Regularity of solutions to anisotropic nonlocal equations

REGULARITY OF SOLUTIONS TO ANISOTROPIC NONLOCAL EQUATIONS

arXiv:1607.08135v1 [math.PR] 27 Jul 2016

JAMIL CHAKER

Abstract. We study harmonic functions associated to the system of stochastic differential equations dXti = Ai1 (Xt− )dZt1 + ⋯ + Aid (Xt− )dZtd , i ∈ {1, . . . , d}, where Zt1 , . . . , Ztd are independent one-dimensional symmetric stable processes with indices αi ∈ (0, 2). We prove H¨ older regularity of bounded harmonic functions with respect to solutions of the above system of stochastic differential equations.

AMS 2010 Mathematics Subject Classification: Primary 60J75; Secondary 60H10, 31B05, 60G52 Keywords: Jump processes, Harmonic functions, H¨ older continuity, Support theorem

1. Introduction In the theory of integro-differential operators and jump processes regularity results gain in importance. One of the reasons to consider jump processes is that they provide realistic models in natural sciences or economics, see for instance [CT04]. In [KS79] Krylov and Safonov developed a technique of proving H¨ older regularity and the Harnack inequality for harmonic functions corresponding to non-divergence form elliptic operators. Their main tool was a support theorem, which gives information about the topological support of a solution to the martingale problem associated to the operator. This technique was also used in [BL02] to prove similar results for nonlocal operators of the form [f (x + h) − f (x) − 1{∣h∣≤1} h ⋅ ∇f (x)]a(x, h)dh, Lf (x) = ∫

R

d ∖{0}

under suitable assumptions on a. Next, in [BC10] Bass and Chen followed the same ideas to prove H¨ older regularity for harmonic functions associated to solutions of systems of stochastic differential equations driven by Lévy processes with highly singular Lévy measures. In this work we extend the results obtained by Bass and Chen to a larger class of driving Lévy processes. A one-dimensional Lévy process (Yt )t≥0 is called symmetric stable processes of order γ ∈ (0 , 2 ) if its characteristic function is given by EeiξYt = e−t∣ξ∣ , γ

ξ∈

The Lévy measure of such a process is given by ν(dh) = cγ ∣h∣

R.

−1−γ

N

) / ∣Γ (− γ2 )∣. dh, where cγ = 2γ Γ ( 1+γ 2

Let d ∈ and d ≥ 2. We assume that Zti , i = 1, . . . , d, are independent one-dimensional symmetric stable processes with indices αi ∈ (0, 2). We set Z = (Zt )t≥0 = (Zt1 , . . . , Ztd )t≥0 . 1

2

JAMIL CHAKER

The Lévy-measure of this process is supported on the coordinate axes and given by ⎛ ⎞⎞ ⎛ cαk dwk ∏ δ{0} (dwj ) . 1+α k ⎝j≠k ⎠⎠ k=1 ⎝ ∣wk ∣ d

ν(dw) = ∑

R

Therefore ν(A) = 0 for every set A ⊂ d , which has an empty intersection with the coordinate axes. The generator L of Z is given for f ∈ Cb2 ( d ) by the formula d

Lf (x) = ∑ ∫ k=1

R∖{0}

R

(f (x + hek ) − f (x) − 1{∣h∣≤1} ∂k f (x)h)

cαk dh. ∣h∣1+αk

For a deeper discussion on Lévy processes and their generators we refer the reader to [Sat13].

R

Let x0 ∈ d and A ∶ differential equations

Rd → Rd×d a matrix-valued function.

We consider the system of stochastic

⎧ d ⎪ ⎪ i ⎪ dX = Aij (Xt− )dZtj , ⎪ ∑ ⎪ t ⎨ (1) j=1 ⎪ ⎪ ⎪ i i ⎪ ⎪ ⎩ X0 = x0 , where Xt− = lim Xs is the left hand limit. s↗t

This system has been studied in the case α1 = α2 = ⋯ = αd = α ∈ (0, 2) by Bass and Chen in [BC06] and [BC10]. With the help of the martingale problem Bass and Chen show in [BC06] that for each x0 ∈ d there exists one and only one weak solution (X = (Xt1 , . . . , Xtd )t≥0 , Px0 ) to (1). Moreover the family {X, Px , x ∈ d } forms a conservative strong Markov process on d whose semigroup maps bounded continuous functions to bounded continuous functions (see Theorem 1.1, [BC06]). Consequently it follows that

R

R

R

d

Lf (x) = ∑ ∫

R

R∖{0}

j=1

(f (x + aj (x)h) − f (x) − h1{∣h∣≤1} ∇f (x) ⋅ aj (x))

cα dh, ∣h∣1+α

coincides on Cb2 ( d ) with the generator for any weak solution to (1), where aj (x) denotes the jth column of the matrix A(x). A standard reference on martingale problems is [EK86]. In the scope of this paper we will not show the existence of a unique solution to (1), but assume there exists one. The following assumptions will be needed throughout the paper. Assumption. (i) For every x ∈ d the matrix A(x) is non-degenerate, that is det(A(x)) ≠ 0. (ii) The functions x ↦ Aij (x) and x ↦ A−1 ij (x) are continuous and bounded for all 1 ≤ i, j ≤ d and x ∈ d . (iii) For each x0 ∈ d , there exists a unique weak solution (X = (Xt1 , . . . , Xtd )t≥0 , Px0 ) to (1). The family {X, Px , x ∈ d } forms a conservative strong Markov process on d and

R

R

R

R

d

(2)

Lf (x) = ∑ ∫ j=1

R∖{0}

(f (x + aj (x)h) − f (x) − h1{∣h∣≤1} ∇f (x) ⋅ aj (x))

R

cαj dh, ∣h∣1+αj

Rd) with the generator for any weak solution to (1).

coincides on Cb2 (


3

Remark. Note that assumption (iii) is equivalent to existence of a unique solution to the martingale problem for L on Cb2 ( d ) for any x0 ∈ d .

R

R

Notation. Let A be the matrix-valued function from (1). Throughout the paper ̟(D) denotes the modulus of continuity of A on a Borel set D and we write Λ(D) for the upper bound of A on D. We set α∗ ∶= min{α1 , . . . , αd } and α∗ ∶= max{α1 , . . . , αd }. For i ∈ we write ci for positive constants and additionally ci = ci (⋅) if we want to highlight all the quantities the constant depends on.

N

In order to deal with the anisotropy of the process we consider a corresponding scale of cubes. Definition 1. Let r ∈ (0, 1] and α1 , . . . , αd ∈ (0, 2). Moreover, let α be the geometric mean of α1 , . . . , αd , that is α = (α1 ⋅ . . . ⋅ αd )1/d . For k > 0, we define Mrk (x) ∶= ⨉ (xi − (krα/αi ), xi + (krα/αi )) . d

i=1

For simplicity we write Mr (x) instead of Mr1 (x). Note that Mrk is increasing in k and r. The advantage of using these sets is the fact that they reflect the different jump intensities of the process Z and compensate them. The purpose of this paper is to prove the following result.

R

R

Theorem 2. Let r ∈ (0, 1], s > 0 and x0 ∈ d . Suppose h is bounded in d and harmonic in Mr1+s (x0 ) with respect to X. Then there exist c1 = c1 (Λ(Mr1+s (x0 )), ̟(Mr1+s (x0 ))) > 0 and β = β(Λ(Mr1+s (x0 )), ̟(Mr1+s (x0 ))) > 0, independent of h and r, such that ∣x − y∣ ∣h(x) − h(y)∣ ≤ c1 ( α/α ) sup ∣h(z)∣ r ∗ Rd β

for x, y ∈ Mr (x0 ).

Let us compare Theorem 2 with the result in [BC10], where the authors considered the case α1 = ⋯ = αd = α. Theorem (Theorem 2.9, [BC10]). Let r ∈ (0, 1] and γ > 1. Suppose that h is harmonic in B(x0 , γ r) with respect to X and h is bounded in d . There exists positive constants c1 and β that depend on γ, the upper bound of A(x) and A(x)−1 on B(x0 , γ r), and the modulus of continuity of A(x) on B(x0 , γ r) but otherwise is independent of h and r such that

R

∣h(x) − h(y)∣ ≤ c1 (

∣x − y∣ β ) sup ∣h(z)∣. r Rd

We want to emphasize, that in the case α1 = ⋯ = αd the sets Mr (x0 ) reduce to a cube with radius r. Therefore this result is contained in Theorem 2, when choosing cubes instead of balls.

4

JAMIL CHAKER

Structure of the article. This article is organized as follows. In Section 2 we provide definitions and auxiliary results. We constitute sufficient preparation and study the behavior of the solution to the system. In Section 3 we study the topological support of the solution to the martingale problem associated to the system of stochastic differential equations. The aim of this section is to prove a Krylov-Safonov type support theorem. Section 4 contains the proof of Theorem 2. Acknowledgement. This work is part of the author’s Master thesis, written under the supervision of Moritz Kassmann at Bielefeld University. Financial support by the German Science Foundation DFG (SFB 701) is gratefully acknowledged. 2. Definitions and auxiliary results In this section we will provide important definitions and prove auxiliary results associated to the solution of the system (1). Let Aτ (x) denote the transpose of the matrix A(x) and (aτj (x))−1 the j−th row of (Aτ (x))−1 . For a Borel set D, let TD ∶= inf{t ≥ 0∶ Xt ∈ D} and τD ∶= inf{t ≥ 0∶ Xt ∉ D} be the first entrance time and the first exit time of D by X. From now on α stands for the geometric mean of α1 , . . . , αd . Let us first recall the definition of harmonicity with respect to a Markov process.

R

R

is called harmonic with respect to X in a domain Definition 3. A bounded function h ∶ d → D ⊂ d if for every bounded open set U with clos(U ) ⊂ D

R

h (Xt∧τU ) is a Px -martingale for every x ∈ U.

̂ = Ms3 (y). The next Proposition is a pure geometrical For R = Ms (y) we use the notation R statement and not related to the system of stochastic differential equations. Proposition 4. Let r ∈ (0, 1], q ∈ (0, 1) and x0 ∈ a set D ⊂ Mr (x0 ) such that

Rd.

If A ⊂ Mr (x0 ) and ∣A∣ < q, then there exists

̂i such that the interiors of the Ri are pairwise disjoint, (1) D is the union of rectangles R (2) ∣A∣ ≤ ∣D ∩ Mr (x0 )∣ and (3) for each i, ∣A ∩ Ri ∣ > q∣Ri ∣.

Proof. We follow the proof of Proposition V.7.2 in [Bas98]. We construct a collection of rectangles R = {Ri } as follows: d Divide M ∶= Mr (x0 ) into 2d equal rectangles {R(i1 )}2i1 =1 with disjoint interiors. Here two cases can occur for i1 = 1, . . . , 2d : If ∣A ∩ R(i1 )∣ > q∣R(i1 )∣, then we let R(i1 ) be one of the rectangles in R. This rectangles will be left untouched. Note that these sub-rectangle have side length rα/αi in each coordinate i ∈ {1, . . . , d}. If ∣A ∩ R(i1 )∣ ≤ q∣R(i1 )∣, we split R(i1 ) into 2d equal rectangles {R(i2 )}di2 =1 of side length 2−1 rα/αi in each coordinate i ∈ {1, . . . , d}, such that the interiors are disjoint and differ again: If ∣A ∩ R(i2 )∣ >


5

q∣R(i2 )∣, then R(i2 ) will be one of the Ri , otherwise we split R(i2 ) again into sub-rectangles. We continue this procedure and form with this method a suitable collection of rectangles. To be more precise, let Rn ∶= {R′ ⊂ M ∶ R′ is a rectangle with side length 2−n+1 rα/αi in the ith coordinate and has vertices (

nd rα/αi n1 rα/αi , . . . , ) for n1 , . . . , nd ∈ Z and i ∈ {1, . . . , d}}. 2n−1 2n−1

An element R′ ∈ Rn will be in R if ∣A ∩ R′ ∣ > q∣R′ ∣ and R′ is not contained in any R′′ ∈ R0 ∪ R1 ∪ ⋯ ∪ Rn−1 with ∣A ∩ R′′ ∣ > q∣R′′ ∣. Now set ̂i . D = ⋃R i

Then the assertions (i) and (iii) are clear and it remains to prove (ii). By Lebesgue’s density theorem almost every point z ∈ A is a point of density A, that is ∣A ∩ Bε (z)∣ ε→0 Ð→ 1 a.e.. ∣Bε (z)∣ Let z be a point of density of A and Tn the element of Rn with z ∈ Tn , then ∣Tn ∩ A∣ n→∞ Ð→ 1 a.e.. ∣Tn ∣ If z is a point of density of A and z is not on the boundary of some rectangles in Rn for some n, it follows that z must be in some Ri ∈ R. Hence ∣A ∖ D∣ = 0. We now form a new collection d of rectangles S = {Si } as follows. Divide M into 2d equal rectangles {S(j1 )}2j1 =1 with disjoint interiors. If S(j1 ) ⊂ D, it will be one of the rectangles in S; otherwise we split S(j1 ) into 2d equal sub-rectangles with disjoint interiors and continue. To be more accurate, S ′ ∈ Rn will be in S if ̂i , we S ′ ⊂ D but S ′ is not contained in any S ′′ ∈ R0 ∪ R1 ∪ ⋯ ∪ Rn−1 with S ′′ ⊂ D. Since D = ⋃i R get ∣D ∩ M ∣ = ∑ ∣Si ∣. i

Hence ∣A∣ = ∑i ∣Si ∩ A∣. Thus it suffices to show that for each Si ∈ S ∣A ∩ Si ∣ ≤ q∣Si ∣.

Then we can sum over i and the assertion will be proved. Let Si ∈ S. If Si = M , then ∣A∩M ∣ = ∣A∣ ≤ q. d Otherwise Si ∈ Rn for some n ≥ 1 and is contained in a rectangle R′ ∈ Rn−1 . Let {Cj }2j=1−1 denote the other rectangles of Rn that are contained in R′ . Since Si ∈ S, we know 2d −1

′ ⋃ Cj ∪ Si = R ∉ S.

j=1

Moreover, since Si ⊂ D, at least one of the rectangles C1 , . . . , C2d −1 cannot be contained in D. We have Si ∪ C1 ∪ ⋅ ⋅ ⋅ ∪ C2d −1 ⊂ Ŝi and hence Ŝi ⊂/ D. Hence Si ∉ R. All in all we have Si ∪ C1 ∪ ⋅ ⋅ ⋅ ∪ C2d −1 is not contained in D, but Si ∉ R which implies ∣A ∩ Si ∣ ≤ q∣Si ∣.

6

JAMIL CHAKER

We next prove a Lévy system type formula. The proof follows the idea of Proposition 2.3 in [BL02]. Proposition 5. Suppose A and B are two Borel sets with dist(A, B) > 0. Then ∑ 1{Xs− ∈A,Xs ∈B} − ∫

0

s≤t

⎛ ⎞⎞ ⎛ ∣(aτk (Xs ))−1 ∣1+αk cαk dhk ∏ δ{Xsj } (dhj ) ds k 1+α k ∣hk − Xs ∣ B k=1 ⎝ ⎝j≠k ⎠⎠ d

1A (Xs ) ∫ ∑

t

x

is a P -martingale for each x.

Rd) with f = 0 on A and f = 1 on B. Moreover set

Proof. Let f ∈ Cb2 (

Mtf ∶= f (Xt ) − f (X0 ) − ∫

By Assumption (iii) for each x ∈ problem for L. Hence we get

t

0

Lf (Xs )ds.

Rd the probability measure Px is a solution to the martingale ∫

0

t

1A (Xs− )dMsf

is a Px -martingale, since the stochastic integral with respect to a martingale is itself a martingale. Rewriting f (Xt ) − f (X0 ) = ∑s≤t (f (Xs ) − f (Xs− )) yields to ∑ (1A (Xs− )(f (Xs ) − f (Xs− ))) − ∫

t

0

s≤t

1A (Xs− )Lf (Xs )ds

is a Px -martingale. Since Xs ≠ Xs− for only countably many values of s, then ∑ (1A (Xs− )(f (Xs ) − f (Xs− ))) − ∫

(3)

t

0

s≤t

1A (Xs )Lf (Xs )ds

is also a Px -martingale. Recall that (aτj (x))−1 denotes the j−th row of (Aτ (x))−1 . Let w = (w1 , . . . , wd ) and u = (u1 , . . . , ud ). For x ∈ A we have f (x) = 0 and ∇f (x) = 0, and so d

Lf (x) = ∑ ∫ k=1 d

R∖{0}

f (x + ak (x)h)

⎛ ⎞⎞ cαk ⎛ f (x + Aτ (x)w) 1+α δ{0} (dwj ) dwk ∏ ∣w∣ k ⎝j≠k Rd ∖{0} ⎝ ⎠⎠

= ∑∫ k=1 d

= ∑∫ k=1

cαk dh ∣h∣1+αk

R

d ∖{0}

f (u)

⎞ ∣(aτk (x))−1 ∣1+αk cαk ⎛ ∏ δ{xj } (duj ) duk . 1+α k ∣u − x∣ ⎝j≠k ⎠

Note, that cαk /∣h∣1+αk is integrable over h in the complement of any neighborhood of the origin for any k ∈ {1, . . . , d}. Since A and B are a positive distance from each other, the sum in (3) is finite. Hence ∑ (1A (Xs− )(1B (Xs ) − 1B (Xs )))

s≤t

−∫

0

t

⎛ ⎞⎞ ⎛ ∣(aτj (Xs ))−1 ∣1+αj cαk dhk ∏ δ{Xsj } (dhj ) ds ∣hk − Xsk ∣1+αk B k=1 ⎝ ⎝j≠k ⎠⎠ d

1A (Xs ) ∫ ∑

is a Px -martingale, which is equivalent to our assertion.


7

The next Proposition gives the behavior of the expected first exit time of the solution to (1) out of the set Mr (⋅). This Proposition highlights the advantage of Mr (⋅) and shows that the scaling of the cube in the different directions with respect to the jump intensity gives the desired properties.

R

Proposition 6. Let x ∈ d and r ∈ (0, 1]. Then there exists a constant c1 = c1 (Λ(Mr (x)), d) > 0 such that for all z ∈ Mr (x) Ez [τMr (x) ] ≤ c1 rα . Proof. Note

(4)

Ez [τMr (y) ] = Ez [ min inf{t ≥ 0 ∶ Xti ∉ (yi − r(α/αi ) , yi − r(α/αi ) )}] 1≤i≤d

≤

1 1 d z i (α/αi ) , yi − r(α/αi ) )}] =∶ ∑ Ez [Υi ] . ∑ E [inf{t ≥ 0 ∶ Xt ∉ (yi − r d i=1 d i=1 d

Let j ∈ {1, . . . , d} be fixed but arbitrary. We want to show, there exists a c2 > 0 such that

(5)

Ez (Υj ) ≤ c2 rα .

Since we reduced the problem to a one-dimensional one, we may suppose by scaling r = 1. Let κ ∶= inf {∣A(x)ej ∣ ∶ x ∈ M1 (x)} .

By Assumption (i), we have κ > 0. There exists a c3 ∈ (0, 1) with Pz (∃z ∈ [0, 1] ∶ ∆Zsj ∈

R ∖ [−3/κ, 3/κ]) ≥ c3.

The independence of the one-dimensional processes implies that with probability zero at least two of the Z i ’s make a jump at the same time. This leads to 3 (6) Pz (∃s ∈ [0, 1] ∶ ∆Zsj > and ∆Zsi = 0 for i ∈ {1, . . . , d} ∖ {j}) ≥ c3 . κ Our aim is to show that the probability of the process X for leaving M1 (x) in the j-th coordinate after time m is bounded in the following way Pz (Υj > m) ≤ (1 − kj )m

for all m ∈

N.

= 0 for i ∈ {1, . . . , d} ∖ {j}, and Xs− ∈ M1 (x). Suppose there exists s ∈ [0, 1] such that > Then ∣∆Xsj ∣ = ∣∆Zsj ∣ ∣A(Xs− )(ej )∣ > 3. ∆Zsj

3 , κ

∆Zsi

Note, that we leave M1 (x) by this jump. By (6)

Pz (Υj ≤ 1) ≥ c3 ⇔ Pz (Υj > 1) ≤ (1 − c3 ).

Let {θt ∶ t ≥ 0} denote the shift operators for X. Now assume Pz (Υj > m) ≤ (1 − c3 )m . By the Markov property Pz (Υj > m + 1) ≤ Pz (Υj > m; Υj ○ θm > 1)

= Ez [PXm (Υj > 1); Υj > m]

≤ (1 − c3 )Pz (Υj > m) ≤ (1 − c3 )m+1 .

8

JAMIL CHAKER

Assertion (5) follows by Ex [Υj ] = ∫

0

∞

∞

∞

m=0

m=0

Pz (Υj > t)dt ≤ ∑ Pz (Υj > m) ≤ ∑ (1 − c3 )m = c2 ,

where we used the fact that the sum on the right hand side is a geometric sum. Thus the assertion follows by (4) and (5). Next, we give an estimate for leaving a rectangle with a comparatively big jump.

R

Rd), d) > 0,

Proposition 7. Let x ∈ d , r ∈ (0, 1] and R ≥ 2r. There exists a constant c1 = c1 (Λ( such that for all z ∈ Mr (x) r α Pz (XτMr (x) ∉ MR (x)) ≤ c1 ( ) . R Proof. Let

R

Cj ∶= ∖ [xj − Rα/αj , xj + Rα/αj ] and for 1 ≤ j ≤ d let kj = supx∈R ∣(aτj (x))−1 ∣cαj . By Proposition 5 and optional stopping we get for c2 = ∑dj=1 ((2kj 2α )/cαj ) ≤ 8d supx∈R ∣(aτj (x))−1 ∣ ⎤ ⎡ t∧τ d ∣(aτ (X ))−1 ∣c ⎥ ⎢ Mr (x) ⎞ s αj ⎛ j ⎥ i δ (dh ) dh ds Pz (Xt∧τMr (x) ∉ MR (x)) = Ez ⎢ ∏ ∑ i j {X } ∫ ⎥ ⎢∫0 j s MR (x)c j=1 ∣hj − Xs ∣1+αj ⎝ i≠j ⎠ ⎥ ⎢ ⎦ ⎣ ⎤ ⎡ t∧τMr (x) d k j ⎥ ⎢ dhj ds⎥ ≤ Ez ⎢∫ ∑∫ j 1+αj ⎥ ⎢ 0 j=1 Cj ∣hj − Xs ∣ ⎦ ⎣ ⎤ ⎡ t∧τMr (x) d kj ⎥ ⎢ dh ds ≤ Ez ⎢∫ ⎥ ∑ j ∫ 1+α α/α ⎥ ⎢ 0 j j )∣ j=1 Cj ∣hj − (xj + r ⎦ ⎣ d

= Ez [t ∧ τMr (x))] ∑

j=1 d

≤ Ez [t ∧ τMr (x))] ∑

αj

2kj − rα/αj )αj

(Rα/αj

2kj c2 = α Ez [t ∧ τMr (x)) ]. α/α α j j R ) j=1 αj ((R/2)

Using the monotone convergence on the right and dominated convergence on the left, we have for t→∞ c2 r α Pz (Xt∧τMr (x) ∉ MR (x)) ≤ α Ez (τMr (x) ) ≤ c2 c3 ( ) , R R z α where c3 is the constant showing up in the estimate E (τMr (x) ) ≤ c3 r of Proposition 6. 3. The support theorem The aim of this section is to prove the support theorem. It states that sets of positive Lebesgue measure are hit with positive probability. This theorem was first proved in 1979 in [KS79] for the diffusion case dXt = σ(Xt )dWt + bt dt.


9

In the article [BC10], Bass and Chen prove the support theorem in the context of pure jump processes. They consider the system (1) in the case αi = α for all i ∈ {1, . . . , d} and use the KrylovSafonov technique in [BC10] to prove H¨ older regularity with the help of the support theorem. The idea we use to prove the support theorem is similar in spirit to the one in [BC10]. The following Lemma is a statement about the topological support of the law of the stopped process. It gives the existence of a bounded stopping time T such that with positive probability the stopped process stays in a small ball around its starting point up to time T , makes a jump along the k-th coordinate axis and stays afterwards in a small ball.

R

Lemma 8. Let r ∈ (0, 1], x0 ∈ d , k ∈ {1, . . . , d}, vk = A(x0 )ek , γ ∈ (0, rα/α∗ ), t0 > 0 and ξ ∈ [−rα/α∗ , rα/α∗ ]. There exists a constant c1 > 0 = c1 (γ, t0 , ξ, r, Λ(Mr2 (x0 ))), ̟(Mr2 (x0 ))) > 0 and a stopping time T ≤ t0 , such that (7)

Px0 (sup ∣Xs − x0 ∣ < γ and sβ} , Z i = Z i − Z Z s

t

Let (X t )t≥0 be the solution to

t

s

s≤t

d

t

t

dX t = ∑ Aij (X t− )dZ t , X0i = xi0 . i

j

j=1

The continuity of A allows us to find a δ < γ/(6∥A∥∞), such that γ sup sup ∣Aij (x) − Aij (x0 )∣ < . (8) 12d i,j ∣x−x0 ∣ 0, such that

d

[X , X ] ≤ c2 ∑ [Z , Z ] . i

i

t

j=1

j

j

t

10

JAMIL CHAKER

Note, that β ∈ (0, ξ) ⊂ (0, rα/α∗ ) ⊂ (0, 1). Therefore, we get d

d

Ex0 [X , X ] ≤ c2 ∑ Ex0 [Z , Z ] = c2 ∑ ∫ i

i

t

j

j

t

j=1

j=1

t

0

(∫

β −β

By Tschebyscheff’s inequality and Doob’s inequality, we get Px0 [sup ∣X s − X 0 ∣ > δ] ≤ i

i

s≤t0

Choose β ∈ (0, ξ) such that

cαj h2 ∗ dh) dt ≤ c3 tdβ 2−α . 1+α j ∣h∣ ∗

i i 2 i i 2 1 x0 1 c4 t0 dβ 2−α E [sup (X s − X 0 ) ] ≤ 2 4Ex0 [(X t0 − X 0 ) ] ≤ . 2 δ δ δ2 s≤t0

c5 t0 β 2−α ≤ ∗

(9)

δ2 2d

holds. Then by (9), we get (10)

i i 1 Px0 (C) = 1 − Px0 (sup ∣X s − X 0 ∣ > δ) ≥ . 2 s≤t0

̃k to have a single jump before time t0 , and for that jump’s size to be in the interval [ξ, ξ + δ], For Z ̃k must have then up to time t0 Z t (i) no negative jumps, (ii) no jumps whose size lies in [β, ξ), (iii) no jumps whose size lies in (ξ + δ, ∞), (iv) precisely one jump whose size lies in the interval [ξ, ξ + δ].

̃k is a compound Poisson process and use the knowledge about Poisson We can use the fact, that Z random measures. The events descriped in (i)-(iv) are the probabilities that Poisson random variables P1 , P2 .P3 and P4 of parameters λ1 = c6 t0 β −αk , λ2 = c6 t0 (β −αk − ξ −αk ), λ3 = c6 t0 (ξ + δ)−αk , and λ4 = c6 t0 (ξ −αk − (ξ + δ)−αk ), respectively, take the values 0, 0, 0, and 1, respectively. So there exists a constant c7 = c7 (αk , t0 , δ, ξ, β) > 0 such that ̃k has a single jump before time t0 , and its size is in [ξ, ξ + δ]) ≥ c7 . Px0 (Z

̃j does not have a jump before time t0 , is the probability that For all j ≠ k, the probability that Z ̃j for a Poisson random variable with parameter 2c6 t0 β −αj is equal to 0. Using the indepence of Z j = 1, . . . , d, we can find a c8 > 0 such that ̃j = 0 for all s ≤ t0 and all j ≠ k) ≥ c8 . Px0 (∆Z s

Thus we obtain

Px0 (D) ≥ c9

i ̃j ’s for all for a c9 = c9 (α1 , . . . , αd , t0 , δ, ξ, β) > 0. Furthermore the Z ’s are independent of the Z i, j ∈ {1, . . . , d}, so C and D are independent and we obtain

Similary we obtain (11)

Px0 (C ∩ D) ≥ c9 /2.

Px0 (E) ≥ c10

and Px0 (C ∩ E) ≥ c11 .

̃k jumps the first time, i.e. Z k makes a jump greater then β. Then Let T be the time, when Z


11

Zs− = Z s− for all s ≤ T and hence Xs− = X s− for all s ≤ T. So up to time T , Xs does not move away more than δ away from its starting point. Note ∆XT = A(XT − )∆ZT . By (8), we obtain on C ∩ D ∣XT − (x0 + ξvk )∣ ≤ ∣XT − − x0 ∣ + ∣∆XT − ξA(x0 )ek ))∣

= ∣XT − − x0 ∣ + ∣A(XT − )∆ZT − ξA(x0 )ek )∣

≤ ∣XT − − x0 ∣ + ξ∣(A(XT − ) − A(x0 ))ek ∣ + ∣A(XT − )(∆ZT − ξek )∣ ≤δ+

γ 1 γ γ ξdγ + δ∥A∥∞ ≤ ( + + 1) ≤ . 12d 6 ∥A∥∞ 2 2

Appling the strong Markov property at time T , we get by (11) Px 0 (

sup T ≤s≤T +t0

∣Xs − XT ∣ < δ) ≥ PXT (C ∩ E) ≥ c11 .

Note, that ∣XT − (x0 + ξvk )∣ < γ/2 and ∣Xs − XT ∣ < δ for all T ≤ s ≤ t0 imply ∣Xs − (x0 + ξvk )∣ < γ. All in all we get by the strong Markov property Px0 (sup ∣Xs − x0 ∣ < γ and sup ∣Xs − (x0 + ξvk )∣ < γ) ≥ T ≤s≤t0

s 0 the following lemma shows that solutions stay with positive probability in an ε-tube around a given line segment on [0, t1 ]. The case of α1 = ⋯ = αd was considered in [BC10].

R

Lemma 11. Let r ∈ (0, 1], x0 ∈ d , t1 > 0, ε ∈ (0, rα/α∗ ), ξ ∈ (0, ε/4) and γ > 0. Moreover let ψ ∶ [0, t1 ] → d be a line segment of length ξ starting at x0 . Then there exists c1 = c1 (Λ(Mr2 (x0 ))), ̟(Mr2 (x0 ))), t1 , ε, γ) > 0, such that

R

Px0 (sup ∣Xs − ψ(s)∣ < ε and ∣Xt1 − ψ(t1 )∣ < γ) ≥ c1 . s≤t1

12

JAMIL CHAKER

Proof. Note that ε is chosen such that Bε (x0 ) ⊂ Mr (x0 ). Let ρ ∈ (0, 1) be such that the conclusion of Lemma 10 holds for all matrices A = A(x) with x ∈ Mr2 (x). Take γ ∈ (0, ξ ∧ ρ) such that ρ̃ ∶= γ + ρ < 1 and n ≥ 2 sufficiently large, such that (̃ ρ)n < γ. Let v0 ∶= ψ(t1 ) − ψ(0) = ψ(t1 ) − x0 , which has length ξ. By Lemma 9, there exists a k0 ∈ {1, . . . , d} such that if p0 is the projection of v0 onto A(x0 )ek0 , then ∣v0 − p0 ∣ ≤ ρ∣v0 ∣. Note, that ∣p0 ∣ ≤ ∣v0 ∣ = ξ. By Lemma 8 there exists c2 > 0 and a stopping time T0 ≤ t1 /n such that for ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ D1 ∶= ⎨ sup ∣Xs − x0 ∣ < γ n+1 and sup ∣Xs − (x0 + p0 )∣ < γ n+1 ⎬ . ⎪ ⎪ s 0 and x0 ∈ d . Let ϕ ∶ [0, t0 ] → d be continuous with ϕ(0) = x0 and the image of ϕ contained in Mr (x0 ). Then there exists c1 = c1 (Λ(Mr2 (x0 )), ̟(Mr3 (x0 )), ϕ, ε, t0 ) > 0 such that Px0 (sup ∣Xs − ϕ(s)∣ < ε) > c1 . s≤t0

Proof. Let ε > 0. We define

U ∶= {x ∈

Rd ∶ ∃s ∈ [0, t0] such that ∣x − ϕ(s)∣ < ε/2}

and approximate ϕ within U by a polygonal path. Hence we can assume that ϕ is polygonal by changing ε to ε/2 in the assertion. We subdivide [0, t0 ] into n subintervals of the same length for n ≥ 2 such that ε kt0 (k + 1)t0 , ))) < , r ∶= L (ϕ (( n n 4 where L denotes the length of the line segment. Let ⎧ ⎫ ⎪ ⎪ ε ε ⎪ ⎪ sup ∣Xs − ϕ(s)∣ < and ∣Xkt0 /n − ϕ(kt0 /n)∣ < √ ⎬ . Dk ∶= ⎨ ⎪ 2 4 d⎪ ⎪ ⎪ ⎩(k−1)t0 /n≤s≤kt0 /n ⎭ Using Lemma 11, there exists a constant c2 > 0 such that Px0 (D1 ) ≥ c2 .

By the strong Markov property at time t0 /n we get

Px0 (D2 ∣Ft0 /n ) ≥ c2 .

Using the Iteration as in the proof of Lemma 11, we get for all k ∈ {1, . . . , d} Px0 (Dk ∣F(k−1)t0 /n ) ≥ c2

and Px0 ( ⋂ Dk ) ≥ cn2 . n

k=1

Hence the assertion follows by n ε Px0 (sup ∣Xs − ϕ(s)∣ < ) ≥ Px0 ( ⋂ Dk ) ≥ cn2 = c1 . 2 s≤t0 k=1

We state two corollaries, which follow immediately from Theorem 12.

R

Corollary 13. Let r ∈ (0, 1], ε ∈ (0, rα/α∗ /4), k = 1 − (ε/rα/α∗ ), δ ∈ (ε, rα/α∗ /2) and x0 ∈ d . Moreover let Q ∶= Mr (x0 ), Q′ ∶= Mrk (x0 ) and y ∈ d such that R ∶= Mrδ (y) ⊂ Q′ . There exists c1 = c1 (Λ(Q), ̟(Q), ε, δ) > 0 such that

R

Px (TR < τQ ) ≥ c1 ,

x ∈ Q′ .

Proof. Note, that dist(∂Q, ∂Q′ ) = ∣rα/αi − krα/αi ∣ = ∣ε

rα/αi ∣ ≥ ε. rα/α∗


15

R

Let x ∈ Q′ be arbitrary and ϕ ∶ [0, t0 ] → d be a polygonal path such that ϕ(0) = x and ϕ(t0 ) = y and the image of ϕ is contained in Q′ . Then the assertion follows by Theorem 12 and Px (sup ∣Xs − ϕ(s)∣ < ε) ≤ Px (TR < τQ ). s≤to

R

Corollary 14. Let r ∈ (0, 1], x0 ∈ d and ε ∈ (0, rα/α∗ /4). For x ∈ Mr (x0 ), we define R ∶= Ms (x) such that R ⊂ Mr (x0 ) =∶ M and dist(∂R, ∂M ) > ε. Then there exists a ξ = ξ(ε, Λ(M ), r, ̟(M ) ∈ (0, 1) such that Py (TR < τM ) ≥ ξ = ξ(ǫ). for all y ∈ M with dist(y, ∂M ) > ε. Proof. Follows immediately by Corollary 13.

We now prove the main ingredient for the proof of the H¨ older regularity. It states, that sets of positive Lebesgue measure are hit with positive probability. Theorem 15. Let r ∈ (0, 1], x0 ∈ ϕ ∶ (0, 1) → (0, 1) such that

Rd and M ∶= Mr (x0 ). There exists a nondecreasing function Px (TA < τM ) ≥ ϕ(∣A∣)

for all x ∈ Mr (x0 ) and all A ⊂ M with ∣A∣ > 0. 1/2

Proof. We will follow the proof of Theorem V.7.4 in [Bas98]. Set

and

ϕ(ε) = inf {Py (TA < τMR (z0 ) ) ∶ z0 ∈

Rd, R > 0, y ∈ MR1/2(z0), ∣A∣ ≥ ε∣MR(z0)∣, A ⊂ MR(z0).}

q0 ∶= inf{ε ∶ ϕ(ε) > 0}. For ǫ sufficiently large, we know by Corollary 13 ϕ(ǫ) > 0. We suppose q0 > 0, and we will obtain our contradiction. Since q0 < 1, we can choose 1 > q > q0 such that (q + q 2 )/2 < q0 . Moreover let η ∶= (q − q 2 )/2,

⎛ 21−d qr− ∑i=1 (α/αi ) ⎞ q+1 ⎝ ⎠ d

β ∶=

1/d

and ρ = ξ((1 − β)rα/α∗ /6),

where ξ is defined as in Corollary 14. 1/2 Let z ∈ d , R ∈ (0, 1], x ∈ MR (z) and A ⊂ MR (z) such that

R

(13)

q−η
(q − η)∣Mr (x0 )∣,

∣D ∩ Mr (x0 )∣ ≥

Define E = D ∩ Mrβ (x0 ). Since

∣A∣ (q − η)∣Mr (x0 )∣ (q + 1)∣Mr (x0 )∣ > = . q q 2

(q + 1)∣Mrβ (x0 )∣ (q + 1)2d β d r∑i=1 αi = = q, 2 2 we get ∣E∣ > q. By the definition of ϕ, we have Px (TE < τMr (x0 ) ) ≥ ϕ(q). We will first show (14)

d

α

Py (TA < τMr (x0 ) ) ≥ ρϕ(q) for all y ∈ E.

̂i for some Ri ∈ R and dist(y, ∂Mr (x0 )) ≥ (1 − β)rα/α∗ . Define R∗ as the cube Let y ∈ ∂E, then y ∈ R i with the same center as Ri but sidelength half as long. By Corollary 14 By Proposition 4 (3) for all Ri ∈ R and therefore

Py (TR∗i < τMr (x0 ) ) ≥ ρ. ∣A ∩ Ri ∣ ≥ q∣Ri ∣

Px0 (TA∩Ri < τMr (x0 ) ) ≥ ϕ(q) Using the strong Markov property, we have for all y ∈ E Py (TA < τMr (x0 ) ) ≥ Ey [P

XT

≥ ρϕ(q).

R∗ i

for x0 ∈ Ri∗ .

(TA < τRi ); TR∗i < τMr (x0 ) ]

Now we get our contradiction by

Px (TA < τMr (x0 ) ) ≥ Px (TE < TA < τMr (x0 ) )

≥ Ex [PXTE (TA < τMr (x0 ) ); TE < τMr (x0 ) ] ≥ ρϕ(q)Px [TE < τMr (x0 ) ] ≥ ρϕ(q)2 .

4. Proof of Theorem 2 In this section we give the proof of the H¨ older regularity. Proof. Let S ∶= Ms (y) ⊂ Mr (x0 ) and A ⊂ S such that 3∣A∣ ≥ ∣S∣. Those sets will be specified later. Set k = 1 − (ε/rα/α∗ ) and S ′ ∶= Msk (y), where ε is chosen such that 6∣S ∖ S ′ ∣ = ∣S∣. Then 6∣A ∩ S ′ ∣ ≥ ∣S∣.

Let R be a collection of N equal sized rectangles as in Definition 1 with disjoint interiors and R ⊂ S. Moreover let R to be a covering of S ′ . For at least one rectangle Q ∈ R 6∣A ∩ S ′ ∩ Q∣ ≥ ∣Q∣.


17

Let Q′ be the rectangle with the same center as Q but each sidelength half as long. By Corollary 13 there exists a c2 > 0 such that (15)

Px (TQ′ < τS ) ≥ c2 ,

x ∈ Ms1/2 (y).

(16)

Px (TA < τS ) ≥ c3 ,

x ∈ Ms1/2 (y).

Using Theorem 15 and the strong Markov property there exists a constant c3 > 0 with Let R ≥ 2r. By Proposition 7 there exists a c4 > 0 such that r α Pz (XτMr (x0 ) ∉ MR (x0 )) ≤ c4 ( ) for all z ∈ Mr (x0 ). R Let 1/α c3 γ 2 γ 1/α α∗ ln(γ) γ ∶= (1 − c3 ), ρ ∶= ( ) ∧( ) and β ∶= . 4c4 2 α ln(ρ)

R

By linearity it suffices to suppose 0 ≤ h ≤ M on d . We first consider the case r = 1. Let Mi = Mρi (x0 ) and τi = τMi . We will show that for all k ∈ 0

N

osc h ∶= sup h − inf h ≤ M γ k .

(17)

Mk

Mk

Mk

To shorten notation, we set ai = inf h and bi = sup h. Assertion (17) will be will be proved by induction. Let k ∈ to show

Mi

N be arbitrary but fixed.

Mi

We suppose bi − ai ≤ M γ i for all i ≤ k; then we need

bk+1 − ak+1 ≤ M γ k+1 .

(18)

By definition Mk+1 ⊂ Mk and therefore in particular ak ≤ h ≤ bk on Mk+1 . Set A′ = {z ∈ Mk+1 ∶ h(z) ≤ (ak + bk )/2}.

Without loss of generality, assume 2∣A′ ∣ ≥ ∣Mk+1 ∣. If this assumption does not hold, we consider M − h instead of h. Let A ⊂ A′ be compact such that 3∣A∣ ≥ ∣Mk+1 ∣. By (16) there exists a c3 > 0 such that Py (TA < τk ) ≥ c3 for all y ∈ Mk+1 . Let ε > 0 and y, z ∈ Mk+1 such that h(y) ≥ bk+1 − ε and h(z) ≤ ak+1 + ǫ. Since h is harmonic, h(Xt ) is a martingale. We get by optimal stopping h(y) − h(z) =Ey (h(XTA ) − h(z); τk > TA )

+ Ey (h(Xτk ) − h(z); τk < TA , Xτk ∈ Mk−1 ) ∞

+ ∑ Ey (h(Xτk ) − h(z); τk < TA , Xτk ∈ Mk−i−1 ∖ Mk−i ) . i=1

We will now study these three components on the right hand side seperately. Note h(z) ≥ ak ≥ ak−i−1 for all i ∈

N

(1) In the first component, X enters A ⊂ A′ before leaving Mk . Hence ak + b k − ak ; TA < τk ) Ey (h(XTA ) − h(z); TA < τk ) ≤ Ey ( 2 b k − ak y M γk y = P (TA < τk ) ≤ P (TA < τk ). 2 2

18

JAMIL CHAKER

(2) In the component X leaves Mk before entering A. While leaving Mk , X does not make a big jump in the following sense: X is at time τk in Mk−1 . Hence in this case h(Xτk ) ≤ bk−1 . This yields to Ey (h(Xτk ) − h(z); τk < TA , Xτk ∈ Mk−1 ) ≤ Ey (bk−1 − ak−1 ; τk < TA , Xτk ∈ Mk−1 ) = (bk−1 − ak−1 )Py (τk < TA ) ≤ M γ k−1 (1 − Py (TA < τk )).

(3) In the third componentXτk ∈ Mk−i−1 for i ∈ ∞

N. Therefore h(τk ) ≤ bk−i−1 .

y ∑E (h(Xτk ) − h(z); τk < TA , Xτk ∈ Mk−i−1 ∖ Mk−i )

i=1

∞

≤ ∑ Ey ((bk−i−1 − ak−i−1 ); τk < TA , Xτk ∈ Mk−i−1 ∖ Mk−i ) i=1 ∞

∞

= ∑(bk−i−1 − ak−i−1 )Py (Xτk ∉ Mk−i ) ≤ ∑ M γ k−i−1 c4 ( i=1

i=1

∞

= c4 M γ k−1 ∑ ( i=1

α

ρ ρ ) = c4 M γ k−1 γ γ − ρα

≤ 2c4 γ k−2 M ρα ≤ where we used

α i

α

ρk ) ρk−i

c3 M γ k , 2 γ 1/α ρα ρα and ρ ≤ ( ) ⇔ ≤2 . α 2 γ −ρ γ

ρα 1 ≤ γ 2

Note, that the choice of γ = 1 − c3 implies

γ−2 c 2γ 3

1 γ

+

+

c3 2

= γ. Hence

M γk y M γ k−1 c3 M γ k P (TA < τk ) + (1 − Py (TA < τk )) + β 2 r 2 1 c3 γ −2 y k P (TA < τk ) + + ) = M γ (( 2γ γ 2 γ − 2 1 c 3 ≤ M γk ( c3 + + ) 2γ γ 2

h(y) − h(z) ≤

= M γ k+1 . We conclude that

bk+1 − ak+1 ≤ M γ k+1 + 2ε. Since ε is arbitrary, this proves (18) and therefore (17). Let x, y ∈ M1 (x0 ) and choose k ∈ 0 such that Mρ2k (x) is the smallest rectangle with y ∈ Mρk (x). √ ∗ Then ∣x − y∣ ≤ 2 dρ(αk)/α and therefore

N

k≥ Hence

α∗ log ( 2√d ) ∣x−y∣

for y ∈ Mρk (x).

α log(ρ)

∣h(y) − h(x)∣ ≤ M γ k = M ek log(γ)

√ d))((α∗ log(γ))/(α log(ρ)))

≤ M e(log[∣x−y∣/(2


=

M ∣x − y∣(α

∗

log(γ))/(α log(ρ))

√ 2 d

19

= c5 M ∣x − y∣β .

Now let h be harmonic on Mr2 (x0 ). Then h′ (x) ∶= h(rα/α∗ x) is harmonic on M12 (x0 ). Let x, y ∈ M1 (x0 ) and x′ , y ′ ∈ Mr (x0 ) such that x = (x′1 /rα/α1 , . . . , x′d /rα/αd ),

y = (y1′ /rα/α1 , . . . , yd′ /rα/αd ) ∈ M1 (x0 ).

Then ∣x − y∣ ≤ rα/α∗ ∣x′ − y ′ ∣. Set ̃ x = x′ /rα/α∗ and ỹ = y ′ /rα/α∗ . We conclude ∣h(x′ ) − h(y ′ )∣ = ∣h(rα/α∗ x ̃) − h(rα/α∗ ỹ)∣ = ∣h′ (̃ x) − h′ (̃ y)∣ z )∣ ≤ c1 ∣̃ x − ỹ∣β sup ∣h(̃

R

z ̃∈

= c1 ∣

x′ rα/α∗

−

d

y′ rα/α∗

β

∣ sup ∣h(z)∣ = c1 (

R

z∈

d

∣x′ − y ′ ∣ ) sup ∣h(z)∣. rα/α∗ z∈Rd β

References [Bas98] Richard F. Bass. Diffusions and elliptic operators. Probability and its Applications (New York). SpringerVerlag, New York, 1998. [BC06] Richard F. Bass and Zhen-Qing Chen. Systems of equations driven by stable processes. Probab. Theory Related Fields, 134(2):175–214, 2006. [BC10] Richard F. Bass and Zhen-Qing Chen. Regularity of harmonic functions for a class of singular stable-like processes. Math. Z., 266(3):489–503, 2010. [BL02] Richard F. Bass and David A. Levin. Harnack inequalities for jump processes. Potential Anal., 17(4):375–388, 2002. [CT04] Rama Cont and Peter Tankov. Financial modelling with jump processes. Chapman & Hall/CRC Financial Mathematics Series. Chapman & Hall/CRC, Boca Raton, FL, 2004. [EK86] Stewart N. Ethier and Thomas G. Kurtz. Markov processes. Wiley Series in Probability and Mathematical Statistics: Probability and Mathematical Statistics. John Wiley & Sons, Inc., New York, 1986. Characterization and convergence. [KS79] N. V. Krylov and M. V. Safonov. An estimate for the probability of a diffusion process hitting a set of positive measure. Dokl. Akad. Nauk SSSR, 245(1):18–20, 1979. [Sat13] Ken-iti Sato. L´ evy processes and infinitely divisible distributions, volume 68 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 2013. Translated from the 1990 Japanese original, Revised edition of the 1999 English translation. E-mail address: [email protected]

¨ t Bielefeld, Fakulta ¨ t fu ¨ r Mathematik, Postfach 100131, D-33501 Bielefeld, Germany Universita