ and (t)β = ^1−e_β^−βt. By (11) we have

p(t, x, y) −p(t, x, y)b

t = 1

(2π)^d Z

R^d

e^{−(t)β|ξ|α+ie−λtx·ξ}− e^{−t|ξ|α+ix·ξ}

t e^−iξ·ydξ

= Z

R^d

e^{−((t)β−t)|ξ|α−i(1−e−λt)x·ξ}− 1

(2π)^dt e^{−t|ξ|α+i(x−y)·ξ}dξ

Like in the proof of Lemma 5 we may enter with limit under the integral and we obtain the assertion of the Lemma.

For any Borel set E ⊂ R^d and x ∈ R^d we define n(x, E) =R

EN (x, y) dy, hence the function N (x, y) is a kernel of a L´evy system n(x, E). Now we prove Proposition 7.

Proof. Assumption (A1) is satisfied by Proposition 3. Clearly for any Borel set E ⊂ R^d it holds n(x, E) < ∞ if x 6∈ E. Let D be open bounded subset of R^d and let f ∈ C₀(R^d) be such that dist(D, supp f ) ≥ δ > 0. There exists t₀ > 0 such that dist(e^−λtx, supp f ) > δ/2 for all t < t₀ and x ∈ D. Hence by (4) and (5)

sup

x∈D,t0>t>0

t⁻¹P_tf (x) ≤ c₁ sup

x∈D,t0>t>0

Z

R^d

f (y) dy

|e^λtx − y|^d+α ≤ c₂||f ||∞δ^−α < ∞ Next we have sup_x∈D,t≥t0t⁻¹P_tf (x) ≤ ||f ||∞/t₀ < ∞. At the end, when x ∈ D, by bounded convergence

lim

t→0⁺t⁻¹Ptf (x) = lim

t→0⁺t⁻¹ Z

R^d

f (y)p(t, x, y) dy = Z

R^d

f (y)N (x, y) dy

= Z

R^d

f (y)n(x, dy) .

4 The case α < 1

In this section we will deal with α < 1. In this case the properties of the process X_t are diffrent from the case α ≥ 1. It may be seen from esti- mates of the mean first exit time from the ball (Lemma 4). When r → 0 then E^x(τ_B(x,r)) is comparable with r for any x 6= 0, while in the case α ≥ 1 we have asymptotics r^α. As a result of this fact we may construct a contrexemple to show the Theorem 1. To do it we will need some aux- iliary facts. Let a > 0 and x0 = (a, 0, 0, . . . , 0). Denote B = B(x0, r), S = {z ∈ R^d: pz₂²+ . . . + z²_d < r/8} and S− = {z ∈ S : z₁ < a + r/8}.

Lemma 9. Let α < 1. There exists a constant C4 such that P^x0(Xτ_B∩S ∈ S−) ≥ 1 − C₄r^1−α for 0 < r < a/4.

Proof. There exists a constant c1 independent from a such that |e^−λc1r/ax0− x₀| > 2r for r < a/4, which means that B(e^−λc1r/ax₀, r) ∩ B = ∅. Let t = c₁r/a. By L´evy inequality and (5) we have

P⁰(sup

s≤t

| bX_s| > r/16) ≤ 2P⁰(| bX_t| > r/16)

≤ 2c₂ Z

B(0,r/16)^c

t

|y|^d+α ≤ c₃r/a r^α = c₃

ar^1−α. Using the representation (3) we have

P^x0(|X_s− e^λsX₀| < r/8, ∀s ≤ t)

= P⁰(| bXs− λ Z s

0

e^−λ(s−u)Xbudu| < r/8, ∀s ≤ t)

≥ P⁰(| bXs| + λ Z s

0

e^−λ(s−u)| bXu| du < r/8, ∀s ≤ t)

≥ P⁰(sup

s≤t

| bXs|(1 + λ Z s

0

e^−λ(s−u)du) < r/8, ∀s ≤ t)

≥ P⁰(sup

s≤t

| bXs| < r/16) ≥ 1 − c3

ar^1−α Therefore putting C₄ = c₃/a

P^x0(Xτ_B∩S ∈ S−) ≥ P^x0(Xs∈ S−, ∀s ≤ t and Xt6∈ B)

≥ P^x0(|X_s− e^−λsX₀| < r/8, ∀s ≤ t)

≥ 1 − C₄r^1−α.

Denote A = {z ∈ S−∩ B^c: dist(z, ∂B) ≤ r/8}.

Corollary 10. There exists a constant C₅ such that P^x0(X_τB ∈ A) ≥ 1 − C5r^1−α for 0 < r < a/4.

Proof. By Ikeda-Watanabe formula and Lemma 4

P^x0(X_τB ∈ B(x₀, 9r/8)^c) = C_d,α Z

B(x0,9r/8)^c

Z

B

GB(x0, z)

|z − y|^d+αdz dy

≤ c₁ Z

B(x0,9r/8)^c

Z

B

G_B(x₀, z)

|z − x₀|^d+αdz dy

= c₂

r^αE^x0(τ_B) ≤ c₂C₃/|x₀|r^1−α Therefore by Lemma 9 we obtain

P^x0(X_τB ∈ A) ≥ P^x0(X_τB∩S ∈ A)

= P^x0(X_τB∩S ∈ S₋) − P^x0(X_τB∩S ∈ S₋\ A)

≥ P^x0(XτB∩S ∈ S−) − P^x0(XτB∩S ∈ B(x0, 9r/8)^c)

≥ 1 − C₄r^1−α− c₂C₃/|x₀|r^1−α = 1 − (C₄+ c₂C₃/|x₀|)r^1−α

We note that C₅ = c/a, where c is independent from a.

Let d > 1. For any x ∈ R^d we consider the rotation T_x around 0 such that T_xx = (|x|, 0, . . . , 0). Let y₀ ∈ B(x₀, r) be such that |y₀| = a and

|y₀ − x₀| = r/3. We note that T_y⁻¹

0 x₀ = y₀. By the same argument as in Lemma 9 we obtain that there is C₆ such that

P^y0(X(τ_B∩T⁻¹

y0 S) ∈ T_y⁻¹₀ S−) ≥ 1 − C₆r^1−α (13) for 0 < r < a/4.

Lemma 11. For r < 2a/9 we have A ∩ T_y⁻¹₀ S−= ∅.

Proof. Let x_t = (1 − t)x₀ and y_t = (1 − t)y₀. It suffices to show that

|x_t− y_t| > r/4 for 0 ≤ t ≤ 9r/8a. Since |x_t− y_t| is decreasing as a function of

t it suffices to verify inequality |(1 − 9r/8a)x₀− (1 − 9r/8a)y₀| > r/4. Since

|x₀− y₀| = r/3 we have

|(1 − 9r/8a)x₀− (1 − 9r/8a)y₀| = a − 9r/8

a |x₀− y₀| > a − a/4 a

r 3 = r

4.

Now we prove Theorem 1.

Proof of Theorem 1. Let x₀ = (1, 0, . . . , 0), r_n = 1/n, B_n = B(x₀, r_n), S_n = {z ∈ R^d: pz₂²+ . . . + z_d² < r_n/8} and S_n− = {z ∈ S_n: z₁ < 1 + r_n/8}. De- note An= {z ∈ Sn−∩B_n^c: dist(z, ∂Bn) ≤ rn/8}. Let hn(x) = P^x(Xτ_Bn ∈ A_n).

Each function h_n is L-harmonic in B_n. Indeed by strong Markov property for any open set E such that E ⊂ B_n it holds

E^x(h_n(X_τE)) = E^x P^XτE(X_τBn ∈ A_n)
= P^x(X_τBn ∈ A_n) = h_n(x) . Let y_n ∈ B(x₀, r_n) be such that |y_n| = 1 and |y_n− x₀| = r_n/3. By Lemma 11 and (13) we have

h(y_n) = 1−P^yn(X_τBn 6∈ A_n) ≤ 1−P^yn(X(τ_Bn∩T_y⁻¹_n S_n) ∈ T_y⁻¹_n S_n−) ≤ C₆r^1−α_n . But by Corollary 10 it holds

hn(x0) ≥ 1 − C5r^1−α_n .

If n → ∞ then h_n(y_n) → 0 and h_n(x₀) → 1, hence there is no constant c such that h_n(y_n) ≥ ch_n(x₀) for all n ∈ N. This shows that the Harnack inequality (1) does not hold for α < 1 < d.

Now we will show that the process on exiting the ball B(x₀, r) not con- taining the point 0 hits the boundary of B(x₀, r) with the probability bigger than 0. It means that the harmonic measure of the ball not containing 0 has a nontrivial singular part on the boundary of the ball.

Proposition 12. Let 0 6∈ B(x₀, r) and x ∈ B(x₀, r). Then P^x(X(τ_B(x0,r)) ∈

∂B(x0, r)) > 0.

Proof. Since the process is invariant by rotation around 0 we may and do suppose x₀ = (|x₀|, 0, . . . , 0). For any z ∈ R^ddenote δ(z) = dist(z, ∂B(x₀, r)).

Let S^∗ = S− ∩ {z ∈ B(x0, r) : δ(z) < |z|/4, C5(δ(z)/2)^1−α < 1}, where C5

is the constant from Corollary 10 corresponding to a = dist(0, B(x₀, r)). For any x ∈ B(x₀, r) denote

B_x = B(x, δ(x)/2), I_x = {tx : t ∈ [0, 1]},

A_x = {z ∈ B_x^c: dist(z, I_x) < δ(x)/8, dist(z, ∂B_x) < δ(x)/8}.

First suppose that x ∈ S^∗. By Corollary 10 and invariance by rotation around 0 it holds

P^x(X_τBx ∈ A_x) ≥ 1 − C₅(δ(x)/2)^1−α. (14) We note that C₅ depends on dist(0, B(x₀, r)). Define a stopping time

T =

(τ_BX0 if X(τ_BX0) ∈ A_X0,

∞ if X(τ_BX0) 6∈ A_X0,

We use a convention that X∞ = ∂, where ∂ is a cementary point. Let T⁰ = 0, T¹ = T and Tⁿ⁺¹ = T ◦ θ_Tⁿ, where θ is a shift operator acting on the space of trajectories by θt(ω)(s) = ω(t + s). Consider an event

Ω_n= {X_T ∈ A_X0, X_T² ∈ A_XT, . . . , X_Tⁿ ∈ A_X

T n−1} . We claim that for some constant 1/2 < p < 1 it holds

P^x(Ω_n) ≥

n

Y

k=1

1 − C₅



p^k−1δ(x) 2

1−α!

. (15)

Actually p will be equal 5/8. We prove (15) by induction. For n = 1 (15) becomes (14). Now suppose that (15) holds for n. For given x ∈ B(x0, r) we define a set Aⁿ_x = {y ∈ A_z: z ∈ Aⁿ⁻¹_x }, where A¹_x = A_x. We note that sup_z∈Axδ(z) ≤ ⁵₈δ(x). Hence by induction sup_z∈An

xδ(z) ≤ ⁵₈
n

δ(x). Let p = ⁵₈. Since A_{XT n} ⊂ Aⁿ_X

0 by strong Markov property, (14) and (15) we

obtain

P^x(Ωn+1) = E^x(P^{XT n}(XT ∈ AX0)1Ωn)

≥ E^x( inf

z∈A_{XT n} P^z(X_T ∈ A_X0)1Ωn)

≥ E^x( inf

z∈Aⁿ_X0P^z(Xτ_Bz ∈ Az)1Ωn)

= P^x(Ω_n) inf

z∈Aⁿ_xP^z(X_τBz ∈ A_z)

≥

n

Y

k=1

1 − C₅



p^k−1δ(x) 2

1−α!

z∈Ainfⁿ_x(1 − C₅(δ(z)/2)^1−α)

≥

n+1

Y

k=1

1 − C₅



p^k−1δ(x) 2

1−α!

Since sup_z∈An

xδ(z) ≤ ⁵₈
n

δ(x), hence the event Ω_n yields that the process X_t hasn’t exited the ball B(x₀, r), but went out from B(x₀, r − (5/8)ⁿδ(x)).

Therefore

Ω∞ = {XTⁿ ∈ AX_{T n−1}, ∀n ≥ 0} ⊂ {Xτ_B(x0,r) ∈ ∂B(x0, r)}

So we obtain that

P^x(Xτ_B(x0,r) ∈ ∂B(x0, r)) ≥ P^x(Ω∞) ≥

∞

Y

k=1

1 − C5



p^k−1δ(x) 2

1−α!

It remains to prove that the last expression is bigger than 0. We calculate the logarithm of it. Since C₅(δ(x)/2)^α−1 < 1 it holds

ln

∞

Y

k=1

1 − C₅



p^k−1δ(x) 2

1−α!!

=

∞

X

k=1

ln 1 − C₅



p^k−1δ(x) 2

1−α!

≥

∞

X

k=1

−2C₅



p^k−1δ(x) 2

1−α

= −δ(x)^1−αp^∗ > −∞ , where p^∗ = 2^αC₅/(1 − p^1−α). Therefore for x ∈ S^∗ it holds

P^x(X_τB(x0,r) ∈ ∂B(x₀, r)) > e^{−δ(x)1−αp∗} > 0. (16)

Now let x 6∈ S^∗. Let ρ = sup_z∈S∗δ(z). By strong Markov property and (16)

P^x(X_τB ∈ ∂B) ≥ E^x(P^X(τB\S∗)(X_τB ∈ ∂B); X_τB\S∗ ∈ S^∗)

> e^{−ρ1−αp∗}P^x(X_τB\S∗ ∈ S^∗) > 0 .

Remark 13. In fact we proved that the process, on exiting a ball B(x0, r), hits the set ∂B(x₀, r) ∩ T_x⁻¹

0 {z ∈ R^d: |z₁| ≤ |x₀|} with probability bigger than 0.

Denote σD = τD^c the first hitting time of the set D of the process Xt. Corollary 14. Let d = 1 > α. If y 6= 0 then P^x(σ{y} < ∞) > 0.

Proof. Since X_t is invariant by rotation around 0 we may and do assume that y > 0. If x > y then by Proposition 12 we have

P^x(σ{y} < ∞) ≥ P^x(σ{y} < τB(x,2|x−y|)) ≥ P^x(X_τB(x,x−y) = y) = c₁ > 0 , Now let x < y. Let c₂ = 2C₅^1/(α−1). We note that if z ∈ (y, 5y/4 ∧ y + c₂) then z − y < |z|/4 and C₅((z − y)/2)^1−α < 1. Hence by strong Markov property and (16)

P^x(σ{y} < τB(x,2|y−x|))

≥ E^x

P^{XτB(x,y−x)}(X_τ

B(X0,|X0−y|) = y); X_τB(x,y−x) ∈ (y, 5y/4 ∧ y + c₂)

≥ e^{−c1−α2 p∗}P^x(Xτ_B(x,y−x) ∈ (y, 5y/4 ∧ y + c2)) > e^{−c1−α2 p∗}c3 > 0 , where p^∗ > 0 is defined in the proof of Proposition 12.

From Corollary 14 we deduce that for d = 1 > α the process X_t is point recurrent except the point 0. We may prove that the process X_t hits 0 in finite time with probability 0 by considering the λ-potentials.

Harnack inequality for d = 1 > α.

Now we consider the case d = 1 > α. In dimension 1 there is no such geometric construction like in the case d ≥ 2 and we may prove the Harnack inequality (1). We will adapt the proof of Theorem 7.1 (transient case) in [4].

Althougth our process is point recurrent we cannot adapt the method given in [4] for recurrent case because P^x(σ{y} < τB(x,r|x−y|)) → 0 for any r > 0, when y → x. First we will need some lemmas

Lemma 15. Let r < 1. There exists a constant C7 such that for all x, y ∈ B(x₀, r/2) it holds

E^xτ_B(x0,2r) ≤ C₇E^yτ_B(x0,2r) (17) Proof. By Lemma 4

E^xτ_B(x0,2r) ≤ E^xτ_B(x,5r/2) ≤ C₂ 5r

2|x| ∧ (5r/2)^α



≤ c₁ r

|x| ∧ r^α



. (18) Since x, y ∈ B(x₀, r/2) we have |y| ≤ |x| + r < |x| + r^1−α ≤ 2(|x| ∨ r^1−α).

Hence r

|y| ∧ r^α ≥ r

2(|x| ∨ r^1−α) ∧ r^α ≥ r 2|x|∧ r^α

2 = 1 2

 r

|x| ∧ r^α

 . By (18) and Lemma 4 we obtain

E^xτ_B(x0,2r) ≤ 2c₁ r

|y| ∧ r^α



≤ c₂ 3r

2|y| ∧ (3r/2)^α



≤ c₂C₂E^yτ_B(y,3r/2)

≤ c₂C₂E^yτ_B(x0,2r)

For any open set D ⊂ R^d we define the function P_D(x, y) by P_D(x, y) =

Z

D

G_D(x, z)

|y − z|^d+αdz , x ∈ D, y ∈ Int(D^c) . If y ∈ D then we put P_D(x, y) = 0.

Lemma 16. Let d = 1 > α. There exists a constant C₈ such that for all r ∈ (0, 1), x ∈ B(x₀, r/2) and y ∈ B(x₀, r)^c it holds

P_B(x0,r)(x, y) ≤ C₈E^x(τ_B(x0,r))r⁻¹δ_B(x0,r)(y)^−α.

Proof. Without loss of generality we may suppose that x ≥ 0. Let y₀ be the one of the points x₀ + r, x₀ − r lying on interval xy. Denote B = B(x₀, r) and B^∗ = B ∩ B(y0, r/8). We note that

1 2C2

r

x ∧ r^α

≤ E^x(τ_B(x,r/2)) ≤ E^x(τ_B) ≤ E^x(τ_B(x,3r/2)) ≤ 3C₂ 2

r

x ∧ r^α . For z ∈ B \ B^∗ we have |y − z| ≥ r/8 hence

Z

B\B^∗

GB(x, z)

|z − y|^1+α dz ≤ 1 δ_B(y)^αr/8

Z

B

G_B(x, z) dz = 8E^x(τ_B)r⁻¹δ_B(y)^−α. So it suffices to show that G_B(x, z) ≤ c₁E^x(τ_B)r⁻¹ for z ∈ B^∗. We estimate the integral R∞

0 p_B(t, x, z) dt. Let c be some positive constant whose value will be determined later. In the estimates below we will often use the density estimates (4), (5). First we estimate

Z ∞ cr^α

p_B(t, x, z) dz = Z ∞

cr^α

Z

B

p_B(t − cr^α, x, u)p_B(cr^α, u, z) du dt

≤ Z ∞

cr^α

Z

B

p_B(t − cr^α, x, u)p(cr^α, u, z) du dt

≤ c₂ Z ∞

cr^α

P^x(τ_B > t − cr^α) 1 − e^αλcrα αλ

^−1/α dt

≤ c₃E^x(τ_B)r⁻¹.

where the constant c₃ depends on c. For t ∈ (0, cr^α) we will estimate pB(t, x, z) ≤ p(t, x, z). We will consider two cases.

Let x < 2r^1−α. For t < r/(8λx) it holds e^−λtx ∈ B(x, r/4) (see i.e.

Lemma 11 in [11]), hence |xe^−λt− z| > r/8. Since 2r/x > r^α, for c = 1/(16λ) we have

Z cr^α 0

p_B(t, x, z) dz ≤ Z cr^α

0

t

|xe^−λt− z|^1+α dt ≤ c₄r^2αr^−1−α ≤ c₅E^x(τ_B)r⁻¹. Now we suppose that x ≥ 2r^1−α. We may and do assume that y < x.

For t > 3r/(λx) it holds e^−λtx 6∈ B(x, 3κr/2) (see Lemma 11 in [11]), with a constant κ > 1. Hence e^−λtx 6∈ B(x0, κr) and we have for c > 3/(2λ)

Z cr^α 3r/(λ|x|)

p_B(t, x, z) dt ≤

Z cr^α 3r/(λ|x|)

c₆ t

|xe^−λt− x|^1+α dt ≤ c₇ x^1+α

Z cr^α 0

t^−αdt

≤ c8

r^α(1−α) x^1+α ≤ c8

r

2^αxr⁻¹ ≤ c9E^x(τB)r⁻¹.

It remains to estimate R3r/(λx)

0 p(t, x, z) dz. Let t₀ = _λ¹ln(x/z) so e^−λt0x = z.

We note that c10r/x ≤ ¹_λln(x/z) ≤ c11r/x hence t0 = a_x^r, where c10≤ a ≤ c11

and c₁₀, c₁₁ do not depend on x, z, r. Let t₁ = a2^1/α−1(r/x)^1/αx⁻¹. We note that t₁ ≤ _2x^ar = ^t₂⁰. We estimate

Z t0+t1

t0−t₁

p(t, x, z) dz ≤ c12

Z t0+t1

t0−t₁

t^−1/αdt ≤ c13(r/x)^−1/αt1 ≤ c14E^x(τB)r⁻¹. Next we estimate

Z t0−t1

0

p(t, x, z) dt ≤ c₁₅

Z t0−t1

0

(t0− t1)dt x^1+α(e^−λt− e^−λt0)^1+α

≤ c16

r x^2+α

Z t0−t₁ 0

dt (t₀− t)^1+α

≤ c₁₆ r x^2+α

Z ∞ t1

ds s^1+α

= c₁₆r

αx^2+α(t₁)^−α ≤ c₁₇E^x(τ_B)r⁻¹. In a similar way we estimateR3r/(λx)

t0+t1 p(t, x, z) dt. Therefore we obtained that for some constant c₁₆ we have G_B(x, z) ≤ c₁₆E^x(τ_B)r⁻¹, where z ∈ B^∗. Integrating R

B^∗ dz

|z−y|^1+α we get the assertion of the lemma.

Denote by ω^x_D the distribution of the process starting from x and exit- ing the set D. From Ikeda-Watanabe formula ω^x_B on Int(B^c) has a density P_B(x, y) = R

B

GB(x,z)

|y−z|^d+αdz. Since X_t hits the boundary of B_t = B(x₀, t) on exiting B_t the measure ω_B^x_t has Lebesgue decomposition,

ω_B^x_t(dy) = PBt(x, y)dy + ptδx0−t(dy) + qtδx0+t(dy) , (19) where p_t = P^x(X_τ

B(x0,t) = x₀ − t) and q_t = P^x(X_τ

B(x0,t) = x₀ + t). In next lemma we will show that if 0 ∈ B(x₀, t) then p_t= q_t= 0.

Lemma 17. If 0 ∈ B(x₀, r) then P^x(X_τB(x0,r) ∈ {x₀+ r, x₀− r}) = 0.

Proof. Since ω^x_D is absolutely continuous with respect to Lebesgue measure on Int(D^c) the process hits x₀+ r on exiting B(x₀, r) only if the event

Ω₀ = {X_τ((x0+r)/2,(x0+r+X0)/2) ∈ [(x₀+ r + X₀)/2, x₀+ r) , X₀ > 0}

happens infinitely many times. It means that the process X_t jumps out from every B(x₀, t), t < r before it exits B(x₀, r). By the same method as in he proof of Proposition 12 we show that P^x(Xτ((x0+r)/2,(x0+r+X0)/2) ∈ (0, (x + r)/2)) > c₁ > 0 for all x ∈ (0, x₀ + r) and some constant c₁ in- dependent of x. Hence P^x(Ω₀) < 1 − c₁ for all x ∈ (0, x₀ + r). Therefore P(Ω⁰ happens infinitely many times) = 0 and we get pr = 0. We use the same reasoning to obtain q_r = 0.

Corollary 18. From Lemma 17 and Proposition 12 we deduce that for x₀ >

t > 0 we have q_t = 0 and p_t > 0 in (19).

Using argument from [4] (proof of Theorem 7.1 in recurrent case) we prove (1) for |y| < |x|.

Lemma 19. Let d = 1 > α. There exists a constant C₉ such that for all functions h nonnegative on R^dand L-harmonic in the ball B(x₀, r) satisfying M > x₀ > 4r it holds

h(x) ≥ C₉h(y), x ∈ (x₀− r, x₀), y ∈ (x₀− r, x) .

Proof. Denote δ_t(x) = dist(x, ∂B(x₀, t)). At first let r < c₁ = 2C₅^1/(α−1). Since x₀ > 4r then for x ∈ (x₀− t, x₀) and 0 < t ≤ r we have |x|/4 > δ_t(x) hence from (16) it holds

P^x(X_τ

B(x0,t) = x₀− t) > e^{−δt(x)1−αp∗} > e^{−c1−α1 p∗} > 0 , x ∈ (x₀− t, x₀) . Therefore for y ∈ (x₀− r, x) we have

h(x) = E^x(h(X_{τB(x,|x−y|)})) ≥ E^x(h(X_{τB(x,|x−y|)}); X_{τB(x,|x−y|)} = y)

≥ e^{−c1−α1 p∗}h(y) .

Now suppose r > c₁. Let n ∈ N be such c1n ≥ r > c₁(n − 1). By the chain argument h(x) ≥ e^{−c1−α1 np∗}h(y) for y ∈ (x0 − r, x). Since r < M we have n ≤ M/c₁ + 1. Therefore we obtain assertion of the lemma with C₉ = e^{−c−α1 (M +c1)p∗}.

Now we may prove Theorem 2 for d = 1 > α.

proof of Theorem 2 for d = 1 > α. Denote B = B(x₀, 2r). Let h be a non- negative function on R^d, L-harmonic in B(x₀, 2r) and x, x^∗ ∈ B(x₀, r/2). We assume that x0 > 0. Let

h₁(z) = E^z(h(X_τB); X_τB ∈ B(x₀, 3r)), h₂(z) = E^z(h(X_τB); X_τB 6∈ B(x₀, 3r)).

We have h = h₁ + h₂ and h₁, h₂ are L-harmonic in B. Hence by Ikeda- Watanabe formula and Lemma 15

h₂(x) = Z

B(x0,3r)^c

h₂(y)P_B(x, y) dy

≤ c₁E^x(τ_B) Z

B(x0,3r)^c

h2(y)

|x − y|^1+α dy

≤ c₂E^x

∗(τ_B) Z

B(x0,3r)^c

h₂(y)

|x^∗− y|^1+α dy

≤ c₃ Z

B(x0,3r)^c

h₂(y)P_B(x^∗, y) dy = c₃h₂(x^∗) .

It is enough to prove an analogous inequality for h₁. Let R = B(x₀, 3r) \ B(x₀, r). We note that supp h₁ ⊂ B(x₀, 3r) and for x ∈ B(x₀, r/2) and y ∈ R we have |x − y| ≤ 4r. Denote B_t = B(x₀, t). By (19) and Corollary 18 we have

ω^x_Bt(dy) = P_Bt(x, y)dy + p_tδ_x0−t(dy) . with p_t = P^x(X_τ

B(x0,t) = x₀ − t). We note that if 0 ∈ B(x₀, t) then p_t = 0.

Hence we get the inequality h₁(x^∗) = E^x∗h₁(X_τB(x0,3r/4))

= Z

B(x0,3r/4)^c

P_B(x0,3r/4)(x^∗, y)h₁(y) dy + p_3r/4h₁(x₀− 3r/4)

≥ c₄ Z

R

Z

B(x0,3r/4)

G_B(x0,3r/4)(x^∗, z)

|z − y|^1+α dz h₁(y)dy + p_3r/4h₁(x₀− 3r/4)

≥ c₅E^x(τ_B(x0,r))r^−1−α Z

R

h₁(y) dy + p_3r/4h₁(x₀− 3r/4) . (20) In the last line we use Lemma 4 to estimate E^x(τ_B(x0,3r/4)) ≥ cE^x(τ_B(x0,r)).

Since ω_B^x_t(A) = ω^x_Bt(A ∩ B_t^c) for all measurable A we have for all t ∈ [r, 2r]

h₁(x) = Z

B(x0,r)^c

h₁(y)ω_B^x_t(dy) .