Harmonic functions for a class of integro-differential operators
aa r X i v : . [ m a t h . P R ] D ec Harmonic functions for a class of integro-differential operators.
Mohammud Foondun
Abstract
We consider the operator L defined on C ( R d ) functions by L f ( x ) = 12 d X i,j =1 a ij ( x ) ∂ f ( x ) ∂x i ∂x j + d X i =1 b i ( x ) ∂f ( x ) ∂x i + Z R d \{ } [ f ( x + h ) − f ( x ) − ( | h |≤ h · ∇ f ( x )] n ( x, h ) dh. Under the assumption that the local part of the operator is uniformly elliptic and with suitableconditions on n ( x, h ), we establish a Harnack inequality for functions that are nonnegative in R d and harmonic in a domain. We also show that the Harnack inequality can fail without suitableconditions on n ( x, h ). A regularity theorem for those nonnegative harmonic functions is alsoproved. Subject Classification:
Primary 60J75; Secondary 60H60
Keywords:
Harnack inequality, Harmonic functions, jump processes, integro-differential operators
Author’s address:
Department of Mathematics, The University of Utah, 155 S. 1400 E. Salt LakeCity, UT 84112–0090, USA.
Email: [email protected] Introduction
Researchers are increasingly using integro-differential operators (or equivalently, processes withjumps) to model problems from economics and the natural sciences. For instance, geometricBrownian motion is a standard model for a stock price. But this model is sometimes not satis-factory because it does not take into account sudden shifts of the stock price. To model this, onewould like to use a process with some jumps, to represent the stock price. So understanding theproperties of those operators is very important.The purpose of this paper is to consider functions that are harmonic with respect to theoperator L , where L f ( x ) = 12 d X i,j =1 a ij ( x ) ∂ f ( x ) ∂x i ∂x j + d X i =1 b i ( x ) ∂f ( x ) ∂x i + Z R d [ f ( x + h ) − f ( x ) − ( | h |≤ h · ∇ f ( x )] n ( x, h ) dh (1.1)is defined on C ( R d ) functions. This is a typical example of a non-local operator, in the sensethat the behavior of the harmonic function at a point depends on values of the function at pointssome distance away rather than just at nearby points. In probabilistic terms, the local part of L corresponds to the continuous part of the process while the non-local part controls the jumps ofthe process. The jump kernel n ( x, h ) represents the intensity of jumps from a point x to the point x + h and will be assumed to be nonnegative.We prove a Harnack inequality as well as a regularity theorem for harmonic functions withrespect to the operator L without assuming any continuity of the coefficients a ij , b i and of thekernel n ( x, h ). We say that a function u is harmonic with respect to L in a domain D if L u = 0in D ; we give a precise definition in Section 2. Roughly speaking, the Harnack inequality statesthat the values of a non-negative harmonic function are comparable in a region. In other words,for all x and y lying away from the boundary of D , there exists a constant C not depending on u such that u ( x ) ≤ Cu ( y ) . We show, with the aid of an example, that if n ( x, h ) does not satisfy some suitable conditions, thena Harnack inequality fails, while under mild conditions on n ( x, h ), a Harnack inequality holds.Since the fundamental work of Moser on Harnack inequalities for second order elliptic[20]and parabolic[ ? ] partial differential equations with bounded and measurable coefficients, theseinequalities have become increasingly important. Major contributions to this area have also beenmade by Krylov-Safonov[16] and Fabes-Stroock[10]. While there has been a lot of research onHarnack inequalities for functions that are harmonic with respect to differential operators, notmuch have been done for non-local operators. It is only recently that these results have beenobtained for harmonic functions associated with purely non-local operators; see [6], [4] and [8]. he techniques we use to prove the Harnack inequality in this paper are similar to those in [6] andin [4], but have their roots in [16] where a non-divergence form elliptic operator was considered.As for the regularity theorem, we show that there exist α ∈ (0 ,
1) and a positive constant C not depending on u such that for all x and y lying away from the boundary of D , the followingholds | u ( x ) − u ( y ) | ≤ C k u k ∞ | x − y | α . Continuity estimates of the above type have a long history. Morrey[19] proved such an es-timate for second order elliptic partial differential operators in divergence form with boundedcoefficients. His result, which was proved in two dimensions only, was independently extended tohigher dimensions by DeGiorgi[11] and Nash[23]. Another proof was later given by Moser[20]. Thecorresponding result for operators in non-divergence form was established by Krylov-Safonov[16].In [7] and [5], the authors considered purely non-local operators and proved a regularity theoremusing probabilistic methods. It is also interesting to compare our result with the one obtainedby Mikulevicius-Pragarauskas [22]. They considered a parabolic integro-differential operators andobtained a continuity estimate. However, their result, when specialized to the elliptic case, is abit weaker than our regularity theorem. In that paper, the jump kernel n ( x, h ) satisfies a strongercondition than in our paper. Moreover, our techniques are different.Another paper which is related to our work here is that of Song-Vondracek [26]. Their result isa Harnack inequality for some discontinuous process. However, the jump kernel considered thereis that of a α -stable process. Our result thus holds for a much wider class of processes. Relatedwork also include a Harnack inequality for subordinate Brownian motion which has been obtainedin [24].The local part of our operator L is of non-divergence form. In a forthcoming paper [9], weconsider an operator whose local part is of divergence form and whose jump kernel is symmetric.In that paper, the problem will be framed in terms of Dirichlet forms and a Harnack inequalitytogether with a regularity theorem will be given.After stating the results in Section 2, we prove some preliminary estimates in Section 3. InSection 4, we prove a support theorem which is essential to our method. The proof of the Harnackinequality and regularity theorem are given in Section 5 and 6 respectively. In Section 7, weshow that if the jump kernel n ( x, h ) does not satisfy some suitable conditions, then the Harnackinequality fails. We begin this section with some notations and preliminaries. We use B ( x, r ) for the open ballof radius r with center x . We also use | · | for the Euclidean norm of points in R d , for the normof vectors and for the norm of matrices. The letter c with subscripts will denote positive finiteconstants whose exact values are unimportant. The Lebesgue measure of a Borel set A will bedenoted by | A | . e consider the operator L defined by (1.1) and make the following assumptions: Assumption 2.1
We assume that the diffusion part of the operator is symmetric and uniformlyelliptic and that the b i s are uniformly bounded. In other words, there exist positive constants Λ and Λ such that(a) the diffusion coefficients a ij satisfy the following Λ | y | ≤ d X i,j =1 y i a ij ( x ) y j ≤ Λ − | y | , y ∈ R d , x ∈ R d , (b) sup i k b k ∞ ≤ Λ . We let N (Λ , Λ ) denote the set of operators of the form (1.1) satisfying Assumption 2.1.Besides nonnegativity, the following assumptions will also be imposed on n ( x, h ). Assumption 2.2 (a) There exists a positive constant K such that Z R d ( | h | ∧ n ( x, h ) dh ≤ K, ∀ x ∈ R d . (b) For any r ∈ (0 , , any x ∈ R d , any x, y ∈ B ( x , r/ and z ∈ B ( x , r ) c , we have n ( x, z − x ) ≤ k r n ( y, z − y ) , where k r satisfies < k r ≤ kr − β with k and β being posi-tive constants. n ( x, h ) can be thought of as the intensity of the number of jumps from x to x + h . n ( x, z − x )thus represents the intensity of the number of jumps from x to z . So Assumption 2.2(b) says thatthe probability of jumping to a point z is comparable if x, y are relatively far from z but relativelyclose to each other. In Section 7, we show that such an assumption is needed for the Harnackinequality to hold.Since our method is probabilistic, we need to work with the Markov process associated with L .Let Ω = D ([0 , ∞ )) denote the set of paths that are right continuous with left limits, endowed withthe Skorokhod topology. Let X t ( ω ) = ω ( t ) for ω ∈ Ω and F t be the right continuous filtrationgenerated by the process X . We say a strong Markov process ( P x , X t ) is associated with L if foreach x , we have P x ( X = x ) = 1 and for each x and for each u ∈ C that is bounded and withbounded first and second partial derivatives, u ( X t ) − u ( X ) − R t L u ( X s ) ds is a local martingaleunder P x . This is equivalent to saying that P x solves the martingale problem for L started at x . e assume that the martingale problem is well posed. In other words, we assume that the a i s and b i s are continuous so that there exists a unique solution to the martingale problem. We makesure that none of our estimates are dependent on the modulus of continuity of the a i s and b i s sothat one can then use an approximation procedure to remove the continuity assumptions.For any Borel set A , let T A = inf { t : X t ∈ A } , τ A = inf { t : X t / ∈ A } , be the first hitting time and first exit time, respectively, of A . We say that the function u isharmonic in a domain D if u ( X t ∧ τ D ) is a P x -martingale for each x ∈ D . If u satisfies someregularity conditions and L u = 0 in D , it is easy to see that u is harmonic in D . Since ouroperator contains a non-local part, our process will be have discontinuities. We write X t − = lim s ↑ t X s , ∆ X t = X t − X t − . Our first result concerns the continuity of harmonic functions. Note that our hypotheses donot require Assumption 2.2(b) to hold.
Theorem 2.3
Suppose Assumptions 2.1 and 2.2 (a) hold. Let z ∈ R d and R ∈ (0 , . Suppose u is a function which is bounded in R d and harmonic in B ( z , R ) with respect to L . Then thereexist α ∈ (0 , , C > depending only on the Λ i ′ s and K such that | u ( x ) − u ( y ) | ≤ C k u k ∞ (cid:18) | x − y | R (cid:19) α , x, y ∈ B ( z , R/ . Our main result is the following Harnack inequality.
Theorem 2.4
Suppose Assumptions 2.1 and 2.2 hold. Let z ∈ R d and R ∈ (0 , . Suppose u isnonnegative and bounded on R d and harmonic in B ( z , R ) with respect to L . Then there exists apositive constant C depending on the Λ i ′ s , k , β , R and K but not on z , u , or k u k ∞ such that u ( x ) ≤ Cu ( y ) , x, y ∈ B ( z , R/ . Remark 2.5
A chaining argument shows that both results above hold if
R > with C = C ( R ) depending on R. Theorem 2.3 does not hold for R > with a constant which is independent of R . Remark 2.6
For the Harnack inequality, it is essential that u be nonnegative everywhere. Kass-mann [13] has shown that a Harnack inequality can fail for functions u that are harmonic withrespect to symmetric stable processes of index α and where u fails to be nonnegative everywhere. Some Estimates
We start off this section with a proposition which allows us to assume Λ = 0 when necessary.The proof is very similar to that of Theorem VI 1.2 in [2]. See also [25]. Define˜ L f ( x ) = 12 d X i,j =1 a ij ( x ) ∂ f ( x ) ∂x i ∂x j + Z R d [ f ( x + h ) − f ( x ) − ( | h |≤ h · ∇ f ( x )] n ( x, h ) dh. Proposition 3.1
Suppose
L ∈ N (Λ , Λ ) . If there exists a solution, say ˜ P , to the martingaleproblem for ˜ L started at x where ˜ L is defined as above, then there exists a solution P to the mar-tingale problem for L started at x . Proposition 3.2
There exist constants c and c not depending on x such that if r ≤ , then P x ( τ B ( x ,r ) ≤ c t ) ≤ tr − for x ∈ B ( x , r ) and hence P x ( τ B ( x ,r ) ≤ c r ) ≤ . Proof.
Let u be a nonnegative C function that is equal to | x − x | for | x − x | ≤ r , which isequal to r for | x − x | ≥ r and such that its first and second derivatives are bounded by cr and c respectively. Since P x solves the martingale problem, we have E x u ( X t ∧ τ B ( x ,r ) ) − u ( x ) = E x Z t ∧ τ B ( x ,r ) L u ( X s ) ds. (3.1)Let us write the operator L as L = L c + L d where L c u ( x ) = 12 d X i,j =1 a ij ( x ) ∂ u ( x ) ∂x i ∂x j + d X i =1 b i ( x ) ∂u ( x ) ∂x i , and L d u ( x ) = Z [ u ( x + h ) − u ( x ) − ( | h |≤ h · ∇ u ( x )] n ( x, h ) dh. Since the first and second derivatives of u ( x ) are bounded, we have L c u ( x ) ≤ c for x ∈ B ( x , r )and hence (cid:12)(cid:12)(cid:12) Z t ∧ τ B ( x ,r ) L c u ( X s ) ds (cid:12)(cid:12)(cid:12) ≤ c t. (3.2) ow let us look at L d u ( x ) for x ∈ B ( x , r ) |L d u ( x ) | = (cid:12)(cid:12)(cid:12) Z [ u ( x + h ) − u ( x ) − | h |≤ h · ∇ u ( x )] n ( x, h ) dh (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) Z | h |≤ [ u ( x + h ) − u ( x ) − ( | h |≤ h · ∇ u ( x )] n ( x, h ) dh (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) Z | h |≥ [ u ( x + h ) − u ( x )] n ( x, h ) dh (cid:12)(cid:12)(cid:12) = I + I . By our assumptions and the fact that the second derivatives of u ( x ) are bounded, we get I = (cid:12)(cid:12)(cid:12) Z | h |≤ [ u ( x + h ) − u ( x ) − ( | h |≤ h · ∇ u ( x )] n ( x, h ) dh (cid:12)(cid:12)(cid:12) ≤ c Z | h |≤ | h | k D u k ∞ n ( x, h ) dh ≤ c ,I = (cid:12)(cid:12)(cid:12) Z | h |≥ [ u ( x + h ) − u ( x )] n ( x, h ) dh (cid:12)(cid:12)(cid:12) ≤ k u k ∞ Z | h |≥ n ( x, h ) dh ≤ c . Hence we have (cid:12)(cid:12)(cid:12)R t ∧ τ B ( x ,r ) L d u ( X s ) ds (cid:12)(cid:12)(cid:12) ≤ c t . This together with (3.1) and (3.2) yield E x u ( X t ∧ τ B ( x ,r ) ) ≤ c t, and so from r P x ( τ B ( x ,r ) ≤ t ) ≤ E x u ( X t ∧ τ B ( x ,r ) ), we get the first part of the proposition. Thesecond part is obtained by choosing t = r . (cid:3) We have the following L´evy system formula:
Proposition 3.3 If A and B are disjoint Borel sets, then for each x , X s ≤ t ( X s − ∈ A,X s ∈ B ) − Z t Z B A ( X s ) n ( X s , u − X s ) duds (3.3) is a P x -martingale. The proof is identical to that of the purely non-local operator and can be found in [6]. emma 3.4 There exist c and c such that if r ≤ ,(a) E x τ B ( x ,r ) ≥ c r for x ∈ B ( x , r/ and(b) E x τ B ( x ,r ) ≤ c r for x ∈ B ( x , r ) . Proof.
By Proposition 3.2, there exists c such that P x ( τ B ( x ,r ) ≤ c r ) ≤ P x ( τ B ( x,r/ ≤ c r ) ≤ . The first inequality follows by writing E x τ B ( x ,r ) ≥ c r P x ( τ B ( x,r ) ≥ c r ) , and using the above. Now let us look at the proof of the second inequality. For simplicity weassume that L ∈ N (Λ , P x solves the martingale problem, we have E x u ( X t ∧ τ B ( x ,r ) ) − u ( x ) = E x Z t ∧ τ B ( x ,r ) L u ( X s ) ds. (3.4)As before, let us write L = L c + L d . Let us choose a bounded smooth function u ( x ) so that u ( x ) = | x − x | for x ∈ B ( x ,
2) and u ( x ) equals some constant greater than 4 outside theball B ( x , P di,j =1 ∂ ij u ( x ) = P di =1 ∂ ii u ( x ) and is a constant for x ∈ B ( x , L implies that there exists apositive constant c such that L c u ( X s ) ≥ c whenever X s ∈ B ( x , r ).To deal with the non-local part, we write L d u ( x ) = Z | h |≤ [ u ( x + h ) − u ( x ) − h · ∇ u ( x )] n ( x, h ) dh + Z | h | > [ u ( x + h ) − u ( x )] n ( x, h ) dh = I + I . Note that for | h | ≤
1, we have x + h ∈ B ( x , /
2) for x ∈ B ( x , r ), so by convexity and the factthat n ( x, h ) ≥
0, we obtain I ≥
0. As for the second term, we have I ≥ Z | h | > [ | x + h − x | − | x − x | ] n ( x, h ) dh ≥ . The facts that x ∈ B ( x , r ) and | h | > x + h / ∈ B ( x , r ) which means that the integrandis always non-negative. Combining the above, we have L d u ( X s ) ≥ X s ∈ B ( x , r ) andhence E x Z t ∧ τ B ( x ,r ) L u ( X s ) ds ≥ c E x ( t ∧ τ B ( x ,r ) ) . ow the left hand side of (3.4) satisfies: E x u ( X t ∧ τ B ( x ,r ) ) − u ( x ) ≤ r . Combining the above and letting t → ∞ , we get the second inequality. (cid:3) Corollary 3.5
For each p ≥ , there exists c depending on p such that for r ≤ and x ∈ B ( x , r ) , E x ( τ pB ( x ,r ) ) ≤ c r p . Proof.
Note that E x ( τ B ( x ,r ) ) ≥ t P x ( τ B ( x ,r ) ≥ t ) . Letting t = 2 r and using Lemma 3.4, we obtain P x ( τ B ( x ,r ) ≥ r ) ≤ . If θ t is the shift operatorfrom the theory of Markov processes, then by the Markov property P x ( τ B ( x ,r ) ≥ ( m + 1) r ) ≤ P x ( τ B ( x ,r ) ≥ mr , τ B ( x ,r ) ◦ θ mr ≥ r )= E x [ P X mr ( τ B ( x ,r ) ≥ r ); τ B ( x ,r ) ≥ mr ] ≤ P x ( τ B ( x ,r ) ≥ mr ) . By induction P x ( τ B ( x ,r ) ≥ mr ) ≤ ( ) m . The required result then follows easily from this. (cid:3) The following result was first proved in the continuous case by Krylov in [15]. Since then, thisinequality has been extended for diffusions with jumps. See for instance Theorem III of [17].What follows is a consequence of Corollary 2 of [22]. Proposition 3.6
Consider
L ∈ N (Λ , Λ ) and suppose that Assumption 2.2(a) holds. Let x ∈ R d and R ∈ (0 , . If ( P x , X t ) is a solution to the martingale problem associated with L ∈ N (Λ , Λ ) ,then for any bounded measurable function f , the following holds: E x Z τ B ( x ,R ) f ( X s ) ds ≤ N R k f k L d ( B ( x ,R )) , (3.5) where N depends on the Λ ′ i s , and K . Remark 3.7
In the above, the constant N depends on the non-local part only through the constant K . It also does not depend on the radius R but the upper bound does depend on the radius as shownin (3.5) . he following gives a lower bound on the probability that our process hits a set A beforeleaving a ball. However, the set A should fill most of the ball. We will extend this result in thenext section. Proposition 3.8
Let ǫ > and r ∈ (0 , / . If x ∈ B ( x , r ) , A ⊂ B ( x , r ) and | B ( x , r ) − A | ≤ ǫ ,then P x ( T A ≤ τ B ( x ,r ) ) ≥ ρ ( r, ǫ ) , where there exists some ǫ such that ρ ( r, ǫ ) > for < ǫ ≤ ǫ . Proof.
From inequality (3.5), we have (cid:12)(cid:12)(cid:12)(cid:12) E x Z τ B ( x ,r ) A c ( X s ) ds (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) E x Z τ B ( x ,r ) ( B ( x ,r ) − A ) ( X s ) ds (cid:12)(cid:12)(cid:12)(cid:12) ≤ N r | B ( x , r ) − A | /d ≤ N rǫ /d . So we can write E x τ B ( x ,r ) ≤ E x ( τ B ( x ,r ) ; T A ≤ τ B ( x ,r ) ) + E x Z τ B ( x ,r ) A c ( X s ) ds ≤ (cid:16) E x τ B ( x ,r ) (cid:17) / (cid:0) P x ( T A ≤ τ B ( x ,r ) ) (cid:1) / + N rǫ /d . From Lemma 3.4 and Corollary 3.5 we have E x τ B ( x ,r ) ≥ c r and E x τ B ( x ,r ) ≤ c r . So the aboveyields P x ( T A ≤ τ B ( x ,r ) ) ≥ c r − N rǫ /d c r ! . The proposition is then proved with ρ ( r, ǫ ) = ( c r − Nrǫ /d c r ) . (cid:3) The following will be used only in the proof of the Harnack inequality. So far, this is the onlyplace where we use Assumption 2.2(b).
Proposition 3.9
Under Assumption 2.2, there exists a constant c which depends on K , suchthat if r ≤ / , z ∈ B ( x , r ) and H is a bounded non-negative function supported in B ( x , r ) c ,then E x H ( X τ B ( x , r ) ≤ c k r E z H ( X τ B ( x , r ) . (3.6) Proof.
By linearity and a limit argument, it suffices to consider only H ( x ) = 1 C ( x ) for a set C contained in B ( x , r ) c . From Assumption 2.2(b), we have n ( w, v − w ) ≤ k r n ( y, v − y ) for all w, y ∈ B ( x , r ) and v ∈ B ( x , r ) c . Hence we have,sup y ∈ B ( x , r ) n ( y, v − y ) ≤ k r inf y ∈ B ( x , r ) n ( y, v − y ) . (3.7) y optional stopping and the L´evy system formula, we have E z ( X t ∧ τB ( x , r ∈ C ) = E z X s ≤ t ∧ τ B ( x , r ( | X s − X s − |≥ r ,X s ∈ C ) . = E z Z t ∧ τ B ( x , r Z C n ( X s , v − X s ) dvds. ≥ E z ( t ∧ τ B ( x , r ) ) Z C inf y ∈ B ( x , r ) n ( y, v − y ) dv. Letting t → ∞ and using the dominated convergence theorem on the left and monotone conver-gence on the right, we obtain P z ( X τ B ( x , r ∈ C ) ≥ E z τ B ( x , r ) Z C inf y ∈ B ( x , r ) n ( y, v − y ) dv. Since E z τ B ( x , r ) ≥ E z τ B ( z, r ) , we have P z ( X τ B ( x , r ∈ C ) ≥ E z τ B ( z, r ) Z C inf y ∈ B ( x , r ) n ( y, v − y ) dv. (3.8)Similarly we have P x ( X τ B ( x , r ∈ C ) ≤ E x τ B ( x , r ) Z C sup y ∈ B ( x , r ) n ( y, v − y ) dv. (3.9)Combining inequalities (3.7), (3.8) and (3.9) and using Lemma 3.4, we get our result. (cid:3) Our process is a discontinuous one consisting of small jumps as well as big jumps. In manycases it is more convenient to discard the big jumps and add them later. This can be done byusing a construction which is due to Meyer [18]. We will use this in the next section for the proofof the support theorem.
Meyer’s construction:
Suppose that we have two jump kernels n ( x, h ) and n ( x, h ) with n ( x, h ) ≤ n ( x, h ) and such thatfor all x ∈ R d , N ( x ) = Z R d ( n ( x, h ) − n ( x, h )) dh ≤ c. Let L and L be the operators corresponding to the kernels n ( x, h ) and n ( x, h ) respectively. If X t is the process corresponding to the operator L , then we can construct a process X t correspondingto the operator L as follows. Let S be an exponential random variable of parameter 1 independent f X t , let C t = R t N ( X s ) ds , and let U be the first time that C t exceeds S . At the time U ,we introduce a jump from X U − to y , where y is chosen at random according to the followingdistribution: n ( X U − , h ) − n ( X U − , h ) N ( X U − ) dh. This procedure is repeated using an independent exponential variable S . Since N ( x ) is finite,this procedure adds only a finite number of big jumps on each finite time intervals. In [18], it isproved that the new process corresponds to the operator L . The main result of this section is the support theorem. Before stating and proving this result,we present some ideas which will be crucial for its proof. More precisely, we will represent thesolution of the martingale problem as a solution to a stochastic differential equation. We begin byrepresenting the jumps of our discontinuous process as a function of a Poisson point process.Suppose that P x is a solution to the martingale problem associated with L ∈ N (Λ , Λ ) startedat x . Let Y s be the point process associated with X s , that is, Y s = ∆ X s if ∆ X s = 0 and 0otherwise. Then there exists a measurable function F ( x, z ) such that Y s = F ( X s − , ˆ Y s ) where ˆ Y s is a Poisson point process with intensity measure ˜ λ . To make this statement more precise, let F ( x, A ) = { F ( x, z ) : z ∈ A } and define N t ( A ) = X s ≤ t (∆ X s ∈ F ( X s − ,A )) . Then under P x , N t ( · ) is a Poisson point process with intensity measure ˜ λ . Moreover the measure˜ λ satisfies the following(a) Z ( | z | ∧
1) ˜ λ ( dz ) < ∞ , (4.1)(b) Z A ( h ) n ( x, h ) dh = Z A ( F ( x, z ))˜ λ ( dz ) . (4.2) Remark 4.1
The second condition above gives the relationship between the jump kernel n ( x, h ) and the Poisson process N t ( · ) . Moreover, the indicator function in (4.2) can be replaced by a largerclass of functions. For a more precise statement and proof of the above, see Theorem 12 in [14].See also Chapter XIV of [12]. e now relate ( P x , X ), the solution of the martingale problem to that of a stochastic differentialequation. Set µ ([0 , t ] × A ) = N t ( A ) and ν ([0 , t ] × A ) = t ˜ λ ( A ). Let W t be a Brownian motion withrespect to the filtration F t . Then X t solves the following stochastic differential equation dX t = σ ( X t ) dW t + b ( X t ) dt + Z | F ( X t − ,z ) |≤ F ( X t − , z )( µ − ν )( dz, dt )+ Z | F ( X t − ,z ) | > F ( X t − , z ) µ ( dz, dt ) , X = x, (4.3)where σσ T has a ij as entries and σ T denotes the transpose of σ . The above has been taken from[17]. Chapter XIV of [12] contains more information about this relation. In fact, according toTheorem II of [17], this representation holds under a more stringent condition on the big jumpsof the process ˆ Y t (see Property M in [17]). Since the proof of the theorem below involves dealingwith small jumps only and then adding the big jumps later, this does not affect our result (see theproof below). Here is our support theorem: Theorem 4.2
Suppose
L ∈ N (Λ , Λ ) and P x is a solution to the martingale problem for L started at x . Let ǫ > and suppose that φ : [0 , t ] → R d is differentiable with φ (0) = x . Thereexist constants c , c and c depending on Λ , Λ , t , K and sup t ≤ t | φ ′ ( t ) | but not on ǫ such thatfor all λ > , P x (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c h − exp (cid:2) − λǫ + λ t c + c e | λ | ) (cid:3)i . (4.4)The support theorem says that the graph of X s stays inside an ǫ -tube about φ . In other words,if G ǫφ = { ( s, y ) : | y − φ ( s ) | < ǫ, s ≤ t } , then { ( s, X s ) : s ≤ t } is contained in G ǫφ with a positiveprobability. Proof of theorem 4.2.
We will assume
L ∈ N (Λ , L is defined by n ( x, h ) = n ( x, h )1 ( | h | < and denote the corresponding process by X t . We will later use Meyer’s constructionto remove this restriction. We now use the stochastic differential equation representation of thesolution to the martingale problem. In other words, we use the fact that X t satisfies (4.3). Sinceour process do not have jumps of size greater than 1, the last term of (4.3) can be taken to beidentically zero. More precisely, replacing 1 A ( h ) by | h | ( | h | > and n ( x, h ) by n ( x, h ) in (4.2), weobtain R | F ( x,z ) | > | F ( x, z ) | ˜ λ ( dz ) = 0. Define a new measure Q by d Q d P x = exp h − Z t φ ′ ( s ) σ − ( X s − ) dW s − Z t | φ ′ ( s ) σ − ( X s − ) | ds i . (4.5)Let Z t = X t − Z t Z | F ( X s − ,z ) |≤ F ( X s − , z )( µ − ν )( dz, ds ) . e see that (cid:28) − Z t φ ′ ( s ) σ − ( X s − ) dW s , Z t (cid:29) = (cid:28) − Z t φ ′ ( s ) σ − ( X s − ) dW s , Z t σ ( X s − ) dW s (cid:29) = − Z t φ ′ ( s ) ds = − φ ( t ) + φ (0) . So by Girsanov’s theorem, under Q , each component of Z t is a semi-martingale. If c W t = Z t σ − ( X s − ) dX s − Z t Z | F ( X s − ,z ) |≤ σ − ( X s − ) F ( X s − , z )( µ − ν )( dz, ds ) − Z t σ − ( X s − ) φ ′ ( s ) ds, then c W t is a continuous martingale and d h c W it , c W jt i = δ ij dt under Q . Hence c W t is a d -dimensionalBrownian motion under Q . Note d ( X t − φ ( t )) = σ ( X t − ) d c W t + Z | F ( X t − ,z ) |≤ F ( X t − , z )( µ − ν )( dt, dz ) . (4.6)If we prove the following: Q (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c , (4.7)then the theorem will be proved, for if A is the event { sup s ≤ t | X s − φ ( s ) | < ǫ } , then c ≤ Q ( A ) = Z A ( d Q /d P x ) d P x ≤ ( E x ( d Q /d P x ) ) ( P x ( A )) . (4.8)The theorem then follows easily by noting the d Q /d P x has a finite second moment which isbounded by a constant depending on t , Λ and sup t ≤ t | φ ′ ( t ) | ; see page 188 of [2]. Now let uslook at the proof of (4.7). Let us write the left hand side of (4.6) as dD t i.e, D t := X t − φ ( t ).Let λ be a constant to be chosen later. Define N t = λD t − λ Z t | σ ( X s − ) | ds − Z t Z | z |≤ ( e λz − − λz ) n ( X s − , z ) dzds. (4.9)Set Y λt = e N t . Then, by Ito’s formula (for processes with jumps), we obtain Y λt = 1 + λ Z t e N s − dD t − Z t Z | z |≤ e N s − ( e λz − − λz ) n ( X s − , z ) dzds + X s ≤ t [ e N s − e N s − − e N s ∆ N s ] . rom Assumptions 2.1(a) and 2.2(a), there exist constants c and c such that Z t | σ ( X s ) | ds ≤ c t, and Z | z |≤ ( e λz − − λz ) n ( X s − , z ) dz ≤ λ e | λ | Z | z |≤ | z | n ( X s − , z ) dz ≤ c λ e | λ | . By noting that ∆ N s = λ ∆ D s , and using Theorem 10 of [17] together with the above, we see that Y λt is a martingale. The above bounds, together with (4.9) also yield Q (sup t ≤ t | D t | ≥ ǫ ) ≤ Q (sup t ≤ t e N t ≥ exp[ λǫ − λ c t − λ e | λ | c t ]) . Since Y λt = e N t , we can apply Doob’s inequality as follows: Q (sup t ≤ t | D t | ≥ ǫ ) ≤ Q (sup t ≤ t Y λt ≥ exp[ λǫ − λ c t − λ e | λ | c t ]) ≤ E Q Y λt exp[ − λǫ + λ c t + λ e | λ | c t ]= E Q Y λ exp[ − λǫ + λ c t + λ e | λ | c t ] ≤ exp[ − λǫ + λ c t + λ e | λ | c t ] . (4.10)From the above we conclude that Q (sup t ≤ t | D t | < ǫ ) ≥ − exp (cid:2) − λǫ + λ t c + c e | λ | ) (cid:3) . (4.11)We now use Meyer’s construction to recover the process X t so that ( P x , X t ) is a solution tothe martingale problem associated with the operator L whose jump kernel satisfies the weakerAssumption 2.2(a). The trajectories of X t now have jumps greater than 1. Recall that U is thefirst time that C t exceeds S where S is an exponential random variable with parameter 1. Moreprecisely, we have Q (sup t ≤ t | X t − φ ( t ) | < ǫ ) = Q (sup t ≤ t | X t − φ ( t ) | < ǫ ; U ≤ t ) + Q (sup t ≤ t | X t − φ ( t ) | < ǫ ; U > t ) ≥ Q (sup t ≤ t | X t − φ ( t ) | < ǫ ) Q ( U > t ) . (4.12) sing the fact that Q ( U ≤ t ) ≤ Q ( S ≤ (sup N ) t ) = 1 − e − (sup N ) t , inequality (4.12) reduces to Q (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c Q (sup t ≤ t | X t − φ ( t ) | < ǫ ) for somepositive constant c . This inequality, together with (4.11) and (4.8) complete the proof. (cid:3) Remark 4.3
By taking λ = ǫ in inequality (4.4) , we obtain upon choosing ǫ small enough, P x (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c [1 − e − ǫ / ] . We now use the fact that − e − x ≥ (1 − e − ) x whenever ≤ x ≤ to obtain P x (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c ǫ , (4.13) for some positive constant c not depending on ǫ . However, c does depend on φ via sup t ≤ t | φ ′ ( t ) | ;see (4.8) and the discussion following it. We now present a corollary of the above support theorem.
Corollary 4.4
Suppose
L ∈ N (Λ , Λ ) and P x is a solution to the martingale problem for L started at x . Let ǫ > and suppose that φ : [0 , t ] → R d is continuous with φ (0) = x . Thereexist constants c , c and c depending on ǫ , Λ , Λ , t and the modulus of continuity of φ suchthat for all λ > , P x (sup t ≤ t | X t − φ ( t ) | < ǫ ) ≥ c h − exp (cid:2) − λǫ + λ t c + c e | λ | ) (cid:3)i . (4.14) Proof.
Let us choose a differentiable function φ d with derivative bounded by say c and suchthat sup s ≤ t | φ ( s ) − φ d ( s ) | ≤ ǫ . Moreover, we can choose φ d ( s ) such that k φ ′ d k ∞ depends only on t , ǫ and the modulus of continuity of φ ; see Page 60 of [1]. Hence proving the following P x (sup t ≤ t | X t − φ d ( t ) | < ǫ ) ≥ c h − exp (cid:2) − λǫ + λ t c + c e | λ | ) (cid:3)i will imply (4.14) but with 2 ǫ instead of ǫ . But the above inequality follows from Theorem 4.2.Hence the corollary is proved. (cid:3) Remark 4.5
The above corollary only requires the function φ to be continuous but the downsideof this generalization is that we can longer keep track of the dependence of the constants on ǫ . et Q ( x, r ) denote the cube of side length r centered at x . If R i denotes a cube with side length r , then ˆ R i also denotes a cube with the same center but with side length r/
3. The next resultis not a probabilistic result. It enables us to decompose Q (0 ,
1) into smaller subcubes such thata subset A of Q (0 ,
1) fills a percentage of each of the smaller subcubes. Since this is PropositionV.7.2 of [2], we do not include a proof here.
Proposition 4.6
Let q ∈ (0 , . If A ⊆ Q (0 , and | A | ≤ q , then there exists D such that (i) D isthe union of cubes ˆ R i such that the interiors of the R i are pairwise disjoint, (ii) | A | ≤ q | D ∩ Q (0 , | ,and (iii) for each i, | A ∩ R i | ≥ q | R i | . A corollary of the support theorem is the following:
Corollary 4.7
Let r ∈ (0 , R ) and R ∈ (0 , . Let y ∈ Q (0 , R ) with dist ( y, ∂Q (0 , R )) ≥ r , L ∈ N (Λ , Λ ) , and P be the solution to the martingale problem started at y . If Q ( z, r ) ⊆ Q (0 , R ) ,then P ( T Q ( z,r ) ≤ τ Q (0 ,R ) ) ≥ ζ ( r ) where ζ ( r ) > depends only on r , K and the Λ i s . The above two results together with the Proposition 3.8 are the main ingredients in obtaining theestimate below. The proof is essentially the same as that of Theorem V7.4 in [2] so we omit ithere.
Proposition 4.8
There exists a non-decreasing function ψ : (0 , → (0 , such that if B ⊆ Q (0 , R ) , | B | > , R ∈ (0 , and x ∈ Q (0 , R/ , then P x ( T B ≤ τ Q (0 ,R ) ) ≥ ψ ( | B | /R d ) . We now give a different version of the above proposition. This will allow us to use balls instead ofcubes.
Corollary 4.9
There exists a non-decreasing function φ , such that if B ⊆ B (0 , R ) , | B | > , R ∈ (0 , and x ∈ B (0 , R/ , then P x ( T B ≤ τ B (0 ,R ) ) ≥ φ ( | B | /R d ) . Proof.
For simplicity, we assume d = 2. Higher dimensional cases differ only in notation. Let k be a large positive integer and let R ij be squares of the form [( i − R/k, iR/k ] × [( j − R/k, jR/k ],where i, j ∈ {− k + 1 , ..., − , , , ..., k } .Take ǫ > k sufficiently large so that C = { R ij : | R ij ∩ B | > , and R ij ⊂ B (0 , (1 − ǫ ) R ) } is nonempty. Let M be the number of elements in C . Let R ∗ ij be the cube with the same centeras R ij but side length half as long. Let D = ∪ R ij ∈C R ∗ ij . Pick z ∈ R ∗ ij , where R ij satisfies R ij ∩ B | ≥ | B | M . We can choose k larger if necessary so that we can find such a cube. Then usingProposition 4.8 and the fact that | R ij | = R k , we have P z ( T R ij ∩ B < τ R ij ) ≥ ψ ( | B | k /M R ) ≥ ψ ( | B | /R ) , where the last inequality is obtained by noting that M ≤ k . Since dist( x, ∂B (0 , R )) ≥ R/ D ⊂ B (0 , R ), we can use Corollary 4.7 to obtain P x ( T D < τ B (0 ,R ) ) ≥ c , where c is a constant. Using the Markov property and the above inequalities, we obtain P x ( T B < τ B (0 ,R ) ) ≥ E x [ P X TR ∗ ij ( T B < τ R ij ); T R ∗ ij < τ B (0 ,R ) ] ≥ c ψ ( | B | /R ) . (cid:3) Now we are ready to prove the regularity theorem.
Proof of Theorem 2.3.
Let us suppose u is bounded by M in R d and z ∈ B ( z , R/ r n = θ ρ n , s n = θ a n , for n ∈ N , where a < ρ < /
2, and θ ≥ M are constants to be chosen later. We choose θ small enoughthat B ( z , r ) ⊂ B ( z , R/ B n = B ( z , r n ) and τ n = τ B n . Set M n = sup x ∈ B n u ( x ) , m n = inf x ∈ B n u ( x ) . We will use induction to show that M n − m n ≤ s n for all n . The H¨older continuity at z follows fromthis. Let n be a positive number to be chosen later. Suppose M i − m i ≤ s i for all i = 1 , , ..., n ,where n ≥ n ; we want to show M n +1 − m n +1 ≤ s n +1 . Let ǫ > z, y ∈ B n +1 such that u ( y ) ≤ m n +1 + ǫ and u ( z ) ≥ M n +1 − ǫ . We willshow that u ( z ) − u ( y ) ≤ s n +1 and since ǫ > M n +1 − m n +1 ≤ s n +1 asdesired.Let A n = { x ∈ B n : u ( z ) ≤ ( M n + m n ) / } . e may suppose that | A n | / | B n | ≥ /
2, for if not, we can look at M n − u instead. Let A be acompact subset of A n such that | A | / | B n | ≥ /
3. Corollary 4.9 gives the following P x ( T A ≤ τ n ) ≥ c , (5.1)where c is a constant and x ∈ B n +1 . Let z, y ∈ B n +1 . By optional stopping, u ( z ) − u ( y ) = E z [ u ( X T A ) − u ( y ); T A ≤ τ n ]+ E z [ u ( X τ n ) − u ( y ); τ n ≤ T A , X τ n ∈ B n − ]+ n − X i =1 E z [ u ( X τ n ) − u ( y ); τ n ≤ T A , X τ n ∈ B n − i − − B n − i ]+ E z [ u ( X τ n ) − u ( y ); τ n ≤ T A , X τ n / ∈ B ]= I + I + I + I . (5.2)By the L´evy system formula, and Lemma 3.4 (see the proof of Proposition 3.5 of [6]) , there exist c and c such thatsup y ∈ B n +1 P y ( X τ n / ∈ B n − i ) ≤ sup y ∈ B n +1 E y τ n Z | h | >r n − i − r n +1 n ( y, h ) dh = sup y ∈ B n +1 E y τ n [ Z | h | > n ( y, h ) dh + Z ≥| h | >r n − i − r n +1 n ( y, h ) dh ] ≤ c r n + c (cid:18) ρ i − ρ i (cid:19) . (5.3)The first term on the right of (5.2) is bounded as follows I ≤ (cid:18) M n + m n − m n (cid:19) P y ( T A ≤ τ n ) ≤ s n P y ( T A ≤ τ n ) . (5.4)As for the second term, we have I ≤ ( M n − − m n − ) P y ( τ n ≤ T A ) ≤ s n − (1 − P y ( T A ≤ τ n )) . (5.5)To bound the third term, we choose ρ = √ a ∧ q c a c and note that n − X i =1 s n − i − = s n − n − X i =1 a − i ≤ s n − [ a a n (1 − a ) ] nd n − X i =1 s n − i − ρ i ≤ s n − [ ρ /a − ρ /a ] . Using (5.3) and the above, the third term is bounded by n − X i =1 ( M n − i − − m n − i − ) P y ( X τ n / ∈ B n − i ) ≤ c r n n − X i =1 s n − i − + c n − X i =1 s n − i − ρ i ≤ s n − [ c a θ ρ n a n (1 − a ) + c ρ /a − ρ /a ] . By our choice of ρ , we obtain 1 − ρ /a ≥ / ρ n /a n ≤ / n and ρ /a ≤ c c so that the abovereduces to n − X i =1 ( M n − i − − m n − i − ) P y ( X τ n / ∈ B n − i ) ≤ s n − [ a θ c − a + 4 c ρ /a ≤ s n − [ a θ c − a + c
32 ] . We also choose θ smaller if necessary so that θ ≤ s c (1 − a )2 a c and obtain I ≤ s n − c . (5.6)Using (5.3) again, we see that the fourth term is bounded by2 M P y ( X τ n / ∈ B ) ≤ M [ c r n + c ρ n − ] ≤ θ [ c a n θ + c a n − ] . By choosing n bigger if necessary and recalling that a <
1, we obtain for n ≥ n , I ≤ s n − c . (5.7)Inequalities (5.1)-(5.7) give the following: u ( y ) − u ( z ) ≤ as n − P y ( T A ≤ τ n ) + s n − (1 − P y ( T A ≤ τ n )) + s n − [ c
16 + c . sing the fact that a is less than one, we obtain u ( z ) − u ( y ) ≤ s n a h − P y ( T A < τ n )2 + c
16 + c i ≤ s n a [1 − c
16 ] . We now choose a as follows: a = r − c . This yields u ( z ) − u ( y ) ≤ s n a = s n +1 . (5.8)The continuity estimate now follows from [20]. (cid:3) Proof of Theorem 2.4.
By looking at u + ǫ and letting ǫ ↓
0. We may suppose that u isbounded below by a positive constant. Also, by looking at au , for a suitable a , we may supposethat inf B ( z ,R/ u ∈ [1 / , u above in B ( z , R/
2) by a constant not dependingon u . Our proof is by contradiction.Since u is continuous, we can choose z ∈ B ( z , R/
2) such that u ( z ) = . Let r i = r Ri − where r < is a chosen constant so that P i =1 r i < R/ . Recall that from Proposition 3.9, thereexists c such that if r < , y ∈ B ( x, r/
4) and H is a bounded non-negative function supportedin B ( x, r ) c , then E x H ( X τ B ( x,r/ ) ≤ c k r E y H ( X τ B ( x,r/ ) . (6.1)For inequality (6.1) to hold, we need Assumption 2.2(b). Let η be a constant to be chosen later.Also, let ξ be a constant defined as follows ξ = 12 ∧ ηc . Let c , c and c be positive constants to be chosen later. Once these constants have been chosen,we suppose that there exists x ∈ B ( z , R/
2) with h ( x ) = K for some K large enough so thatthe following is satisfied: ξK e c j r β +6 j c c k ≥ , (6.2)for all j . This is possible because of the fact that r j = r Rj − . The constants k and β are takenfrom Assumption 2.2(b). e will show that there exists a sequence { ( x j , K j ) } with x j +1 ∈ B ( x j , r j ) ⊂ B ( x j , r j ) ⊂ B ( z , R/
4) with: K j = u ( x j ) and K j ≥ K e c j . (6.3)This would imply that K j → ∞ as j → ∞ contradicting the fact that u is bounded. Supposethat we already have x , x , ..., x i such that (6.3) is satisfied. We will show that there exists x i +1 ∈ B ( x i , r i ) ⊂ B ( x i , r i ) such that K i +1 = u ( x i +1 ) and K i +1 ≥ K e c ( i +1) . Then by induction,(6.3) will hold for all j . Define A = { y ∈ B ( x i , r i / u ( y ) ≥ ξK i r βi k } . We are going to show that | A | ≤ | B ( x i , r i / | . To prove this fact, we suppose the contrary.Choose a compact set A ′ ⊂ A with | A ′ | > | B ( x i , r i / | . Note that upon choosing r smaller ifnecessary, we can use (4.13) to obtain P z ( T B ( x i ,r i / < τ B ( z ,R ) ) ≥ c r i , where c is independent of r i . To see this, consider (4.13) with ǫ = r i / φ be a linesegment joining z and x i ( φ (0) = z and φ ( t ) = x i ). Since z , x i ∈ B ( z , R ), | φ ′ ( t ) | is boundedby a constant which is independent of i . Hence c is also independent of i ; see (4.8) and thediscussion following it.Hence, using the strong Markov property, we can write P z ( T A ′ < τ B ( z ,R ) ) ≥ E z [ P X TB ( xi,ri/ ( T A ′ < τ B ( x i ,r i ) ); T B ( x i ,r i / < τ B ( z ,R ) ] ≥ φ (cid:18) | A ′ || B ( x i , r i ) | (cid:19) P z ( T B ( x i ,r i / < τ B ( z ,R ) ) ≥ φ ((1 / d +1 ) c r i . We now take c = φ (cid:0) (1 / d +1 (cid:1) and c = c . By optional stopping, the above inequality andthe fact that u ( X t ∧ T A ′ ) is right continuous, we obtain13 = u ( z ) ≥ E z [ u ( X T A ′ ∧ τ B ( z ,R ) ); T A ′ < τ B ( z ,R ) ] ≥ ξK i r βi k P z ( T A ′ < τ B ( z ,R ) ) ≥ ξK e c i r β +6 i c c k ≥ . his is a contradiction. Therefore | A | ≤ | B ( x i , r i / | . So we can find a compact set E such that E ⊂ B ( x i , r i / − A and | E | ≥ | B ( x i , r i / | . Let us write τ r i for τ B ( x i ,r i / . From Corollary 4.9,we have P x i ( T E < τ r i ) ≥ c where c is some positive constant.Let M = sup B ( x i ,r i ) u ( x ). We then have K i = u ( x i ) = E x i [ u ( X T E ∧ τ ri ); T E < τ r i ]+ E x i [ u ( X T E ∧ τ ri ); T E > τ r i , X τ ri ∈ B ( x i , r i )]+ E x i [ u ( X T E ∧ τ ri ); T E > τ r i , X τ ri / ∈ B ( x i , r i )]= I + I + I . (6.4)Writing p i = P x i ( T E < τ r i ), we see that the first two terms are easily bounded as follows: I ≤ ξK i p i r βi k , and I ≤ M (1 − p i ) . To bound the third term, we prove E x i [ u ( X τ ri ); X τ ri / ∈ B ( x i , r i )] ≤ ηK i . If not, then by using(6.1), we will have, for all y ∈ B ( x i , r i / u ( y ) ≥ E y u ( X τ ri ) ≥ E y [ u ( X τ ri ); X τ ri / ∈ B ( x i , r i )] ≥ c k r i E x i [ u ( X τ ri ); X τ ri / ∈ B ( x i , r i )] > ηK i c k r i > ξK i r βi k , contradicting the fact that | A | ≤ | B ( x i , r i / | . Hence I ≤ ηK i . So (6.4) becomes K i ≤ ξK i p i r βi k + M (1 − p i ) + ηK i or MK i ≥ − η − ξp i r βi /k − p i = 1 + (1 − ξr βi /k ) p i − η − p i . (6.5)Choosing η = c and using the definition of ξ together with the fact that p i ≥ c and r βi /k <
1, wesee that there exists a positive L , such that inequality (6.5) reduces to M ≥ K i (1 + L ). Thereforethere exists x i +1 ∈ B ( x i , r i ) with u ( x i +1 ) ≥ K i (1 + L ) . Setting K i +1 = u ( x i +1 ), we see that K i +1 ≥ K i (1 + L )= K i e log(1+ L ) . he condition (6.3) is thus satisfied provided we choose c = log(1 + L ). Finally, note that thefact that P i =1 r i < R implies that B ( x i , r i ) ⊂ B ( z , R/ (cid:3) In this section, we show that if an assumption along the lines of Assumption 2.2(b) does not hold,then the Harnack inequality can fail. This example is very similar to the one in [4]. But sincewe need some modifications and for the sake of completeness, we give a proof of the followingproposition:
Proposition 7.1
There exists a function n ( x, h ) which satisfies Assumptions 2.2(a) but not (b)and for which the Harnack inequality fails for functions harmonic with respect to the correspondingoperator. Proof.
Let B = B (0 , y = (1 / ,
0) and for m ≥
4. let x m = ( − / , − m ), z m = (16 , − m ), C m = B ( x m , − m − ), and E m = B ( z m , − m − ). Define n ( x, h ) = ∞ X m =4 C m ( x )1 E m ( x + h ) . Note that n ( x, h ) satisfies Assumption 2.2(a) and not 2.2(b). Now we show that P y ( T C m < τ B )is small when m is large. We see that from Lemma 3.4, E y τ B ≤ c < ∞ . As before we are goingto write L = L c + L d . Now fix m , let ǫ = 2 − m − , let g ( x ) = | x − x m | − β , where β ∈ (0 ,
1) let φ bea non-negative C ∞ function with support in B (0 , /
2) whose integral is 1, let φ ǫ ( x ) = ǫ − d φ ( x/ǫ )and let f ǫ = g ∗ φ ǫ . Hence f ǫ ∈ C ∞ and we see that f ǫ ≥ c ǫ − β on C m . Since the local part isuniformly elliptic, we have |L c f ǫ ( x ) | ≤ c . From the definition of n ( x, h ), we have |L d f ǫ ( x ) | ≤ c .Hence |L f ǫ ( x ) | ≤ c . Since P y is a solution to the martingale problem for L , E y f ǫ ( X T Cm ∧ τ B ) − f ǫ ( y ) = E y Z T Cm ∧ τ B L f ǫ ( X s ) ds ≤ c E y τ B ≤ c . Hence c ǫ − β P y ( T C m < τ B ) ≤ E y f ǫ ( X T Cm ∧ τ B ) ≤ c + f ǫ ( y ) ≤ c . Thus P y ( T C m < τ B ) will be small if m is large. Now suppose that the Harnack inequality doeshold for non-negative functions that are harmonic in B , that is, suppose there exists c such that u ( x ) ≤ c u ( y ) x, y ∈ B (0 , / , or any nonnegative bounded function u which is harmonic in B . Let u m ( x ) = E x [1 E m ( X τ B )] . Then u m ( x ) is bounded. nonnegative, and harmonic in B . Note that the only way that X τ B canbe in E m is if X τ B − is in C m . We then have, using the assumption that the Harnack inequalityholds, u m ( y ) = E y [1 E m ( X τ B ); T C m < τ B ]= E y [ E X TCm [1 E m ( X τ B )]; T C m < τ B ]= E y [ u m ( X T Cm ); T C m < τ B ] ≤ c u m ( x m ) P y ( T C m < τ B ) . Then, u m ( x m ) u m ( y ) ≥ c P y ( T C m < τ B )which can be made arbitrary large if we take m large enough. This is a contradiction and thereforethe Harnack inequality cannot hold. (cid:3) Acknowledgments
The author wishes to thank his thesis advisor, Prof. Richard F. Bass, for all his help during thepreparation of this work. The author also thanks an anonymous referee and the associate editorfor pointing out several mistakes in earlier drafts of the paper.
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