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CH-Heat Kernel for Perturbation of by gradient operator


Heat Kernel for Perturbation of ? + ?α/2 by gradient operator
Zhen-Qing Chen and Eryan Hu?

Abstract For d ≥ 2, α ∈ (0, 2) and M > 0, we consider the gradient perturbation of the operator {? + aα ?α/2 , a ∈ (0, M ]}. We establish the existence of the continuous heat kernel of the operator La,b = ? + aα ?α/2 + b · ?, where b is in Kato class Kd,1 on Rd . Furthermore, we give the sharp two-sided estimates of the heat kernel of La,b .

AMS 2010 Mathematics Subject Classi?cation: Primary 35K08, 60J35, 47G20; Secondary 47D07 Keywords and phrases: heat kernel, transition density, Feller semigroup, perturbation, positivy, L? evy system, gradient operator, Kato class

1

Introduction

Markov processes may have continuous paths called di?usion, or discontinuous paths called jump process. Brownian motion is a di?usion with in?nitesimal generator ?. Perturbation of ? by gradient operator b · ? has been studied by Cranston and Zhao [11] and Zhang [17, 18], where b belongs to some appropriate Kato class. The authors in [17, 18] proved that the transition density corresponding to the operator ? + b · ? has Gaussian Bounds. The rotationally symmetric process with generator (fractional Laplacian) ?α/2 , α ∈ (0, 2) is a very important example of jump process. Bogdan and Jakubowski [3] studied the perturbation of ?α/2 by gradient operator b · ? for α ∈ (1, 2) and b ∈ Kd,α?1 . The authors established the existence and the two-sided bounds of the heat kernel of ?α/2 + b · ?. Chen, Kim and Song [5] further obtained the dirichlet heat kernel estimates of ?α/2 + b · ?. For more results, we refer the reader to the work of Kim and Song [14, 15], Jakubowski and Szczypkowski [13]. Markov process with both di?usion and jump components has also received much attention in recent years. Song and Vondraˇ cek [16] showed the sharp bounds of the heat kernel of α/ 2 ? + ? . Chen and Kumagui [4] generalized this results with di?erent method. See [6, 7, 8] for more results. In this paper, we assume d ≥ 2 to be an integer and α ∈ (0, 2). We consider the perturbation of ? + aα ?α/2 for a ∈ (0, M ] by b · ? for some constant M > 0 and b ∈ Kd,1 . We establish the sharp two-sided bounds of the corresponding heat kernel La,b , which can recover the results in Zhang [17, 18] by letting a → 0. Let X 0 be a Brownian motion in Rd with in?nitesimal generator ? and Y be the rotationally symmetric α-stable process in Rd which is the L? evy process such that Ex [eiξ(Yt ?Y0 ) ] = e?t|ξ| , for every x, ξ ∈ Rd and t > 0. The generator of Y is ?α/2 := ?(??)α/2 , which is a prototype of nonlocal operator and can be written in the form f (y ) ? f (x) 2 dy, f ∈ Cc (Rd ), (1.1) ?α/2 f (x) = lim A(d, ?α) ε→0 |x?y |≥ε |x ? y |d+α
? α

supported partially by NFSC (11071138).

1

where A(d, ?α) := α2α?1 π ?d/2 Γ((d + α)/2)Γ(1 ? α/2)?1 . Here Γ is the Gamma function ∞ de?ned by Γ(λ) := 0 tλ?1 e?t dt, λ > 0. Assuming that X 0 and Y are independent, for a > 0, a := X 0 + aα Y . We call the process X a the independent sum of the we de?ne X a by Xt t t Brownian motion X 0 and the symmetric α-stable process Y with weight a. Obviously, the in?nitesimal generator of X a is ? + aα ?α/2 . It is well known that the symmetric process X a has L? evy intensity function J a (x, y ) = aα A(d, ?α)|x ? y |?(d+α) . (1.2)

The function J a (x, y ) determines a L? evy system for X a , which describes the jumps of the process X a . We denote the fundamental solution of ? + aα ?α/2 by pa (t, x, y ) = pa t (x ? y ), where pa ( x ) is the smooth function determined by t
iz ·ξ pa dz = e?t(|ξ| t (z )e
2 +aα |ξ |α )

,

Rd

ξ ∈ Rd .

(1.3)

which is also the transition density of X a , and denote the fundamental solution of La,b by pa,b (t, x, y ). The informal idea of constructing pa,b (t, x, y ) is as follows. Since La,b is the perturbation of ? + aα ?a,b by b · ?, pa,b (t, x, y ) and pa (t, x, y ) should satisfy the Duhamel’s formula:
t

pa,b (t, x, y ) = pa (t, x, y ) +
0 Rd

pa,b (t ? s, x, z )b(z )?z pa (s, z, y )dzds,

(1.4)

for t > 0 and x, y ∈ Rd . As in [3, 18], we may apply (1.4) repeatedly and expect that a,b a,b a pa,b (t, x, y ) = ∞ k=0 pk (t, x, y ), if the sum is convergent, where p0 (t, x, y ) = p (t, x, y ) and pa,b k (t, x, y ) =
t 0 Rd a pa,b k?1 (t, x, z )b(z )?z p (s, z, y )dzds, for k ≥ 1.

(1.5)

The estimate of each pa,b k (t, x, y ) is mainly based on Theorem 3.2, which is an analogy of so called 3P Theorem such as [3, Lemma 13] or [18, Lemma 3.1]. However, we received many di?culties. Unlike in [18], there is only Gaussian term, or unlike in [3], there is only polynomial term, we have to deal with the mixture of them, since the upper bound of pa (t, x, y ) has both Gaussian term and polynomial term. To simply the computation, we ?nd a proper upper bound of the Gaussian term in Theorem 3.2. On the other hand, it is not as easy to verify a,b ∞ k=0 pk (t, x, y ) is non-negative as the author did in [3]. Fortunately, we may apply HilleYosida-Ray theorem(See [1, Theorem 3.5.1]) to get that pa,b (t, x, y ) ≥ 0 for continuous b. For a,bn (t, x, y ) → pa,b (t, x, y ) in some regular b, we choose proper continuous {bn }∞ n=1 such that p a,b sense to get the positivity of p (t, x, y ). We Denote C∞ (Rd ) the space of continuous functions on Rd vanishing at the in?nity with k (Rd ) the space of all continuous functions supremum norm. For an integer k > 0, we denote Cc d on R with compact supports that have continuous derivatives up to and including k -order, ∞ (Rd ) = ∩∞ C k (Rd ). and let Cc k=1 c The following two theorems are the main results of this paper. Theorem 1.1. Suppose b ∈ Kd,1 . For every a ∈ (0, M ], there is a positive jointly continuous function pa,b (t, x, y ) on (0, ∞) × Rd × Rd satisfying (1.4) and the followings hold, (1) There is a constant t? (d, α, M, b) > 0 depending on b only via the rate at which Mb (r) goes to zero, such that


p

a,b

(t, x, y ) =
k=0

d d pa,b k (t, x, y ), on (0, t? ] × R × R ,

where pa,b k (t, x, y ) is de?ned by (1.5). 2

(2) (semigroup property): pa,b (t + s, x, y ) =
Rd

pa,b (t, x, z )pa,b (s, z, y )dz,

for t > 0, x, y ∈ Rd .

(3) (conservativeness property): pa,b (t, x, y )dy = 1,
Rd ∞ (Rd ) and g ∈ C (Rd ), (4) for every f ∈ Cc ∞

for t > 0, x ∈ Rd .

t→0 Rd

lim

Pta,b f (x) ? f (x) g (x)dx = t

La,b f (x)g (x)dx,
Rd

(1.6)

where Pta,b f (x) =

Rd

pa,b (t, x, y )f (y )dy .

Here and after, the meaning of the phrase ”depending on b only via the rate at which Mb (r) goes to zero” is that the statement is true for any real valued or Rd valued function ? b on Rd with M? b (r ) ≤ Mb (r ), for all r > 0. Before we state the second result, we make some notation for simplicity. For positive real numbers a1 , a2 , a1 ∧ a2 denotes min{a1 , a2 } and a1 ∨ a2 denotes max{a1 , a2 }. For constants a, c > 0, we de?ne
a qd,c (t, x, y ) = t?d/2 exp ?

c|x ? y |2 t

+ t?d/2 ∧

aα t , for t > 0, x, y ∈ Rd . |x ? y |d+α

Theorem 1.2. For every M > 0 and T > 0, there are constants Ci = Ci (d, α, M ), i = 1, 2 and Cj = Cj (d, α, M, T, b), j = 3, 4 depending on b only via the rate at which Mb (r) goes to zero, such that for all a ∈ (0, M ] and (t, x, y ) ∈ (0, T ] × Rd × Rd ,
a a C4 qd,C (t, x, y ) ≤ pa,b (t, x, y ) ≤ C3 qd,C (t, x, y ). 2 1

It is more di?cult to get the lower bound of pa,b (t, x, y ) than the upper bound. We will mainly use the L? evy system of the Feller process X a,b associated with {Pta,b , t ≥ 0} to get the polynomial part in the lower bound (see Lemma 5.7), and use the technology in [12](see also [17]) to get the Gaussian part (see Lemma 5.8). The rest of this paper is organized as follows. In section 2, we recall some results of pa (t, x, y ) and give the estimates of ?x pa (t, x, y ). Also, we introduce the Kato class Kd,1 and some property of the functions in Kd,1 . In section 3, we construct pa,b (t, x, y ) through the series of pa,b k (t, x, y ) and verify Theorem 1.1 through series of Lemmas except the positivity of pa,b (t, x, y ). Besides, we give the upper bound of |pa,b (t, x, y )|. The positivity of pa,b (t, x, y ) is shown in section 4, where we will use the result that {Pta,b , t ≥ 0} is a strongly continuous semigroup in C∞ (Rd ). In section 5, we give the lower bound of the heat kernel mainly through the L? evy system of X a,b which is the Feller process associated with {Pta,b , t ≥ 0}. Finally, we show the proof of Theorem 1.2. Throughout this paper, unless stated otherwise, the capital letters C1 , C2 , · · · will be ?xed. The lower case constants c1 , c2 , · · · will change from one appearance to another. The exact values of which are not important. Besides, c = c(d, α, · · · ) means that the constant c depends only on d, α, · · · . We use the symbol := to denote a de?nition. For two non-negative functions
c

f, g , the notation f g means that f ≤ cg on their common domain of de?nition while f g means that c?1 g ≤ f ≤ cg . We also write mere and if c is unimportant or understood.

c

3

2

Preliminaries

The following theorem is shown in [6, Corollary 1.2]. See also [4, Theorem 1.4] and [16, Theorem 2.13]. Theorem 2.1. For any M > 0 and T > 0, there exist constants Ci , i = 5, 6 and Cj = Cj (d, α, M, T ), j = 7, 8 such that for all a ∈ (0, M ] and (t, x, y ) ∈ (0, T ] × Rd × Rd ,
a a (t, x, y ). (t, x, y ) ≤ pa (t, x, y ) ≤ C7 qd,C C8 qd,C 5 6

For a positive constant β > 0 we de?ne, for simplicity, gβ (t, x, y ) = t?d/2 exp ? β |x ? y |2 t , (t, x, y ) ∈ (0, ∞) × Rd × Rd .

Easily, for any θ > 0, there is a positive constant c1 = c1 (d, β, θ) such that gβ (t, x, y ) ≤ t?d/2 ∧ c1 tθ , |x ? y |d+2θ t > 0 and x, y ∈ Rd , (2.1)

which will be frequently used in the rest of this paper. a (t, x, y ). In fact, there is a constant C = C (α, M, T, β ) such Recall the de?nition of qd,β 9 9 d d that for all a ∈ (0, M ] and all (0, T ] × R × R ,
a (t, x, y ) qd,β C9

gβ (t, x, y ) +

aα t 1 (t, x, y ). 2 |x ? y |d+α {|x?y| ≥t}

(2.2)

Indeed, note that if |x ? y |2 < t, then t?d/2 ∧ If |x ? y |2 ≥ t, then t?d/2 ∧ aα t |x ? y |d+α ≤ aα t . |x ? y |d+α aα t |x ? y |d+α ≤ t?d/2 ≤ eβ gβ (t, x, y ).

Adding up the last two inequalities, we have for all a > 0, t > 0 and x, y ∈ Rd ,
a qd,β (t, x, y ) eβ +1

gβ (t, x, y ) +

aα t 1 . 2 |x ? y |d+α {|x?y| ≥t}

(2.3)

On the other hand, for all a ∈ (0, M ] and t ∈ (0, T ], if |x ? y |2 ≥ t, aα t ≤ M α t?d/2+1?α/2 ≤ M α T 1?α/2 t?d/2 , |x ? y |d+α and so, aα t 1 (t, x, y ) = (M α T 1?α/2 t?d/2 ) ∧ 2 |x ? y |d+α {|x?y| ≥t} aα t 1 (t, x, y ) 2 |x ? y |d+α {|x?y| ≥t} aα t . ≤ (M α T 1?α/2 ∨ 1) t?d/2 ∧ |x ? y |d+α

Adding gθ (t, x, y ) to both sides of the above inequality we have gβ (t, x, y ) +
M aα t 1 2 ≥t} (t, x, y ) {| x ? y | |x ? y |d+α
α T 1?α/2 ∨1

a qd,β (t, x, y ).

(2.4)

Therefore, (2.2) follows from (2.3) and (2.4) for C9 = (M α T 1?α/2 ∨ 1) ∨ (eβ + 1). The following theorem shows the estimate of |?x pa (t, x, y )|. In this paper, we only use the upper bound. 4

Theorem 2.2. For any M > 0 and T > 0 ,there is a positive constant C10 = C10 (d, α, M, T ) such that for all a ∈ (0, M ] and (t, x, y ) ∈ (0, T ] × Rd × Rd
a a a 2πC8 qd +2,C6 (t, x, y )|x ? y | ≤ |?x p (t, x, y )| ≤ C10 qd+1,3C5 /4 (t, x, y ). ?d/2 e?|x| /(4t) be the Gaussian Proof. By (1.3), it su?ces to estimate |?pa t |. let gt (x) = (4πt) (α) kernel and pt be the heat kernel of Y . Since X 0 and Y are independent, we have for any x ∈ Rd ,
2

pa t (x) =

Rd

gt (x ? y )paα t (y )dy.

(α)

(2.5)

See the second paragraph of [16, Section 2] for the case a = 1. By (2.5), pt attains is maximum d at 0. Then we have ?pa t (0) = 0, t > 0. Now, We assume x ∈ R \{0}. Let St be the α/2-stable a (u) be the density function of a2 S . Since B a subordinator at time t and ηt t t+a2 St and Xt have the same distribution, by the subordination formula:
+∞ +∞

pa (t, x) =
t

gu (x)P(t + a2 St ∈ du) =
t

a gu (x)ηt (u ? t)du,

we have


?x pa (t, x) = ?x
t

a gu (x)ηt (u ? t)du.

Letej = (0, · · · , 0, 1, 0, · · · , 0), where 1 is on j th place. Set s ∈ (?|x|/2, |x|/2). By the mean value theorem, there exists ξ ∈ (?|s|, |s|) such that gu (x + sej ) ? gu (x) xj + ξ ? gu (x + ξej ) = gu (x + ξej ) = s ?xj 2u |x|gu (x/2) ≤ c(d)|x|?d?1 , ≤ u where c(d) is a positive constant depending only on d. Since by dominated convergence theorem we have
∞ ∞ a ?x gu (x)ηt (u ? t)du = t t ∞ ?d?1 η (u ? t)du t t c(d)|x|

< ∞,

?x pa (t, x) =

?

xgu (x) a ηt (u ? t)du = ?2πxpa ?), (2.6) (d+2) (t, x 2u

where x ? ∈ Rd+2 is such that |x ?| = |x| and p(d+2) (t, x ?) is the heat kernel of X a in dimension d + 2. Thus, by Theorem 2.1, we have
a a a 2πC8 qd +2,C6 (t, x, y )|x ? y | ≤ |?x p (t, x, y )| ≤ 2πC7 qd+2,C5 (t, x, y )|x ? y |.

Note that for all t > 0 and x, y ∈ Rd , C5 |x ? y |2 |x ? y | t |x ? y | 3C5 |x ? y |2 C5 |x ? y |2 =t?(d+1)/2 exp ? · 1/2 exp ? 4 t 4 t t 2 ?(d+1)/2 3C5 |x ? y |2 ≤ t exp ? , C5 e 4 t t?(d+2)/2 exp ? and by (2.2), we ?nish the proof for C10 := 2πC7 C9 2/(C5 e) ∨ 1 .

d d d For a vector ?eld f = (fj )d j =1 : R → R or a function f : R → R, we de?ne

Mf (r) = sup
x∈Rd |x?y |<r

|f (y )| dy, |x ? y |d?1 5

for r > 0.

d d d De?nition 2.3. We say that a vector ?eld f = (fj )d j =1 : R → R or a function f : R → R belongs to Kato class Kd,1 , if lim Mf (r) = 0. r↓0

It’s easy to see that if f ∈ Kd,α?1 , then for any r > 0, Mf (r) < ∞. Indeed, by de?nition we α (r ) < ∞. For any r > 0, since M (r ) is decreasing see that there exists r0 > 0 such that Mf 0 f in r, we need only to consider the case r > r0 . Note that B (x, r) can be covered by ?nite number of balls {B (xj , r0 )}n j =0 with radius r0 and x0 = x, then we have ? ? n ? |b(y )| dy Mf (r) ≤ sup ? + |y ? x|d?1 B (x,r0 ) B (x,r0 )c ∩B (xj ,r0 ) x∈Rd
j =1 n

≤ Mf (r0 ) + sup
x∈Rd j =1 n B (x,r0 )c ∩B (xj ,r0 )

|b(y )| |y ? xj |d?1 dy |y ? xj |d?1 |y ? x|d?1

≤ Mf (r0 ) + sup
x∈Rd j =1 B (xj ,r0 )

|b(y )| dy |y ? xj |d?1

≤ (n + 1)Mf (r0 ) < ∞, where in the last second inequality we have used the fact that for any y ∈ B (x, r0 )c ∩ B (xj , r0 ), |y ? x| ≥ r > |y ? xj |. Particularly, for any compact set K ? Rd we have |f (y )|dy < ∞
K

(2.7)

For β >

1 2

d d d and a vector ?eld f = (fj )d j =1 : R → R or a function f : R → R, we de?ne

rβ 1 ∧ , r > 0, x, y ∈ Rd |x ? y |d?1 |x ? y |d?1+2β 1 rβ β (r, x) = |f (y )| Hf ∧ dy, r > 0, x ∈ Rd . |x ? y |d?1 |x ? y |d?1+2β Rd H β (r, x, y ) =
1 . There is a constant C11 = C11 (d, β ) such that Lemma 2.4. Assume β > 2

√ β Hf (r, x) ≤ C11 Mf ( r),

(2.8)

d d d for every r > 0, x ∈ Rd and vector ?eld f = (fj )d j =1 : R → R or function f : R → R. And consequently, f ∈ Kd,1 if and only if β lim sup Hf (r, x) = 0. r↓0 x∈Rd

The proof of Lemma 2.4 is almost the same to [3, Lemma 11,corollary 12], if we replace α by 2 in their proofs. Here we omit the details. Since there is an exponential term in the bounds of pa (t, x, y ), we introduce another function class. For β > 0, we de?ne
r

N β (r, x, y ) =
0

s?(d+1)/2 exp ?

β |x ? y |2 s

ds,

r > 0, x, y ∈ Rd .

The following de?nition appears in [17] and [18, de?nition 1.1], etc.
d d d De?nition 2.5. We say that a vector ?eld f = (fj )d j =1 : R → R or a function f : R → R satis?es condition K , if

lim sup

r↓0 x∈Rd

|f (z )|N β (r, x, y )dy = 0, for all β > 0.
Rd

6

The Claim following the de?nition of condition K in [17] shows that if f ∈ L1 (Rd ) and f ∈ Kd,1 , then f satis?es condition K (see also [18, Proposition 2.1]). However we can verify a stronger conclusion. Lemma 2.6. f ∈ Kd,1 if and only if f satis?es condition K . Proof. Changing the variable s in the expression of N β (r, x, y ) by β |x ? y |2 /s, we have
r 0

s?(d+1)/2 exp ?

β |x ? y |2 s



ds =
β |x?y |2 /r

s(d?3)/2 e?s ds. β (d?1)/2 |x ? y |d?1

(2.9)

Then, for every 0 < r < 1, sup
x∈Rd Rd

|f (z )|N β (r, x, y )dy ≥ sup

s(d?3)/2 e?s |f (y )|dsdy β (d?1)/2 |x ? y |d?1 x∈Rd |x?y |<r β |f (y )| = c(d, β ) sup dz, d?1 x∈Rd |x?y |<r |x ? y | On the other hand, by (2.9),



(2.10)

where c(d, β ) =

∞ (d?3)/2 ?s e ds/β (d?1)/2 . β s ∞

N β (r, x, y ) =
β |x?z |2 /r

s(d?3)/2 e?s ds β (d?1)/2 |x ? z |d?1 Γ( d?1 1 ) 2 |x ? z |d?1 ∧ Γ( r d+1 ) 2 β |x ? z |d+1 (2.11)



1 β (d?1)/2

≤ c(d, β )H 1 (r, x, y ),

d+1 1 (d+1)/2 ∧ 1). Combining (2.10), (2.11) and Lemma 2.4, where c(d, β ) = (Γ( d? 2 ) ∨ Γ( 2 ))/(β we ?nish the proof.

3

Construction of the heat kernel and its upper bound

By [18, Lemma 3.1] and its proof we have the following lemma. Lemma 3.1. For any 0 < β1 < β2 < ∞, there exist constants Cg = Cg (d, β1 /β2 ) and Cβ = min{β2 ? β1 , β1 /2} such that for all t > 0 and x, y ∈ Rd ,
t 0

gβ1 (t ? s, x, z )gβ2 (s, z, y )s?1/2 ds ≤ Cg (N Cβ (t, x, z ) + N Cβ (t, z, y ))gβ1 (t, x, y ).

In the rest of this paper, we assume b ∈ Kd,1 and let γ = (1 + α ∧ 1)/2. The following lemma plays an important role in this paper and it is an analogy of [3, Lemma 13] or [18, Lemma 3.1]. Lemma 3.2. Suppose M > 0 and T > 0. For any 0 < β1 < β2 < ∞, there is a positive constant C12 = C12 (d, α, M, T, β1 , β2 ) such that for all a ∈ (0, M ] and (t, x, y, z ) ∈ (0, T ] × Rd × Rd × Rd ,
t 0 a a qd,β (t ? s, x, z )qd +1,β2 (s, z, y )ds 1 C12 a (H γ (t, x, z ) + H γ (t, z, y ))qd,β (t, x, y ), 1

(3.1)

and consequently, there is a positive constant C13 = C13 (d, α, M, T ) such that for all a ∈ (0, M ] and (t, x, y ) ∈ (0, T ] × Rd × Rd ,
t 0 Rd a a qd,β (t ? s, x, z )|b(z )|qd +1,β2 (s, z, y )dzds 1 C13

√ a Mb ( t)qd,β (t, x, y ), 1

(3.2)

7

Proof. We ?rst verify (3.1). By (2.2), for all (t, x, y ) ∈ (0, T ] × Rd × Rd , there is a constant c1 = c1 (α, M, T, β1 , β2 ) such that
t

I :=
0 c1 t

a a (t ? s, x, z )qd qd,β +1,β2 (s, z, y )ds 1

gβ2 (s, z, y ) aα (t ? s) aα s 1 + 1 ds 2 2 |x ? z |d+α {|x?z | ≥t?s} |z ? y |d+1+α {|z ?y| ≥s} s1/2 0 t gβ (s, z, y ) ds = gβ1 (t ? s, x, z ) 2 1/2 s 0 Rd t aα s + ds gβ1 (t ? s, x, z ) 1 2 |z ? y |d+1+α {|z ?y| ≥s} 0 t gβ (s, z, y ) aα (t ? s) + 1{|x?z |2 ≥t?s} 2 1/2 ds d + α s 0 |x ? z | t aα (t ? s) aα s + 1 1 ds 2 ≥t?s} 2 {| x ? z | d+α |z ? y |d+1+α {|z ?y| ≥s} 0 |x ? z | =: I1 + I2 + I3 + I4 . gβ1 (t ? s, x, z ) + We will treat I1 , · · · , I4 separately. Firstly, it follows from Lemma 3.1 that there are two constants c2 = c2 (d, β1 /β2 ) and c3 = c3 (β2 ? β1 ), β1 /2) such that I1 ≤ c2 (N c3 (t, x, z ) + N c3 (t, z, y ))gβ1 (t, x, y ). Next, we estimate I2 ,
t/2

I2

=
0 2d+α M α

gβ1 (t ? s, x, z ) t?d/2
0 t/2∧|z ?y |2

aα s 1 ds + 2 |z ? y |d+1+α {|z ?y| ≥s}

t

· · · ds
t/2

s ds |z ? y |d+1+α
t t/2

+t?d/2 1{|z ?y|2 ≥t/2}
2

s?d/2 exp ?

β1 |x ? z |2 s

t1?α/2 ds |z ? y |

t?d/2

1 t2 ∧ |z ? y |d?3 |z ? y |d+1+α
t/2 0

ds β1 |x ? z |2 s ds

+t?d/2+1?α/2
T 1?α/2

s?(d+1)/2 exp ?

t?d/2

1 t(1+α)/2 ∧ |z ? y |d?1 |z ? y |d+α

+ t?d/2 N β1 (t, x, z ) (3.3)

=

t?d/2 N β1 (t, x, z ) + H (1+α)/2 (t, z, y ) .

On the other hand, if |x ? z | ≥ |z ? y |, then, 2|x ? z | ≥ |x ? z | + |z ? y | ≥ |x ? y |, and so
c4 (d,α,β1 ) t∧|z ?y |2 0 c5 (d,α)

I2

(t ? s)α/2 aα s ds |x ? z |d+α |z ? y |d+1+α
2

2



t∧|z ?y | aα t s tα/2?1 ds d + α |x ? y | |z ? y |d+1+α 0 2 4 aα t α/2?1 t ∧ |z ? y | t |x ? y |d+α |z ? y |d+1+α α a t H 1+α/2 (t, z, y ). |x ? y |d+α

(3.4)

If |x ? z | < |z ? y |, then, 2|z ? y | ≥ |x ? y |, and so
2d+α

I2

aα |x ? y |d+α

t 0

√ gβ1 (t ? s, x, z ) sds 8

√ t √ s t?s aα t β1 |x ? z |2 ?(d+1)/2 = ( t ? s ) exp ? t t?s |x ? y |d+α 0 aα t ≤ N β1 (t, x, z ). |x ? y |d+α Combining (3.3), (3.4) and (3.5), we have
c6 (d,α,β1 ,M,T )

ds (3.5)

I2

t?d/2 ∧

aα t |x ? y |d+α

N β1 (t, x, z ) + H (1+α)/2 (t, z, y ) .

We treat I3 similar to I2 , and have
c7 (d,α,β2 ,M,T )

I3

t?d/2 ∧

aα t |x ? y |d+α

H 1+α/2 (t, x, z ) + N β2 (t, z, y ) .

I’s left to estimate I4 . If |x ? z |2 ≥ t ? s and |z ? y |2 ≥ s, then √ |x ? z | ∨ |z ? y ≥ t ? s ∨ s ≥ t/2. Combining that |x ? z | ∨ |z ? y | ≥ |x ? y |/2, we have 1 √ |x ? z | ∨ |z ? y | ≥ ( t ∨ |x ? y |). 2 And so, t?s s |x ? z |d+α |z ? y |d+1+α 1 1 (t ? s)s d + α d + α |z ? y | (|x ? z | ∨ |z ? y |) (|x ? z | ∧ |z ? y |) (t ? s)s 1 d + α (|x ? z | ∧ |z ? y |)d+1+α ( t ∨ |x ? y |) t (t ? s)s 1 1 t?d/2+1?α/2 ∧ + d + α d +1+ α t |x ? y | |x ? z | |z ? y |d+1+α √ t?d/2 ∧ t |x ? y |d+α t?s s + d +1+ α |x ? z | |z ? y |d+1+α

=
2d+α


(T ∨1)1?α/2

Thus,
t

I4

=
c8 (d,α,T )

a2α
0

1{|x?z |2 ≥t?s} 1{|z ?y|2 ≥s} t |x ? y |d+α
t

t?s s ds d + α |x ? z | |z ? y |d+1+α

a2α t?d/2 ∧ ×

1{|x?z |2 ≥t?s} 1{|z ?y|2 ≥s}
0

s t?s + d +1+ α |x ? z | |z ? y |d+1+α

ds

M α (M ∨1)α

t?d/2 ∧ ×

t aα t 1{|x?z |2 ≥t?s} 1{|z ?y|2 ≥s} |x ? y |d+α 0 t?s s + ds |x ? z |d+1+α |z ? y |d+1+α

(3.6)

Notice that
t 0

t?s 1 t2 ∧ |x ? z |4 1 ds = 2 2 |x ? z |d+1+α |x ? z |d+1+α {|x?z | ≥t?s} 9

≤ 2?1 t1?α/2

1 t1+α/2 ∧ |x ? z |d?1 |x ? z |d+1+α (3.7)

≤ 2?1 T 1?α/2 H 1+α/2 (t, x, z ). Similarly, s 1 ds ≤ 2?1 T 1?α/2 H 1+α/2 (t, z, y ). 2 d+1+α {|z ?y | ≥s} | z ? y | 0 Combining (3.6), (3.7) and (3.8), we have I4 ≤ c9 (d, α, M, T ) t?d/2 ∧ Hence, by (2.11) we have
C12 (d,α,β1 ,β2 ,M,T ) t

(3.8)

aα t |x ? y |d+α

H 1+α/2 (t, x, z ) + H 1+α/2 (t, z, y ) .

I

(H γ (t, x, z ) + H γ (t, z, y )) gβ1 (t, x, y ) + t?d/2 ∧

aα t |x ? y |d+α

,

where we have used the fact that for ?xed t, x, y , the function β → H β (t, x, y ) is decreasing. Thus, we have proved (3.1). Multiplying the both sides of (3.1) by |b(z )| and integrating it over Rd against dz , we have
t 0 C12 Rd 2 Rd a a qd,β (t ? s, x, z )|b(z )|qd +1,β2 (s, z, y )dzds 1

a qd,β (t, x, y )|b(z )| (H γ (t, x, z ) + H γ (t, z, y )) dz 1

γ a qd,β (t, x, y ) sup Hb (t, x) 1 x∈Rd

C11

√ a (t, x, y ). Mb ( t)qd,β 1

Let C13 = 2C12 C11 . Then, we ?nish the proof. Similar to the arguments following Lemma 13 in [3], we can inductively de?ne a sequence d d d of functions |p|a,b n : R × R × R → R. For t > 0 and x, y ∈ R , we de?ne
a |p|a,b 0 (t, x, y ) = p (t, x, y ), t 0 Rd

|p|a,b k (t, x, y ) =

a |p|a,b k?1 (t ? s, x, z )|b(z )||?z p (s, z, y )|dzds, for k ≥ 1.

For every M > 0 and T > 0, we can verify by induction that √ k a |p|a,b a ∈ (0, M ], (t, x, y ) ∈ (0, T ] × Rd × Rd . (3.9) k (t, x, y ) ≤ C7 (C10 Mb ( t)) qd,C5 /2 (t, x, y ), Indeed, (3.9) holds for k = 0. Assume (3.9) holds for k . Then by assumption and (3.2),
t √ k a a |p|a,b qd,C (t ? s, x, z )|b(z )|qd +1,3C5 /4 (s, z, y )dzds k+1 (t, x, y ) ≤C7 (C10 C13 Mb ( t)) C10 5 /2 0 Rd √ a √ k ≤C7 (C10 C13 Mb ( t)) C10 C13 Mb ( t)qd,C5 /2 (t, x, y ) √ a ≤C7 (C10 C13 Mb ( t))k+1 qd,C (t, x, y ). 5 /2 d d Now, we can de?ne functions pa,b k : R+ × R × R → R by a pa,b 0 (t, x, y ) = p (t, x, y ),

pa,b k (t, x, y ) =

t 0 Rd

a pa,b k?1 (t ? s, x, z )b(z ) · ?z p (s, z, y )dzds,

k ≥ 1.

(3.10)

And by (3.9), for every k ≥ 1, t ∈ (0, T ], √ k a a,b |pa,b ( t, x, y ) | ≤ | p | ( t, x, y ) ≤ C ( C C M ( t)) qd,C5 /2 (t, x, y ) < ∞. 7 10 13 b k k 10 (3.11)

Lemma 3.3. Suppose M > 0. For all a ∈ (0, M ] and every k ≥ 0, pa,b k (t, x, y ) is jointly d d continuous on (0, ∞) × R × R . Proof. We will use induction to prove this lemma. Obviously, pa,b 0 (t, x, y ) is jointly continuous a,b d d on (0, ∞) × R × R . We assume pk (t, x, y ) is jointly continuous. Note that by (2.1) with θ = 1,
a (t, x, y ) ≤ t?d/2 ∧ qd,C 5 /2

c1 t aα t ?d/2 + t ∧ , |x ? y |d+2 |x ? y |d+α

t > 0, x, y ∈ Rd ,

(3.12)

for some positive constant c1 depending only on d. Suppose T > 0 and ε < 1/(2T ). For t ∈ [T ?1 , T ] and s ∈ [ε, t ? ε], we have by (3.11) and (3.12), there is a constant c2 = c2 (d, α, M, T, b) such that √ k a ?d/2 |pa,b ≤ 2c2 ε?d/2 , k (t ? s, x, z )| ≤ C7 (C10 C13 Mb ( t)) qd,C5 /2 (t ? s, x, z ) ≤ 2c2 (t ? s) (3.13)
a ?(d+1)/2 |?z pa (s, z, y )| ≤ C10 qd ≤ 2C10 ε?(d+1)/2 . +1,3C5 /4 (s, z, y ) ≤ 2C10 s

(3.14)

and for any R > 1, |pa,b k (t ? s, x, z )| ≤ 2c2 Then,
t?ε ε |x?z |≥R a |pa,b k (t ? s, x, z )||b(z )||?z p (s, z, y )|dzds

1 , |x ? z |d+α

if |x ? z | ≥ R.

≤4c2 C10 ε?(d+1)/2 T
|x?z |≥R

|b(z )| dz, |x ? z |d+α

which goes to zero as R → ∞. Besides, for any m > 0, by (3.13-3.14), (2.7) and dominated convergence theorem, we have
t?ε

x→
ε |x?z |<R

a pa,b k (t ? s, x, z )b(z )?z p (s, z, y )dzds

is continuous on B (0, m). Thus, we can conclude that
t?ε

(t, x, y ) →
ε Rd

a |pa,b k (t ? s, x, z )||b(z )||?z p (s, z, y )|dzds

(3.15)

is jointly continuous on [T ?1 , T ] × B (0, m) × B (0, m). Since m is arbitrary, (3.15) is jointly continuous on [T ?1 , T ] × Rd × Rd . On the other hand, by (3.14) and (3.11),
t

sup
t∈[1/T,T ]

sup
x,y ∈Rd t?ε Rd

a |pa,b k (t ? s, x, z )||b(z )||?z p (s, z, y )|dzds t

≤ ≤

sup

sup sup

sup

|?z pa (s, z, y )|
t?ε ε Rd

t∈[1/T,T ] x∈Rd s∈[t?ε,t],z,y ∈Rd

|pa,b k (t ? s, x, z )||b(z )|dzds

|?z pa (s, z, y )| sup
x∈Rd ε 0 Rd

s∈[1/(2T ),T ],z,y ∈Rd

|pa,b k (s, x, z )||b(z )|dzds
ε

≤2C10 (2T )(d+1)/2 sup
x∈Rd 0 Rd



ε|b(z )|

gC5 /2 (s, x, z ) s1/2

dzds +
0 |x?z |≥s

|b(z )| aα
Rd

aα s dzds |x ? z |d+α

≤2C10 (2T )(d+1)/2



ε sup
x∈Rd Rd

|b(z )|N C5 /2 (ε, x, z )dz + sup
x∈Rd

ε 2 ∧ | x ? z |4 dz |x ? z |d+α

11

√ √ √ ≤2C10 (2T )(d+1)/2 CN C11 εMb ( ε) + aα ε(3?α)/2 C11 Mb ( ε) , which goes to zero as ε → 0. Similarly, by (3.13),
ε

sup

sup
0 Rd

√ √ ≤2C2 (2T )d/2 CN C11 Mb ( ε) + aα ε1?α/2 C11 Mb ( ε) , which goes to zero as ε → 0. Therefore, pa,b k+1 (t, x, y ) =
t 0 Rd a pa,b k (t ? s, x, z )b(z )?z p (s, z, y )dzds

t∈[1/T,T ] x,y ∈Rd

a |pa,b k (t ? s, x, z )||b(z )||?z p (s, z, y )|dzds

is jointly continuous on [T ?1 , T ] × Rd × Rd . Since T > 0 is arbitrary, we ?nish the proof. Lemma 3.4. Suppose M > 0, there are two positive constants t? (d, α, M, b) > 0 depending on b only via the rate at which Mb (r) goes to zero and C14 = C14 (d, α, M ) such that for all t ∈ (0, t? ] and x, y ∈ Rd ,
∞ ∞

pa,b k (t, x, y )
k=0


k=0

a |pa,b k (t, x, y )| ≤ C14 qd,C5 /2 (t, x, y ).

(3.16)

And for all |x ? y |2 < t with t ∈ (0, t? ]
∞ ?1 ?d/2 . pa,b k (t, x, y ) ≥ C14 t k=0

(3.17)

Proof. By (3.11) with T = 1, there is a constant 0 < t? < 1 such that for all t ∈ (0, t? ] √ 1 C8 e?C6 C10 C13 Mb ( t) ≤ ∧ , 2 4 and so,


|pa,b k (t, x, y )|
k=1

√ C10 C13 Mb ( t) a √ q (t, x, y ) ≤C7 1 ? C10 C13 Mb ( t) d,C5 /2 √ a ≤2C7 C10 C13 Mb ( t)qd,C (t, x, y ), x, y ∈ Rd . 5 /2

(3.18)

Thus, by Lemma 2.1 with T = 1 and (3.18), we have for all (t, x, y ) ∈ (0, t? ] × Rd × Rd ,
∞ ∞

pa,b k (t, x, y )
k=0


k=0

a |pa,b k (t, x, y )| ≤ 2C7 qd,C5 /2 (t, x, y ),

which gives (3.16). On the other hand, if |x ? y |2 < t ≤ t? , then
a pa (t, x, y ) ≥ C8 e?C6 t?d/2 , and qd,C (t, x, y ) ≤ 2t?d/2 . 5 /2

Thus, we have
∞ ∞

pa,b k (t, x, y )
k=0

≥p (t, x, y ) ?
k=1

a

?C6 ?d/2 t ? |pa,b k (t, x, y )| ≥ C8 e

C8 e?C6 ?d/2 t 2

=

C8

e?C6 2

t?d/2 ,

(t, x, y ) ∈ (0, t? ] × Rd × Rd |x ? y |2 ≤ t.

12

In the remainder of this paper, we ?x t? . By Lemma 3.4, the series absolutely converges on (0, t? ] × Rd × Rd . For all a ∈ (0, M ], de?ne


a,b ∞ k=0 pk (t, x, y )

p

a,b

(t, x, y ) =
k=0

pa,b k (t, x, y ),

0 < t ≤ t? and x, y ∈ Rd .

(3.19)

Lemma 3.5. Suppose M > 0. For all a ∈ (0, M ], pa,b (t, x, y ) is jointly continuous on (0, t? ] × Rd × Rd . Proof. For any 0 < t1 < t? , we have sup
[t1 ,t? ]×Rd ×Rd a qd,C (t, x, y ) ≤ 2t1 5 /2 ?d/2

< ∞.

a,b ∞ By Lemma 3.4 and inequality (3.11), the series k=0 pk (t, x, y ) converges uniformly on d d [t1 , t? ] × R × R . Since t1 is arbitrary, the result follows from Lemma 3.3.

The following lemma is the semigroup property of pa,b (t, x, y ) at small time. We can prove it directly from the de?nition of pa,b (t, x, y ) (3.19). Theorem 3.6. Suppose M > 0. For all a ∈ (0, M ] and every 0 < s, t ≤ t? with s + t ≤ t? and x, y ∈ Rd , we have pa,b (t + s, x, y ) =
Rd

pa,b (t, x, z )pa,b (s, z, y )dz.

(3.20)

Proof. Note that for s, t > 0 with s + t ≤ t? ,
∞ ∞

pa,b (t, x, z )pa,b (s, z, y ) =
m=0 ∞

pa,b m (t, x, z )
k=0 k

pa,b k (t, z, y )

=
k=0 m=0

a,b pa,b m (t, x, z )pk?m (t, z, y ).

So, it su?ces to prove that for any k ≥ 0,
k

pa,b k (t

+ s, x, y ) =
d m=0 R

a,b pa,b m (t, x, z )pk?m (t, z, y )dz.

(3.21)

a We will prove it inductively. When k = 0, (3.21) is true since pa,b 0 (t, x, y ) = p (t, x, y ). Now, suppose (3.21) holds when k = l and we have

pa,b l+1 (t + s, x, y ) = =

t+s 0 s 0 s Rd Rd

a,b pa,b l (t + s ? τ, x, w )b(w )?w p0 (τ, w, y )dwdτ t+s

a,b pa,b l (t + s ? τ, x, w )b(w )?w p0 (τ, w, y )dwdτ + l

· · · dwdτ
s Rd

=
0 m=0 t+s s l Rd Rd Rd

a,b a,b pa,b m (t, x, z )pl?m (s ? τ, z, w )dzb(w )?w p0 (τ, w, y )dwdτ a,b pa,b 0 (τ ? s, w, z )p0 (s, z, y )dzdwdτ

+ =
m=0 Rd

pa,b l (t + s ? τ, x, w )b(w )?w
s

Rd

pa,b m (t, x, z )
0 t

Rd

a,b pa,b l?m (s ? τ, z, w )b(w )?w p0 (τ, w, y )dwdτ dz

+
Rd 0 Rd

a,b a,b pa,b l (t ? τ, x, w )b(w )?w p0 (τ, w, z )dwdτ p0 (s, z, y )dz

13

l

=
m=0 l+1 Rd

a,b pa,b m (t, x, z )pl+1?m (s, z, y )dz +

Rd

a,b pa,b l+1 (t, x, z )p0 (s, z, y )dz

=
d m=0 R

km (t, x, z )kl+1?m (s, z, y )dz,

where in the last third equality, we can change the order of the integral by the fact that for all (t, x, y ) ∈ (0, t? ] × Rd × Rd and any m, l ∈ Z+ , by (3.9) and lemma 3.2,
s 0 Rd Rd a,b a |pa,b m (t, x, w )||pl (s ? τ, w, z )||b(z )||?z p (τ, z, y )dwdzdτ s 0 Rd a |pa,b l (s ? τ, w, z )||b(z )||?z p (τ, z, y )dzdτ dw

=
Rd

|pa,b m (t, x, w )|


Rd

a,b |pa,b m (t, x, w )||p |l+1 (s, w, y )dw < ∞.

And we can change the order of ?w and the integral by the fact that Lemma 2.1 and dominated convergence theorem implies that for any 0 < s < τ < ∞ and z, y ∈ Rd , ?z pa (τ, z, y ) =
Rd

?z pa (τ ? s, z, w)pa (s, w, y )dw.

In view of Theorem (3.6), the de?nition of pa,b (t, x, y ) can be uniquely extended to t > 0 by using Chapman-Kolmogorov equation (3.20) as follows. Suppose pa,b (t, x, y ) has been well de?ned on (0, 2k t? ] × Rd × Rd for k > 0. For t ∈ (2k t? , 2(k+1) t? ], we de?ne pa,b (t, x, y ) =
Rd

pa,b (t/2, x, z )pa,b (t/2, z, y )dz,

x, y ∈ Rd .

(3.22)

Theorem 3.7. Suppose M > 0. For every a ∈ (0, M ], pa,b (t, x, y ) is continuous on (0, ∞) × Rd × Rd and Rd pa,b (t, x, y )dy = 1 for every t > 0 and x ∈ Rd . Proof. The continuity of pa,b (t, x, y ) for all t > 0 follows from Lemma 3.5, (3.22) and dominated convergence theorem. By (2.6) we easily get Rd ?x pa (t, x, y )dy = 0 for all t > 0 and x ∈ Rd . Hence, for every k ≥ 1, by Lemma 3.2, (3.10-3.11) and Fubini’s theorem we have for each k ≥ 1 pa,b k (t, x, y )dy = =
Rd 0 t Rd Rd t 0 a pa,b k?1 (t ? s, x, z )b(z ) · ?z p (s, z, y )dsdzdy

Rd

pa,b k?1 (t ? s, x, z )b(z ) ·

?z pa (s, z, y )dydsdz = 0.
Rd

By (3.11) and dominated convergence theorem, we have for all t ∈ (0, t? ] and x ∈ Rd ,


pa,b (t, x, y )dy =
Rd k=0 Rd

pa,b k (t, x, y )dy =

pa (t, x, y )dy = 1,
Rd

which extends to all t > 0 by (3.22) and Fubini’s theorem. For bounded measurable function f on Rd , t > 0 and x ∈ Rd , we de?ne the operator Pta,b Pta,b f (x) = pa,b (t, x, y )f (y )dy.
Rd

14

Lemma 3.8. Suppose M > 0. For all a ∈ (0, M ] and s, t > 0, we have
a,b a,b Ps Pt = Pta,b +s .

(3.23)

and pa,b (t + s, x, y ) =
Rd

pa,b (t, x, z )pa,b (s, z, y )dz,

x, y ∈ Rd .

(3.24)

Proof. Repeat the proof of [3, Lemma 23] and we get (3.23). By (3.23), we have for all s, t > 0 and x ∈ Rd , pa (t + s, x, ·) =
Rd

pa (t, x, z )pa (s, z, ·)dz

a.s.,

which follows (3.24), since pa,b (t, x, ·) is continuous on Rd by Theorem 3.7. The following theorem shows that the generator of {Pta,b , t ≥ 0} is La,b in the weak sense. The proof is almost the same to part of the proof of [3, Theorem 1]. Here we give the details of the proof for completeness. For any compact set K ? Rd and r > 0, let K r = {y ∈ Rd : ?x ∈ K such that |x ? y | < r} be the 1-neighborhood of K .
∞ (Rd ), g ∈ C (Rd ), Theorem 3.9. Suppose M > 0. For every a ∈ (0, M ] and for all f ∈ Cc ∞

t→0 Rd

lim

Pta,b f (x) ? f (x) g (x)dx = t

La,b f (x)g (x)dx.
Rd

(3.25)

Proof. Note that for all t ∈ (0, t? ], Pta,b f (x) ? f (x) 1 g (x)dx = t t + pa,b 0 (t, x, y )f (y )dy ? f (x) g (x)dx
∞ Rd Rd

Rd

Rd

Rd

1 t

pa,b 1 (t, x, y ) +
k=2

pa,b k (t, x, y ) f (y )g (x)dydx.

a Since pa,b 0 (t, x, y ) = p (t, x, y ) we have

t→0

lim

1 t

Rd

Rd

pa,b 0 (t, x, y )f (y )dy ? f (x) g (x)dx =

? + aα ?α/2 f (x)g (x)dx.
Rd Rd

For t ∈ (0, t? ], let I (t) = t?1 Rd Rd pa,b 1 (t, x, y )f (y )g (x)dx. We will show I (t) → ?f (x)dx as t → 0. By (3.10), Fibini’s theorem and integration by parts we have
t

b(x) ·

I (t) =
Rd Rd Rd 0

1 a p (t ? s, x, z )pa (s, z, y )dsb(z ) · ?y f (y )g (x)dzdydx. t

Let K be the support of ?f . Then, the function (x, y ) → ?f (y )g (x) is uniformly continuous ∞ and g ∈ C (Rd ). Denote M = sup and bounded, since f ∈ Cc ∞ 0 x,y ∈Rd |?f (y )g (x)|. Fix an arbitrary small ε > 0, and there is δ > 0 such that for every z ∈ Rd and (x, y ) ∈ B (z, δ )×B (z, δ ) we have |?f (y )g (x) ? ?f (z )g (z )| < ε. Recall K 1 is the 1-neighborhood of K . We have |I (t) ?
Rd t

b(z ) · ?f (z )g (z )dz | 1 a p (t ? s, x, z )pa (s, z, y )ds|b(z )||?f (y )g (x) ? ?f (z )g (z )|dxdydz t
t t t


Rd Rd Rd 0

=
(K 1 )c K Rd 0

+
K1 (B (z,δ )×B (z,δ ))c 0

+
K1 B (z,δ ) B (z,δ ) 0

· · · dxdydz

15

= : J1 + J2 + J3 . We will estimate J1 , J2 and J3 separately. Note that if x ∈ K and z ∈ (K 1 )c , then |x ? z | ≥ 1 and so, by Theorem 2.1, for t > 0, x, y ∈ Rd and 0 < s < t,
a (t ? s, x, z ) ≤ C7 c1 pa (t ? s, x, z ) ≤ C7 qd,C 5

t?s . |x ? z |d+α

where c1 is a positive constant depending only on d, α, M . Thus,
t

J1 ≤ 2M0
(K 1 )c K 0 Rd t K 0

pa (s, z, y )dy

1 a p (t ? s, x, z )|b(z )|dsdxdz t

≤ 2M0 C7 c1
(K 1 )c

1 t?s |b(z )|dsdxdz t |x ? z |d+α |b(z )| (3?α)/2 (1, x) → 0, dz = tM0 C7 c1 |K | sup Hb |x ? z |d+α x∈K2

≤ tM0 C7 c1 |K | sup
x∈Rd |x?z |≥1

as t goes to zero. Similarly, if (x, y ) ∈ (B (z, δ ) × B (z, δ ))c , then |x ? z | ≥ δ or |y ? z | ≥ δ . By (2.7) we have J2 ≤ 2tM0 C7 c1
K1 |x?z |≥δ

|b(z )|

1 dxdz ≤ 2tM0 C7 c1 δ ?d?2 |x ? z |d+α

|b(z )|dz → 0,
K1

as t → 0.
t

J3 ≤ ε
Rd Rd Rd 0

1 a p (t ? s, x, z )pa (s, z, y )ds|b(z )|dxdydz ≤ ε t

|b(z )|dz.
K1

Since ε is arbitrary, we have limt→0 I (t) = Rd b(z ) · ?f (z )g (z )dz . By (3.10), Theorem 3.4 and dominated convergence theorem, we have


pa,b k (t, x, y )
k=2

t

∞ a pa,b k (t ? s, x, z ) b(z ) · ?z p (s, z, y )dsdz k=1

=
Rd 0

Similarly to the estimates of I (t), by Fubini’s theorem, integration by parts and (3.18), we have for all t ∈ (0, t? ] 1 t = ≤ 1 t
∞ Rd Rd

pa,b k (t, x, y ) f (y )g (x)dydx
k=2 t ∞ a pa,b k (t ? s, x, z ) p (s, z, y )b(z ) · ?f (y )g (x)dsdzdydx k=1

Rd

Rd

Rd

t 2C7 C10 C13 Mb ( t) a qd,C (t ? s, x, z )pa (s, z, y )|b(z )||?f (y )g (x)|dsdzdydx 5 t d d d R R R 0 √ 2C6 d/2 ?1 ≤2C7 C10 C13 Mb ( t) C8 C5 t 1 2 C6 × pa ( (t ? s), x, z )pa (s, z, y )|b(z )||?f (y )g (x)|dsdzdydx, t Rd Rd Rd 0 C5



0

which goes to zero as t → 0. We ?nish the proof.

4

Positivity of pa,b (t, x, y )

Theorem 4.1. Suppose M > 0. There are constants C15 = C15 (d, α, M ), C16 = C16 (d, α, M, b) such that for all a ∈ (0, M ], pa,b (t, x, y ) ≤ C15 eC16 t pa ( 2C6 t, x, y ), C5 16 t > 0 and x, y ∈ Rd .

and consequently, for any T > 0, there is constant C17 = C17 (d, α, M, T ) such that
a pa,b (t, x, y ) ≤ C17 eC16 t qd,C 2 /(2C ) (t, x, y ), 6
5

t ∈ (0, T ] and x, y ∈ Rd .

a Proof. Note that by the expression of qd,C (t, x, y ) and the lower bound of pa (t, x, y ) in 5 /2 Theorem 2.1 with T = 1,

a (t, x, y ) ≤ qd,C 5 /2

2C6 C5

d/2 a ( qd,C 6

2C6 t, x, y ) ≤ C5

2C6 C5

d/2 ?1 a C8 p (

2 C6 t, x, y ). C5

(4.1)

Recall that t? is the constant in Lemma 3.4. If t < t? , by (3.16) and Lemma (2.1),
a (t, x, y ) ≤ c1 pa ( pa,b (t, x, y ) ≤ C14 qd,C 5 /2 C14 (2C6 )d/2 C8 C5
d/2

2C6 t, x, y ), C5

where c1 =

depending only on d, α, M . It is left to consider the case t > t? . Let

k = [t/t? ] + 1, then t/k ∈ (0, t? ). Combining (4.1), (3.24) and (3.16), we have pa,b (t, x, y ) ≤
Rd(k?1) a = ck 1p ( a ck 1p (

2 C6 t 2C6 t , x, x1 ) · · · pa ( , xk?1 , y )dx1 · · · dxk?1 C5 k C5 k

2 C6 t, x, y ) C5 t t? a 2C6 p ( t, x, y ), ≤ c1 c1 C5 which gives the ?rst conclusion with C15 = c1 and C16 = t1 ln c1 . Furthermore, by the upper ? a bound of p (t, x, y ) in Theorem 2.1, for t ∈ (0, T ] and x, y ∈ Rd pa ( 2C6 2C6 2 C6 C7 a a t, x, y ) ≤ C7 qd,C ( t, x, y ) ≤ q 2 (t, x, y ). 5 /2 C5 C5 C5 d,C5 /(2C6 )

Combining the last two displays, we ?nish the proof by setting C17 = c1 ((2C6 C7 /C5 ) ∨ 1). Lemma 4.2. Suppose M > 0. For all a ∈ (0, M ] and every t > 0, Pta,b maps bounded functions to continuous functions. Furthermore, {Pta,b , t ≥ 0} is a strongly continuous semigroup in C∞ ( R d ) . Proof. By Lemma 4.1 and Lemma 3.7, one can easily verify that Pta,b maps bounded functions to continuous function for every t > 0. For every f ∈ C∞ (Rd ) and t > 0, by Lemma 4.1, lim Pta,b f (x) ≤ lim
a C15 eC16 t qd,C 2 /(2C ) (t, x, y )f (y )dy 6
5

|x|→∞

|x|→∞ Rd |x|→∞ Rd

≤ lim

a C15 eC16 t qd,C 2 /(2C ) (t, 0, y )f (x + y )dy = 0, 6
5

which shows Pta,b f ∈ C∞ (Rd ). Moreover, since f is uniformly continuous on Rd , for every ε > 0, there is a constant δ > 0 such that |f (x) ? f (y )| ≤ ε for all x, y ∈ Rd with |x ? y | ≤ δ . And so, by (3.12), lim sup sup |pa,b (s, x, y )|dy
|x?y |≥δ a C15 eC16 s qd,C 2 /(2C ) (s, x, y )dy 6
5

t→0 s≤t x∈Rd

≤ lim sup sup ≤ lim sup sup

t→0 s≤t x∈Rd t→0 s≤t x∈Rd

|x?y |≥δ

C15 eC16 t c1
|x?y |≥δ

t t + d +2 |x ? y | |x ? y |d+α 17

dy = 0,

where c1 is some positive constant depending only on d, α, M . Thus, we have
t→0 s≤t x∈Rd a,b lim sup sup |Ps f (x) ? f (x)| = lim sup sup t→0 s≤t x∈Rd

pa,b (s, x, y )f (y )dy ? f (x)
Rd

≤ lim sup sup ≤ lim sup sup ≤εC15 ,

t→0 s≤t x∈Rd

|pa,b (t, x, y )||f (x) ? f (y )|dy
|x?y |<δ

t→0 s≤t x∈Rd

C15 eC16 t pa (
|x?y |<δ

2 2 C6 2 s, x, y )|f (x) ? f (y )|dy C5

which shows limt→0 Pta,b f ? f



= 0.

Lemma 4.3. Suppose M > 0 and b ∈ Kd,1 is continuous and bounded on Rd . Then, for all a ∈ (0, M ], pa,b (t, x, y ) ≥ 0, t > 0 and x, y ∈ Rd . Proof. We denote the generator of {Pta,b , t ≥ 0} in C∞ (Rd ) by La,b , which is closed. For every 2 (Rd ), since b is continuous, one can easily shows that La,b f ∈ C (Rd ). Similar to f ∈ Cc ∞ Theorem 3.9, we claim that (Pta,b f ? f )/t uniformly converges to La,b f as t → 0. Indeed, for any t ∈ (0, t? ], (Pta,b f ? f )/t ? La,b f = sup
x∈Rd ∞

1 t 1 t 1 t 1 t

∞ Rd k=0

pa,b k (t, x, y )f (y )dy ? f (x)

? ? + aα ?α/2 + b · ? f (x)

≤ sup
x∈Rd

Rd

α α/2 f (x) pa,b 0 (t, x, y )f (y )dy ? f (x) ? ? + a ? ∞ Rd

+ sup
x∈Rd

pa,b k (t, x, y )f (y )dy ? f (x)
k=1 ∞

? b (x)?f (x)

+ sup
x∈Rd

Rd

pa,b k (t, x, y )f (y )dy ? f (x)
k=2

=I1 + I2 + I3 . If follows that I1 goes to zero as t → 0 since ? + aα ?α/2 is the generator of X a . We next treat I2 as we did with I in the proof of Theorem 3.9. Let M0 = supx∈Rd |b (x)?f (x)|. Since 2 (Rd ), for any ε > 0, there is constant δ > 0 such that |b (z )?f (y ) ? b (x)?f (x)| < ε for f ∈ Cc all x ∈ Rd and (z, y ) ∈ B (x, δ ) × B (x, δ ). while, if (z, y ) ∈ (B (x, δ ) × B (x, δ ))c then |z ? x| ≥ δ or |z ? y | ≥ δ . Then, by (3.12), we have
t

I2 ≤ sup
x∈Rd 0 Rd Rd

1 a p (t ? s, x, z )pa (s, z, y ) b (z )?y f (y ) ? b (x)?x f (x) dzdsdy t
t

≤ sup
x∈Rd B (x,δ )×B (x,δ ) 0

1 a p (t ? s, x, z )pa (s, z, y )|b (z )?y f (y ) ? b (x)?x f (x)|dsdzdy t
t

+ sup
x∈Rd (B (x,δ )×B (x,δ ))c 0 t

· · · dsdzdy c1
|x?z |≥δ 0 ?2 α

≤ε + 2M0 sup
x∈Rd

t?s t?s + |x ? z |d+2 |x ? z |d+α

dsdz

≤ε + M0 c1 ωd (δ

/2 + δ /α)t.

where c1 is some positive constant depending only on d, α, M . Since ε is arbitrary, I2 goes to zero as t → 0. Similar to I2 , we can prove that I3 goes to zero as t → 0. Now, we have proved 18

our claim. Therefore,
2 2 Cc (Rd ) ? D(La,b ) and La,b f = La,b f for all f ∈ Cc (Rd ).

(4.2)

On the other hand, by Theorem 4.1,


sup
x∈Rd 0

e?λt |Pta,b f (x)|dt ≤ sup ≤ f

∞ 0

e?λt
Rd

|pa,b (t, x, y )||f (y )|dydt (4.3)
∞,

x∈Rd ∞

∞ 0

C15 e?(λ?C16 )t dt = cλ f

where cλ = C15 /(λ ? C16 ) for all λ > C16 . Consider the strongly continuous semigroup {e?C16 t Pta,b , t ≥ 0} with its generator La,b ? C16 . By (4.3), the residual set ρ(La,b ? C16 ) of La,b ? C16 contains (0, ∞). Besides, La,b ? C16 satis?es the positive maximum principle by (4.2) and [1, Theorem 3.5.3], since b is continuous. Therefore, {e?C16 t Pta,b , t ≥ 0} is positive preserving semigroup on C∞ (Rd ) by Hille-Yosida-Ray([1, Theorem 3.5.1]). Since {e?C16 t Pta,b , t ≥ 0} has a continuous kernel e?C16 t pa,b (t, x, y ), we have pa,b (t, x, y ) ≥ 0 for all (t, x, y ) ∈ (0, ∞) × Rd × Rd . In the rest of this section, we will show that Lemma 4.3 is still true without the condition ∞ (Rd ) with that b is continuous and bounded on Rd . Let ? be a non-negative function in Cc d supp(?) ? B (0, 1) and Rd ?(x)dx = 1. For n ≥ 1, de?ne ?n (x) := n ?(nx) and bn (x) =
Rd

?n (x ? y )b(y )dy,

x ∈ Rd .

For any compact set K ? Rd and r > 0, recall that K r is the r-neighborhood of K . For any 0 ≤ r1 ≤ r2 ≤ +∞ and β ≥ 0, we have sup
x∈ K |x?y |∈[r1 ,r2 )

|bn (y )| dy ≤ sup |x ? y |d?1+2β x∈ K = sup
x∈ K

|x?y |∈[r1 ,r2 )

Rd

?n (y ? z )|b(z )| dzdy |x ? y |d?1+2β ?n (z )|b(y ? z )| dzdy |x ? y |d?1+2β |b(y )| dydz |x ? z ? y |d?1+2β

|x?y |∈[r1 ,r2 )

|z |<1/n

= sup
x∈ K |z |<1/n

?n (z )
|x?z ?y |∈[r1 ,r2 )


|z |<1/n

?n (z ) sup
x∈ K 1 |x?y |∈[r1 ,r2 )

|b(y )| dydz |x ? y |d?1+2β

= sup
x∈ K 1 |x?y |∈[r1 ,r2 )

|b(y )| dy. |x ? y |d?1+2β (4.4)

In particular, for every r > 0 and n ≥ 1, by setting r1 = 0, r2 = r and β = 0, we have Mbn (r) ≤ Mb (r). Recall that γ = (1 + α ∧ 1)/2.
γ d Lemma 4.4. Hb ?bn (t, x) converges uniformly in every compact subset of (0, +∞) × R to zero as n → ∞.

(4.5)

Proof. Let [t0 , T0 ] × K ? (0, +∞) × Rd be an arbitrary compact set. Then, we have sup
(t,x)∈[t0 ,T0 ]×K γ Hb ?bn (t, x) ≤ sup γ 1 T0 ∧ |x ? y |d?1 |x ? y |d?1+2γ

|b(y ) ? bn (y )|dy · · · dy

x∈ K

Rd

≤ sup
x∈ K |x?y |2 <r

+
r≤|x?y |2 <R

+
|x?y |2 ≥R

19

= : I1 + I2 + I3 , where 0 < r < R < ∞ are undetermined. By setting r1 = 0, r2 = have |b(y )| I1 ≤ 2 sup dy. d?1 x∈K 1 |x?y |2 <r |x ? y | √ r and β = 0 in (4.4), we

Since b ∈ Kd,1 , for any ε > 0, we can choose r small enough such that I1 ≤ 2 · By setting r1 = √ ε ε = . 4 2

R, r2 = ∞ and β = γ in (4.4), we have
γ I3 ≤ 2T0 sup

x∈K 1

|x?y |2 ≥R

|b(y )| dy. |x ? y |d?1+2γ √ R > diam(K 1 ), then

Fix a point x0 ∈ K 1 . Note that if x ∈ K 1 , |x ? y |2 ≥ R and assuming √ |x0 ? y | ≥ |x ? y | ? |x ? x0 | ≥ R ? diam(K 1 ),

|x0 ? y | |x0 ? x| + |x ? y | diamK 1 ≤ ≤ √ + 1 ≤ 2, |x ? y | |x ? y | R and so
γ sup I3 ≤2T0 γ ≤2d+2 T0 √ |x0 ?y |2 ≥( R?diam(K 1 ))2

x∈K 1

|x0 ? y |d?1+2γ |b(y )| dy |x0 ? y |d?1+2γ |x ? y |d?1+2γ

√ |x0 ?y |2 ≥( R?diam(K 1 ))2

|b(y )| dy |x0 ? y |d?1+2γ

By lemma 2.4 and dominated convergence theorem, we can choose R large enough such that ε I3 < . 2 Now, we ?x the above r, R. Note that bn → b, a.s. By dominated convergence theorem
n→∞

lim I2 ≤ lim r?(d?1+2γ )/2
n→∞ |x?y |2 <R

|b(y ) ? bn (y )|dy = 0.

Then, we have
n→∞ (t,x)∈[t ,T ]×K 0 0

lim

sup

γ Hb ?bn (t, x) ≤

ε ε + + 0. 2 2

We ?nish the proof since ε is arbitrary. Lemma 4.5. Suppose M > 0, T > 0. There are constants C18 = C18 (d, α, M, T ), C19 = C19 (d, α, M, T ) > 0 such that for every n ≥ 1, k ≥ 1 and all a ∈ (0, M ], (t, x, y ) ∈ (0, T ] × Rd × Rd , √ a,b n |pa,b k (t, x, y ) ? pk (t, x, y )| ≤C18 C19 Mb ( t)
k?1 γ γ a (Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

(4.6) Proof. We will verify (4.6) inductively. When k = 1, by (3.10) and (3.1),
a,b n |pa,b 1 (t, x, y ) ? p1 (t, x, y )| t


0 Rd

pa (t ? s, x, z )|b(z ) ? bn (z )||?z pa (s, z, y )|dzds

20

C7 C10 0 C12

t Rd a a (t ? s, x, z )|b(z ) ? bn (z )|qd qd,C +1,3C5 /4 (s, z, y )|dzds 5 /2

a (t, x, y ) qd,C 5 /2

(H γ (t, x, z ) + H γ (t, z, y )) |b(z ) ? bn (z )|dz
Rd



γ γ a Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

Let C18 = C7 C10 C12 and C19 = 2d+3 C11 C10 C12 . Now, we assume (4.6) is true for k . Then by (3.10),
a,b n |pa,b k+1 (t, x, y ) ? pk+1 (t, x, y )| t


0 Rd t 0

a,b a n |pa,b k (t ? s, x, z ) ? pk (t ? s, x, z )||bn (z )||?z p (s, z, y )|dzds a |pa,b k (t ? s, x, z )||b(z ) ? bn (z )||?z p (s, z, y )|dzds

+
Rd

= : I1 + I2 . By assumption,
C18 C10 t 0 Rd

I1

√ C19 Mb ( t ? s)

k ?1

γ γ Hb ?bn (t ? s, x) + Hb?bn (t ? s, z )



a a × qd,C (t ? s, x, z )|bn (z )|qd +1,3C5 /4 (s, z, y )dzds 5 /2 √ k?1 γ γ C19 Mb ( t) Hb ?bn (t, x) + Hb?bn (t, z ) |bn (z )| Rd t

×
0 C12

a a qd,C (t ? s, x, z )qd +1,3C5 /4 (s, z, y )ds dz 5 /2 k?1 Rd γ γ γ Hb ?bn (t, x) + Hb?bn (t, z ) |bn (z )|

√ C19 Mb ( t)



a × (H γ (t, x, z ) + H (t, z, y )) qd,C (t, x, y )dz 5 /2 √ k?1 γ γ γ C19 Mb ( t) Hb ?bn (t, x) Hbn (t, x) + Hbn (t, y )

+
Rd

γ γ γ a Hb ?bn (t, z )|bn (z )| (H (t, x, z ) + H (t, z, y )) dz qd,C5 /2 (t, x, y ) k?1



√ C19 Mb ( t)

√ γ 2C11 Mb ( t)Hb ?bn (t, x) +

|b(w) ? bn (w)||bn (z )|

· H γ (t, z, w) (H γ (t, x, z ) + Note that

Rd Rd a H γ (t, z, y )) dzdw] qd,C (t, x, y ) 5 /2

H γ (t, z, w) ∧ H γ (t, x, z ) 1 tγ = ∧ |w ? z |d?1 |w ? z |d?1+2γ
2d+1



1 tγ ∧ |z ? x|d?1 |z ? x|d?1+2γ

1 tγ ∧ |w ? x|d?1 |w ? x|d?1+2γ

= H γ (t, x, w).

Similarly, H (t, z, w) ∧ H (t, y, z ) Thus, |b(w) ? bn (w)||bn (z )|H γ (t, z, w) (H γ (t, x, z ) + H γ (t, z, y )) dzdw
Rd 2d+1 Rd Rd Rd γ γ 2d+1

H γ (t, y, w).

|b(w) ? bn (w)||bn (z )| (H γ (t, x, w) (H γ (t, z, w) + H γ (t, x, z )) 21

+ H γ (t, y, w) (H γ (t, z, w) + H γ (t, y, z ))) dzdw =
Rd γ γ |b(w) ? bn (w)| H γ (t, x, w) Hb (t, w) + Hb (t, x) n n

γ γ + H γ (t, y, w) Hb (t, w) + Hb (t, y ) dw n n 2C11 √ Mb ( t ) |b(w) ? bn (w)| (H γ (t, x, w) + H γ (t, y, w)) dw Rd √ γ γ = Mb ( t ) H b ?bn (t, x) + Hb?bn (t, y ) .

Then,
C18

I1

√ C19 Mb ( t)

k?1

√ 2C11 C10 C12 Mb ( t)
a (t, x, y ) qd,C 5 /2

γ γ γ d+1 · Hb Hb ?bn (t, x) + 2 ?bn (t, x) + Hb?bn (t, y )



√ 2C11 C10 C12 Mb ( t)

k

γ γ a 2(d+2)(k?1) (2d+1 + 1) Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

On the other hand, by (3.11),
C7 C10

I2
C12

√ 2C11 C10 C12 Mb ( t)
t 0 Rd

k 0

t Rd a a qd,C (t ? s, x, z )|b(z ) ? bn (z )|qd (s, z, y )dzds 5 /2 1 ,3C5 /4

a |b(z ) ? bn (z )| (H γ (t, x, z ) + H γ (t, x, z )) qd,C (t, x, y )dz 5 /2

= Therefore,

γ γ a Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

C18

a,b n |pa,b k+1 (t, x, y ) ? pk+1 (t, x, y )| √ k (d+2)(k?1) d+1 2C11 C10 C12 Mb ( t) 2 (2 + 2)

γ γ a Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y )



√ C19 Mb ( t)

k

γ γ a Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

Then, we ?nish the proof. Lemma 4.6. Suppose M > 0 and T > 0. For all a ∈ (0, M ]every 0 < T0 < T and compact set K ? Rd , pa,bn (t, x, y ) converges to pa,b (t, x, y ) uniformly in [T0 , T ] × K × K as n → ∞. Proof. By (3.19) and Lemma 4.5 with T = 1, for all t ∈ (0, t? ] and x, y ∈ K , |pa,bn (t, x, y ) ? pa,b (t, x, y )|



k=1

a,b n |pa,b k (t, x, y ) ? pk (t, x, y )| ∞

≤ C18
k=1

√ C19 Mb ( t)

k

γ γ a Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y ).

Since b belongs to Kd,1 , we can ?nd a constant 0 < T1 < t? such that for all t ≤ T1 , √ 1 C19 Mb ( t) ≤ . 2 Then, for all t ≤ T1 and all x, y ∈ Rd ,


|pa,bn (t, x, y ) ? pa,b (t, x, y )| ≤ C18
k=1

γ γ a 2?(k?1) Hb ?bn (t, x) + Hb?bn (t, y ) qd,C5 /2 (t, x, y )

(4.7)



γ 2C18 Hb ?bn (t, x)

+

γ Hb ?bn (t, y )

a qd,C (t, x, y ). 5 /2

22

a Without loss of generality, we may assume T0 < T1 /2. Then, we have qd,C (t, x, y ) ≤ 2T0 5 /2 for T0 ≤ t ≤ T1 and all x, y ∈ Rd . By (4.7) and Lemma 4.4,

?d/2

lim sup sup ≤ 4T0 = 0.
?d/2

sup |pa,bn (t, x, y ) ? pa,b (t, x, y )|
γ γ Hb ?bn (t, x) + Hb?bn (t, y )

n→∞ t∈[T0 ,T1 ] x,y ∈K

C18 lim sup sup

(4.8)

n→∞ x,y ∈K

For t ∈ [T1 , 3T1 /2], let t1 = T1 /2, then t ? t1 ∈ [T1 /2, T1 ]. By semigroup property (3.24), |pa,bn (t, x, y ) ? pa,b (t, x, y )| ≤
Rd

|pa,bn (t ? t1 , x, z ) ? pa,b (t ? t1 , x, z )||pa,bn (t1 , z, y )|dz +
Rd

|pa,b (t ? t1 , x, z )||pa,bn (t1 , z, y ) ? pa,b (t1 , z, y )|dz

= : I1 + I2 . Since by Theorem 4.1 and (4.5),
C16 T1 a (T1 /2)?d/2 , |pa,bn (t ? t1 , x, z )| ≤ C17 eC16 (t?t1 ) qd,C 2 /(2C ) (t ? t1 , x, z ) ≤ 2C17 e 6
5

|p

a,b

(t ? t1 , x, z )| ≤ 2C17 e

C16 T1

(T1 /2)

?d/2

,

and for any ε > 0, there is a constant R0 > 0 such that for all n ≥ 1 and y ∈ Rd , ε |pa,bn (t1 , z, y )|dz < . C T 16 1 4C17 e (T1 /2)?d/2 |z ?y |≥R0 On the other hand, we can choose n large enough such that sup
x,z ∈K R0

|pa,bn (t ? t1 , x, z ) ? pa,b (t ? t1 , x, z )| <

ε . 2C15 eC16 T

Besides, by Theorem 4.1, for all y ∈ Rd , |pa,bn (t1 , z, y )|dz ≤ C15 eC16 T pa (
Rd

2 C6 t1 , z, y )dz = C15 eC16 T C5

Then, we have I1 ≤
|z ?y |>K R0

|pa,bn (t ? t1 , x, z ) ? pa,b (t ? t1 , x, z )||pa,bn (t1 , z, y )|dz + ε 4C17 eC16 T1 (T1 /2)?d/2 + C15 eC16 T ε 2C15 eC16 T

· · · dz
|z ?y |<K R0

<2C17 eC16 T1 (T1 /2)?d/2 =ε.

Similarly, we can get I2 < ε for large enough n. Then, we have proved that pa,bn (t, x, y ) converges to pa,b (t, x, y ) uniformly in [T1 , 3T1 /2] × K × K as n → ∞. We can ?nish the proof by repeating the above arguments for [2T /T1 ? 2] times. Lemma 4.7. Let M > 0. For all a ∈ (0, M ], pa,b (t, x, y ) ≥ 0, t > 0 and x, y ∈ Rd . Proof. Suppose T > 0. By Lemma 4.3, for every n ≥ 1 and all (t, x, y ) ∈ (0, T ] × Rd × Rd , pa,bn (t, x, y ) ≥ 0, since bn is continuous and bounded. Let K ? Rd be an arbitrary compact set and T0 ∈ (0, T ). By Lemma 4.6, pa,b (t, x, y ) ≥ 0, for all (t, x, y ) ∈ [T0 , T ] × K × K . Since T0 , T and K are arbitrary, we ?nish the proof. Proof of Theorem 1.1. Theorem 1.1 follows from (3.19), (3.24), Lemma 3.7, Lemma 4.7 and Theorem 3.9. 23

5

Lower bound of pa,b (t, x, y )

In this section, we will show the sharp lower bound of the heat kernel pa,b (t, x, y ). Lemma 4.2 and Lemma 4.7 show that P a,b is a Feller semigroup in C∞ (Rd ). Then, there exists a consera,b a,b vative Feller process X a,b = {Xt , t ≥ 0, Px , x ∈ Rd } such that Pa,b t f (x) = Ex [f (Xt )], x ∈ Rd , f ∈ C∞ (Rd ). And, the process X a,b has strong Feller property and has pa,b (t, x, y ) as its transition density. Besides, by Lemma 4.1 and Lemma 4.7, pa,b (t, x, y ) has the following upper bound 2C6 pa,b (t, x, y ) ≤ C15 eC16 t pa ( t, x, y ), t > 0 and x, y ∈ Rd . (5.1) C5 and for every T > 0,
a pa,b (t, x, y ) ≤ C17 eC16 t qd,C 2 /(2C ) (t, x, y ), 6
5

t ∈ (0, T ] and x, y ∈ Rd .

(5.2)

where C15 , C16 and C17 are constants in Lemma 4.1. The following lemmas will be used to derive the L? evy system of X a,b . Lemma 5.1. For every f ∈ Kd,1 , we have
t t→0 x∈Rd

lim sup

a,b Ps |f |(x)ds = 0. 0

Proof. By (5.2) with T = 1 and (2.2), for 0 < t < 1 and x ∈ Rd ,
t a,b Ps |f |(x)ds ≤C17 eC16 t 0 0 t Rd a qd,C 2 /(2C ) (s, x, y )f (y )dyds 6
5

≤C9 C17 eC16 t ≤C9 C17 eC16 t Thus, by lemma (2.6)
t t→0 x∈Rd

√ √

t
Rd

|f (y )|N C5 /(2C6 ) (t, x, y )dy +
0
2

2

t |x?y |2 ≥s

aα s |f |(y )dyds |x ? y |d+α
(1+α)/2

t sup
x∈Rd Rd

|f (y )|N C5 /(2C6 ) (t, x, y )dy + t(3?α)/2 H|f |

(t, x) .

lim sup

a,b Ps |f |(x)ds ≤ lim C9 C17 0 t→0



tN|f 5 |

C 2 /(2C6 )

(t) + t(3?α)/2 sup H|f |
x∈Rd

(1+α)/2

(t, x)

= 0.

Similar to [5, Theorem 2.5], by Lemma 4.2, (2.7), Lemma 5.1 and Theorem 3.9, we have the following lemma. Here we omit the proof.
∞ (Rd ), Theorem 5.2. Suppose M > 0. For all a ∈ (0, M ], every x ∈ Rd and every f ∈ Cc a,b a,b Mtf := f (Xt ) ? f (X0 )? t a,b La,b f (Xs )ds 0

is a martingale under Px . Recall that the function J a (x, y ) is de?ned by (1.2). Lemma 5.3. Suppose A, B are disjoint compact sets in Rd and M > 0. Then, for all a ∈ (0, M ] and every x ∈ Rd , the process
a,b a,b 1A (Xs ? )1B (Xs ) ? s≤ t 0 t a,b 1A (Xs ) B a,b J a (Xs , y )dyds

is a martingale under Px . 24

∞ (Rd ) such that f = 0 on A and f = 1 on B . By theorem 5.2, for Proof. Let f ∈ Cc t a,b a,b a,b x ∈ Rd , Mtf := f (Xt ) ? f (X0 ) ? 0 La,b f (Xs )ds is a martingale under Px , then, Ntf := t a,b f a,b a,b,1 a,b,d , · · · , Xt ) 0 1A (Xs? )dMs is also a martingale. Besides, theorem 5.2 implies that Xt = (Xt is a semi-martingale. By Ito’s formula, we have d a,b f (Xt ) t 0

?

a,b f (X0 )

=
i=1

a,b a,b,i ?i f (Xs + ? )dXs s≤t

1 ηs (f ) + At (f ), 2

(5.3)

where ηs (f ) = and At (f ) =
i,j =1 0 a,b f (Xs )

d

?

a,b f (Xs ?)

?
i=1

a,b a,b,i a,b,i ?i f (Xs ? Xs ? )(Xs ? ),

(5.4)

d

t

a,b a,b,i c ?l ?j f (Xs ) , (X a,b,j )c >s . ? )d < (X

(5.5)

Since f = 0 on A, by (5.3-5.5) and (1.1), we have Ntf =
s≤ t a,b a,b 1A (Xs ? )f (Xs ) ? 0 a,b a,b 1A (Xs ? )f (Xs ) ? s≤ t 0 t a,b 1A (Xs ) Rd a,b f (y )J a (Xs , y )dyds t a,b a,b 1A (Xs ) aα ?α/2 f (Xs ) ds

=

∞ (Rd ) with f = 1 on A, f = 1 on B and f ↓ 1 , Taking a sequence of functions fn ∈ Cc n n n B d we have for x ∈ R and every n ≥ 1, Ntfn is a martingale under Px . By passing the limit, we complete the proof.

By lemma 5.2, we get ? Ex ?
s≤t

?
0

a,b a,b ? 1A (Xs ? )1B (Xs ) = Ex

t Rd a,b a,b a a,b 1A (Xs )1B (Xs )J (Xs , y )dyds .

Using this and a routine measure theoretic arguments, we get ? ? Ex ?
s≤ t a,b a,b ? f (s, Xs ? , Xs ) = Ex 0 t Rd a,b a,b f (s, Xs , y )J a (Xs , y )dyds ,

for any non-negative measurable function f on (0, ∞) × Rd × Rd vanishing on {(x, y ) ∈ Rd × Rd : x = y }. Furthermore, following the same arguments as in [9, Lemma 4.7] and [10, Appendix A], we have Theorem 5.4. For M > 0 and all a ∈ (0, M ], X a,b has the same L? evy system as X a , that d is for any x ∈ R and any non-negative measure function f on R+ × Rd × Rd vanishing on {(s, x, y ) ∈ R+ × Rd × Rd : x = y } and stopping time S ( with respect to the ?ltration of X a,b ), ? ? Ex ?
s≤S a,b a,b ? f (s, Xs ? , Xs ) = Ex 0 S Rd a,b a,b f (s, Xs , y )J a (Xs , y )dyds ,

Lemma 5.5. For each M > 0 and R0 > 0, there is a constant κ = κ(d, α, M, R0 , b) < 1 depending on b only via the rate at which Mb (r) goes to zero such that for all a ∈ (0, M ], r ∈ (0, R0 ] and all x ∈ Rd , 1 a,b 2 Px τB ≤ . (x,r) ≤ κr 2 25

Proof. By the strong Markov property of X a,b (See [2, Exercise (8.17), pp. 43-44]), for x ∈ Rd and t > 0, we have
a,b a,b a,b a,b a,b c Px τB (x,r) ≤ t ≤Px τB (x,r) ≤ t, Xt ∈ B (x, r/2) + Px τB (x,r) ≤ t, Xt ∈ B (x, r/2) ? ?

=Ex ?PX a,b + Px
s≤t x∈Rd

a,b τ B (x,r )

X a,b a,b

t?τB (x,r)

a,b a,b ? ? X0 ≥ r/2, τB (x,r) ≤ t

a,b a,b Xt ? X0 ≥ r/2 a,b a,b Xs ? X0 ≥ r/2 .

≤2 sup sup Px

2 , for t ∈ (0, R2 ], there are positive constants c , i = 1, 2, 3 depending only By (5.2) with T = R0 i 0 on d, α, M, R0 such that

sup sup Px
s≤t x∈Rd
2 c1 ec2 R0

a,b a,b Xs ? X0 ≥ r/2

sup sup
s≤t x∈Rd ωd ∞ |x?y |≥r/2

s?d/2 exp ? e?c3 ρ + 1 ∧
2 2

c3 |x ? y |2 s ρd?1 dρ

+ s?d/2 ∧

aα s |x ? y |d+α

dy

sup
s≤t ∞
r √ 2 s

M α s1?α/2 ρd+α


2 r √ t

e?c3 ρ + 1 ∧

2?α M α R0 ρd+α

ρd?1 dρ.

Setting t = κr2 in the last display, where κ ∈ (0, 1) is undetermined, we have
a,b 2 Px τB ≤2c1 ec2 R0 ωd (x,r) ≤ κr
2


1 √ 2 κ

e?c3 r + 1 ∧

2

2?α M α R0 rd+α

rd?1 dr,

which goes to 0 as κ → 0. Then, we can choose κ < 1 such that the conclusion holds.
a,b a,b For any open set U ? Rd , de?ne σU = inf {t ≥ 0 : Xt ∈ U }.

Lemma 5.6. For each M > 0 and R0 > 0, there is a constant c1 = c1 (d, α, M, R0 , b) depending on b only via the rate at which Mb (r) goes to zero, such that for all r ∈ (0, R0 ] and x, y ∈ Rd with |x ? y | ≥ 2r, aα a,b 2 d+2 Px σB < κr ≥ c r . 1 (y,r) |x ? y |d+α Proof. By Lemma 5.5, Ex By Lemma 5.4, we have
a,b 2 Px σB ≥Px X a,b ∈ B (y, r) a,b κr 2 (x,r) < κr ∧ τB 2 (y,r ) ? 2

κr2 κr2 κr2 a,b a,b ∧ τB ≥ P τ ≥ x (x,r) B (x,r) 2 2 2



κr2 . 4

=Ex ?
0

κr 2

a,b ∧ τB (x,r )

?

a,b J a (Xs , u)duds? B (y,r)

≥2?(d+α) Ex

κr2 a,b ∧ τB (x,r) 2

B (y,r)

aα du |x ? y |d+α

26



ωd aα d+2 κr , 4 · 2d+α |x ? y |d+α

a,b where in the last second inequality, we have used the fact that for u ∈ B (y, r), |u ? Xs |≤ 2r + |x ? y | ≤ 2|x ? y |.

Lemma 5.7. For any M > 0, there is a constant C20 = C20 (d, α, M, b) depending on b only via the rate at which Mb (r) goes to zero, such that for all t ∈ (0, t? ], a ∈ (0, M ] and all x, y ∈ Rd pa,b (t, x, y ) ≥ C20 t?d/2 ∧ aα t |x ? y |d+α .

Proof. By (3.17), for t ∈ (0, t? ] and x, y ∈ Rd , with |x ? y |2 ≤ t
?1 ?d/2 ?1 pa,b (t, x, y ) ≥ C14 t ≥ C14 t?d/2 ∧

aα t |x ? y |d+α

.

It is left to consider √ the case |x ? y |2 > t. For any t ∈ (0, t? ], by Lemma 5.5 √ and Lemma 5.6 √ with R0 = t? , r = t/4 and strong Markov property, we have |x ? y | > t > 2r and √ a,b t/2) Px Xκt/ ∈ B ( y, 16 √ ≥Px X a,b hits B (y, t/2) before time κt/16 and stays there for at least κt/16 units of time
a,b √ ≥Px σB < κt/16 (y, t/4) z ∈B (y, t/4)

inf √ inf √

a,b √ Pz τB ≥ κt/16 (y, t/2) a,b √ Pz τB ≥ κt/16 (z, t/4)

≥Px

a,b √ σB (y, t/4)

< κt/16

z ∈B (y, t/4)

aα ≥c1 t(d+2)/2 , |x ? y |d+α for some constant c1 = c1 (d, α, M, b) > 0. Combining this, Lemma 5.5 and semigroup property (3.24), for t ∈ (0, t? ], we have pa,b (t, x, y ) =
Rd

pa,b (κt/16, x, z )pa,b ((1 ? κ/16)t, z, y )dz pa,b (κt/16, x, z )pa,b ((1 √ B (y, t/2)
z ∈B (y, t/2)

≥ ≥

? κ/16)t, z, y )dz

inf √

√ a,b pa,b ((1 ? κ/16)t, z, y )Px Xκt/ ∈ B ( y, t/2) 16 aα |x ? y |d+α ,

≥c2 t?d/2 t(d+2)/2 =c2

aα t aα t ?d/2 ≥ c t ∧ 2 |x ? y |d+α |x ? y |d+α

where c2 = c2 (d, α, M, b) is a positive √constant and in the last third inequality, we have used the fact that κ < 1 and for z ∈ B (y, t/2), |z ? y |2 < κ t < (1 ? )t. 4 16

Lemma 5.8. Suppose M > 0. For all a ∈ (0, M ], t ∈ (0, t? ] and x, y ∈ Rd , there are constants Ci = Ci (d, α, M ) > 0, i = 21, 22 such that pa,b (t, x, y ) ≥ C21 t?d/2 exp ? C22 |x ? y |2 t .

27

Proof. By (3.17), for all t ∈ (0, t? ] and x, y ∈ Rd with |x ? y |2 < t, we have
?1 ?d/2 pa,b (t, x, y ) ≥ C14 t .

(5.6)

Next, we consider the case |x ? y |2 > t. We ?x x, y ∈ Rd with |x ? y |2 ≥ t, . Let k be 1 the smallest integer such that 9|x ? y |2 /t < k . Set ξj = x + j ? k (y ? x), 1 ≤ j ≤ k ? 1 and A=
√ k?1 √t ). B ( ξ , j j =1 3 k

For any (x1 , · · · , xk?1 ) ∈ A, we have

√ √ t t |x ? x1 | < √ < √ , 3 k k xj ? ξj + ξj ?1 ? xj ?1 + 1<j<k?1 √ √ √ t t t √ + √ =√ , 3 k 3 k k √ t |xk?1 ? y | = |xk?1 ? ξk?1 + ξk?1 ? y | < √ . k max |xj ? xj ?1 | = max 1<j<k?1 √ t < √ + 3 k Hence by Lemma 4.7, semigroup property (3.24) and (5.6), pa,b (t, x, y ) = t t pa,b ( , x, x1 ) · · · pa,b ( , xk?1 , y )dx1 dx2 · · · dxk?1 k k Rd(k?1) t t ≥ pa,b ( , x, x1 ) · · · pa,b ( , xk?1 , y )dx1 dx2 · · · dxk?1 k k A √ d(k?1) t ?dk/2 k?1 t ?k √ ≥ C14 ?d k 3 k
?d/2 k d/2

y?x k

=t =

C14

k d/2 ?d C14 3d k d/2

k ?1

k?1 k d/2 ?d/2 ?d t C14 C14 3d 3d ?d/2 C14 3d 9|x ? y |2 ≥ t exp ? ln C14 ?d t
d

.

Denoting C21 =

3d C14

14 3 and C22 = 9 ln C? and combining (5.6), we ?nish the proof. d

Proof. The upper bound of pa,b (t, x, y ) is shown by (5.2). We need only to show the lower bound. Without loss of generality, we assume T > t? . If t ∈ (0, t? ], by Lemma 5.7 and Lemma 5.8, there is a constant c1 = c1 (d, α, M, b) > 0 such that for x, y ∈ Rd pa,b (t, x, y ) ≥ C22 |x ? y |2 aα t + C20 t?d/2 ∧ t |x ? y |d+α 2 α a t C22 |x ? y | ≥c1 t?d/2 exp ? + t?d/2 ∧ t |x ? y |d+α a =c1 qd,C (t, x, y ). 22 1 2 C21 t?d/2 exp ?

(5.7)

If t ≥ t? , we let k be the smallest integer such that t? k ≥ t(> (k ? 1)t? ). Note that by Theorem 2.1, for t ∈ (0, T ] and x, y ∈ Rd ,
a qd,C (t, x, y ) ≥ 22

C5 C22

d/2 a qd,C ( 5

C5 ? 1 a C5 t, x, y ) ≥ (C5 /C22 )d/2 C7 p ( t, x, y ). C22 C22

Using this, semigroup property (3.24) and (5.7), we have
k pa,b (t, x, y ) ≥ c? 1

t t a a qd,C ( , x, x1 ) · · · qd,C ( , xk?1 , y )dx1 · · · dxk?1 22 22 k k Rd(k?1) 28

k ≥ c? 1 k = c? 1

C5 C22 C5 C22

dk/2 ?k C7 dk/2 ?k a C7 p ( d(k?1)/2 a ( qd,C 6 dt/(2t? )

pa (
Rd(k?1)

C5 t C5 t , x, x1 ) · · · pa ( , xk?1 , y )dx1 · · · dxk?1 C22 k C22 k

C5 t, x, y ) C22 C5 t, x, y ) C22

C5 C8 = c1 C7 ≥ ≥ C5 C8 c2 c1 C7 C5 C8 c2 c1 C7

C5 c1 C7 C5 c1 C7 C5 c1 C7

a (t, x, y ) qd,C 6 C22 /C5 dT /(2t? ) a (t, x, y ). qd,C 6 C22 /C5

where c2 = c2 (d, α, M, b) is a positive constant. Combining this and (5.7), we ?nish the proof. Acknowledgement. The paper is completed when the second author was visiting Department of Mathematics, University of Washington. The authors are grateful to Longmin Wang for valuable comments.

References
[1] Applebaum D., L? evy Processes and Stochastic Calculus. Cambridge University Press, 2004. [2] Blumenthal R. M. and Getoor R. K., Markov Processes and Potential Theory. Academic Press, 1968. [3] Bogdan K. and Sztonyk P., Estimates of heat kernel of fractional Laplacian perturbed by gradient operators, Comm. Math. Phys. 271 (2007), no. 1, 179-198. [4] Chen Z.-Q. and Kumagai T., A priori H¨ older estimate, parabolic Harnack principle and heat kernel estimates for di?usions with jumps, Rev. Mat. Iberoam. 26 (2010) 551-589. [5] Chen Z.-Q., Kim P. and Song R., Dirichlet heat kernel estimates for fractional Laplacian under gradient perturbation. Ann. Probab., to appear, 2011. arXiv:1011.3273. [6] Chen Z.-Q., Kim P. and Song R., Heat kernel estimates for ? + ?α/2 in C 1,1 open sets. To appear in J. London Math. Soc., 2011. [7] Chen Z.-Q., Kim P., Song R. and Vondraˇ cek, Z., Sharp Green function estimates for α/ 2 1 , 1 ?+? in C open sets and their applications. Ill. J. Math., to appear, 2011. [8] Chen Z.-Q., Kim P., Song R. and Vondraˇ cek, Z., Boundary Harnack principle for ?+?α/2 . Trans. Amer. Math. Soc., (2011) to appear. [9] Chen Z.-Q., and Kumagai T., Heat kernel estimates for stable-like processes on d-sets. Stoch. Proc. Appl. 108 (2003), 27-62. [10] Chen Z.-Q., and Kumagai T., Heat kernel estimates for jump processes of mixed types on metric measure spaces. Probab. Theory Relat. Fields, 140 (2008), 277-317. [11] Cranston M. and Zhao Z., Conditional transformation of drift formula and potential theory for 1 2 ? + b(·) · ?. Commun. Math. Phys. 112(4), 613.625 (1987). [12] Fabes, E.B. and Stroock, D.W., A new proof of Mosers parabolic Harnack inequality using the old ideas of Nash. Arch. Rational Mech. Anal. 96, 327-338 (1986).

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[13] Jakubowski T. and Szczypkowski K., Time-dependent gradient perturbations of fractional Laplacian. J. Evol. Equ., 10(2):319-339, 2010. [14] Kim,P. and Song R., Two-sided estimates on the density of Brownian motion with singular drift. Illinois J. Math. 50 (2006), 635-688. [15] Kim,P. and Song R., Boundary Harnack principle for Brownian motions with measurevalued drifts in bounded Lipschitz domains. Math. Ann. 339 (2007), 135-174. [16] Song R. and Vondraˇ cek Z., Parabolic Harnack inequality for the mixture of Brownian motion and stable process, Tohoku Math. J. (2) 59 (2007) 1-19. [17] Zhang Q., A Harnack inequality for the equation ?(a?u) + b?u = 0, when |b| ∈ Kn+1 . Manuscripta Math. 89(1), 61-77 (1996). [18] Zhang Q.-S., Gaussian bounds for the fundamental solutions of ?(A?u) + B ?u ? ut = 0. Manuscripta Math. 93(3), 381.390 (1997). Zhen-Qing Chen Department of Mathematics, University of Washington, Seattle, WA 98195, USA E-mail: zchen@math.washington.edu Eryan Hu Department Of Mathematical Sciences, Tsinghua University, Beijing 100084, China E-mail: huey07@mails.tsinghua.edu.cn

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