BMO estimates for stochastic singular integral operators and its ...

BMO estimates for stochastic singular integral operators and its application to PDEs with Lévy noise Guangying Lva , Hongjun Gaob Jinlong Weic , Jiang-Lun Wud

arXiv:1704.05577v1 [math.PR] 19 Apr 2017

a

d

Institute of Contemporary Mathematics, Henan University Kaifeng, Henan 475001, China [email protected] b Institute of Mathematics, School of Mathematical Science Nanjing Normal University, Nanjing 210023, China [email protected] c School of Statistics and Mathematics, Zhongnan University of Economics and Law, Wuhan, Hubei 430073, China [email protected] Department of Mathematics, Swansea University, Swansea SA2 8PP, UK [email protected]

April 20, 2017

Abstract In this paper, we consider the stochastic singular integral operators and obtain the BMO estimates. As an application, we consider the fractional Laplacian equation with additive noises α 2

dut (x) = ∆ ut (x)dt +

∞ Z X

k=1

˜k (dz, dt), g k (t, x)z N

u0 = 0, 0 ≤ t ≤ T,

Rm

R α α ˜k (t, dz) =: Y k are independent m-dimensional pure jump where ∆ 2 = −(−∆) 2 , and Rm z N t Lévy processes with Lévy measure of ν k . Following the idea of [9], we obtain the q-th order BMO quasi-norm of the qα0 -order derivative of u is controlled by the norm of g. Keywords: Anomalous diffusion; Itˆ o’s formula; BMO estimates. AMS subject classifications (2010): 35K20; 60H15; 60H40.

1

Introduction

For a stochastic process {Xt , t ∈ T }, there are two important facts worth studying. One is its probability density function (PDF) or its probability law, the other is the estimates of moment. But for a stochastic process depending on spatial variable, that is, Xt = X(t, ω, x) (x is the spatial variable), it is hard to consider its PDF or probability law. Fortunately, we can get some estimates of moment. In this paper, we focus on the estimates of solutions of stochastic partial differential equations (SPDEs). For SPDEs, many kinds of estimates of the solutions have been well studied. By using parabolic Littlewood-Paley inequality, Krylov [13] proved that for SPDEs of the type du = ∆udt + gdwt , 1

(1.1)

2 it holds that Ek∇ukpLp ((0,T )×Rd ) ≤ C(d, p)EkgkpLp ((0,T )×Rd ) ,

(1.2)

where wt is a Wiener process and p ∈ [2, ∞). van Neerven et al. [16] introduce a significant extension of (1.2) to a class of operators A which admit a bounded H ∞ -calculus of angle less than π/2. Kim [9] established a BMO estimate for stochastic singular integral operators. And as an application, they considered (1.1) and obtained the q-th order BMO quasi-norm of the derivative of u is controlled by kgkL∞ . Just recently, Kim et al. [11] studied the parabolic Littlewood-Paley inequality for a class of time-dependent pseudo-differential operators of arbitrary order, and applied this result to the high-order stochastic PDE. Recently, Yang [18] considered the following SPDEs α

du = ∆ 2 udt + f dXt , α

u0 = 0, 0 < t < T,

α

where ∆ 2 = −(−∆) 2 , 0 < α < 2, and Xt is a Lévy process. They obtained a parabolic TriebelLizorkin space estimate for the convolution operator. Regarding elliptic and parabolic singular integral operators, the BMO estimates was already established in [4, 6]. In this paper, we consider the stochastic singular integral operator Z tZ K(t, s, ·) ∗ g(s, ·, z)(x)N˜ (dz, ds) Gg(t, x) = 0 Z Z tZ Z K(t, s, x − y)g(s, y, z)dy N˜ (dz, ds). (1.3) = 0

Z

Rd

Our main purpose is to present appropriate conditions on the kernel K for the following estimate:

Z q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz) [Gg]BMO(T,q) ≤ N

κ−q κ

Z L

Z

q/q0

q0

+ |g(·, ·, z)|L∞ (OT ) ν(dz)

Z

Lκ˜ (Ω)

Z

q

+ kg(·, ·, z)kL∞ (OT ) ν(dz)

Z

κ

L κ−q

,

(1.4)

where q ∈ [2, p0 ∧ κ], κ ˜ is the conjugate of a positive constant κ, the constant N depends on q and d, and ν is a measure, see Section 2. As an application of (1.4), we prove that the solution of the following equation α 2


∞ Z X

m k=1 R

gk (t, x)z N˜k (dz, dt),

u0 = 0, 0 ≤ t ≤ T,

satisfies that for q ∈ [2, q0 ] q/q0 [∇β u]BMO(T,q) ≤ N cˆ E[k|g|ℓ2 kqL0∞ (OT ) ] ,

R ˜k (t, dz) =: Y k are independent m-dimensional pure jump Lévy processes with Lévy where Rm z N t k measure of ν , β = α/q0 and cˆ is defined as in (4.4), see Section 4 for details. Moreover, we find if we consider the following stochastic parabolic equation α

dut (x) = ∆ 2 ut (x)dt +

∞ X k=1

hk (t, x)dWtk ,

u0 = 0, 0 ≤ t ≤ T,

3 where Wtk are independent one-dimensional Wiener processes. We have the following estimate, for any q ∈ (0, p], 1/p α [∇ 2 u]BMO(T,q) ≤ N E[k|h|ℓ2 kpL∞ (OT ) ] .

under the condition that h ∈ Lp (T, ℓ2 ), see Theorem 4.2. Specially, taking α = 2, we obtain the result of [9, Theorem 3.4]. Due to the difference between the Brownian motion and Lévy process, it is more difficult to get the BMO estimate for Lévy process. Following the idea of [9], we obtain the BMO estimate of stochastic singular integral operators. We remark that there are many places different from those in [9]. First, the assumptions on the kernel are different from those in [9], see Section 2; Second, the exponent q in [9] do not depend on the properties of kernel but we do. For simplicity, we only consider a simple case, see the discussion in Section 4. This paper is organized as follows. In Section 2, we introduce the main results. The proof of the main results is complete in section 3. Section 4 is concerned with an application of our result. This paper ends with a short discussion, which shows that we can give a simple proof of the result in Section 2 if the function g has high regularity. Before we end this section, we introduce some notations used in this paper. As usual Rd stands for the Euclidean space of points x = (x1 , · · · , xd ), Br (x) := {y ∈ Rd : |x − y| < r} and Br := Br (0). R+ denotes the set {x ∈ R, x > 0}. a ∧ b = min{a, b}, a ∨ b = max{a, b} and Lp := Lp (Rd ). N = N (a, b, · · · ) means that the constant N depends only on a, b, · · · .

2

Known results and Main result

Let (Ω, F, F, P) be a complete probability space such that Ft is a filtration on Ω containing all P -null subsets of Ω and F be the predictable σ-field by (Ft , t ≥ 0). We are given a measure space (Z, Z, ν) and a Poisson measure µ on [0, T ] × Z, defined on the stochastic basis. The compensator ˜ := µ − Leb ⊗ ν of µ is Leb⊗ν, and the compensated measure N Fix γ > 0 and T ∈ (0, ∞]. Denote OT = (0, T ) × Rd . For a measurable function h on Ω × OT , we define the q-th order stochastic BMO (Bounded mean oscillation) quasi-norm of h on Ω × OT as follows: Z Z 1 q |h(t, x) − h(s, y)|q dtdxdsdy, [h]BMO(T,q) = sup 2 E Q Q Q Q where the sup is taken over all Q of the type Q = Qc (t0 , x0 ) := (t0 − cγ , t0 + cγ ) × Bc (x0 ) ⊂ OT , c > 0, t0 > 0. It is remarked that when q = 1, this is equivalent to the classical BMO semi-norm which is introduced by John-Nirenberg [8]. Let K(ω, t, s, x) be a measurable function on Ω × R+ × R+ × Rd such that for each t ∈ R+ , (ω, s) 7→ K(ω, t, s, ·) is a predictable L1loc -valued process. Firstly, we recall the results of [9]. In [9], the following assumptions are needed. Assumption 2.1 There exist a κ ∈ [1, ∞] and a nondecreasing function ϕ(t) : (0, ∞) 7→ [0, ∞) such that

4 (i) for any t > λ > 0 and c > 0,

Z Z 2

t

|K(t, r, x)|dx dr

λ |x|≥c

≤ ϕ((t − λ)c−γ ); Lκ/2 (Ω)

(ii) for any t > s > λ > 0,

Z Z 2

λ

|K(t, r, x) − K(s, r, x)|dx dr

0 Rd

≤ ϕ((t − s)(t ∧ s − λ)−1 ); Lκ/2 (Ω)

(iii) for any s > λ ≥ 0 and h ∈ Rd ,

Z Z 2

λ

dr |K(s, r, x + h) − K(s, r, x)|dx

0 d R

≤ N ϕ(|h|(s − λ)−1/γ ).

Lκ/2 (Ω)

Assumption 2.2 Suppose that Gg is well-defined (a.e.) and the following holds:

Z T

Z T

p0 p0

, kGg(t, ·)kLp0 dt ≤ N0 E k|g(t, ·)|l2 kLp0 dt

0

0

Lκ˜ (Ω)

where κ ˜ is the conjugate of κ, and Gg(t, x) =

∞ Z tZ X k=1 0

Rd

K(t, s, x − y)gk (s, y)dydwsk ,

with wt is a Wiener process. Under the Assumptions 2.1 and 2.2, Kim obtained the BMO estimate of Gg. Comparing with the assumption 2.1, due to the Kunita’s first inequality (see Page 265 of [1]), we need the following assumptions. For the Kunita’s inequality and BDG inequality of Lévy noise, see Lemma 3.1 of [14] and [15] respectively. Assumption 2.3 There exist constants q0 ≥ 2, κ ∈ [1, ∞] and a nondecreasing function ϕ(t) : (0, ∞) 7→ [0, ∞) such that (i) for any t > λ > 0 and c > 0,

Z Z q0

t

≤ ϕ((t − λ)c−γ ); |K(t, r, x)|dx dr

κ/q

λ |x|≥c L

(ii) for any t > s > λ > 0,

Z λ Z q0

dr |K(t, r, x) − K(s, r, x)|dx

d 0

R

0 (Ω)

≤ ϕ((t − s)(t ∧ s − λ)−1 ); Lκ/q0 (Ω)

(iii) for any s > λ ≥ 0 and h ∈ Rd ,

Z λ Z q0

dr |K(s, r, x + h) − K(s, r, x)|dx

d 0

≤ N ϕ(|h|(s − λ)−1/γ ).

Lκ/q0 (Ω)

R

Remark 2.1 The difference between assumptions 2.1 and 2.3 is because the following Kunita’s first inequality. ! ( "Z Z p/2 # T 2 p ≤ N (p) E |H(t, z)| ν(dz)dt E sup |I(t)| 0≤t≤T

+E

Z

T

0

Z

0

Z

p

|H(t, z)| ν(dz)dt Z

,

(2.1)

5 where p ≥ 2 and I(t) =

Z tZ 0

˜ (dz, ds). H(s, z)N

Z

˜ (ds, dz) is replaced by dws dz, the second term of right side hand of (2.1) will disappear. When N Hence, in order to deal with the difficult from the Lévy process, we give the assumption 2.3. Assumption 2.4 Similar Assumption 2.2, suppose that Gg is well-defined (a.e.) and the following holds:

Z T Z

Z T

q0 q 0 kGg(t, ·)kLq0 dt ≤ N0 kg(t, ·, z)kLq0 ν(dz)dt E . (2.2)

0

0

Z

Lκ˜ (Ω)

Our main result is the following.

Theorem 2.1 Let Assumptions 2.3 and 2.4 hold. Assume that the function g satisfies

Z

kg(·, ·, z)k̟∞

ν(dz) < ∞, ̟ = 2 or q0 , (2.3) L (OT )

ς Z

where ς = q0 κ ˜∨

L (Ω)

q0 κ 2(κ−q0 )+

(ς = ∞ if κ ≤ q0 ). Then for any q ∈ [2, q0 ∧ κ], one has

Z q/2

[Gg]BMO(T,q) ≤ N kg(·, ·, z)k2L∞ (OT ) ν(dz)

κ−q κ

Z L

q/q0

Z

q0

+

kg(·, ·, z)kL∞ (OT ) ν(dz)

where N = N (N0 , d, q, q0 , γ, κ, ϕ).

Z

Lκ˜ (Ω)

Z

L κ−q

Z

q

+

kg(·, ·, z)kL∞ (OT ) ν(dz)

κ

,

(2.4)

Remark 2.2 1. Comparing Theorem 2.1 with Theorem 2.4 in [9], it is not hard to find in Theorem 2.4 of [9] the exponent q does not depend on q0 . Actually, the range of exponent q is (0, p0 ∧ κ] and in this paper is [2, q0 ∧ κ]. In other words, the range of exponent q depends on the properties of kernel K. The lower bound of q is because the Kunita’s first inequality holds for q ≥ 2. 2. In Theorem 2.1, we did not write the right hand of (2.4) as a uniform format. The reason R is that Z ν(dz) maybe not exist. If we assume that

Z Z q

q − q0 q0 2 − 20

+ f (z) )ν(dz) < ∞, (z ∧ 1)ν(dz) ≤ N1 and kg(·, ·, z)kL∞ (OT ) (1 + f (z)

Z

Z

Lκ∗ (Ω)

where N1 is a positive constant, then (2.4) can be replaced by

q/q0

Z q

q − q0 q0 − 20

)ν(dz) [Gg]BMO(T,q) ≤ kg(·, ·, z)kL∞ (OT ) (1 + f (z) + f (z)

Z

,

Lκ∗ (Ω)

where

κ∗ = κ ˜∨

z 2 + 1 − |z 2 − 1| κ , f (z) = = z 2 ∧ 1. κ−q 2

3. The condition (2.3) coincides with (4.4) in Section 4. Under the condition (2.3), it is easy to check that

Z

kg(·, ·, z)kq ∞

< ∞.

L (OT ) ν(dz) κ Z

L κ−q (Ω)

6

3

Proof of the main result

In this section, we first estimate the expectation of local mean average of Gg and its difference in terms of the supremum of |g| given a vanishing condition on g. Then we complete the proof of main result. Lemma 3.1 Let q ∈ [2, q0 ], 0 ≤ a ≤ b ≤ T , and Assumption 2.4 hold. Suppose that g vanishes on (a, b) × (B3c )c × Z and (0, a) × Rd × Z. Then E

Z bZ a

q/p0

Z

, |g(·, ·, z)|q0 ν(dz) |Gg(t, x)|q dxdt ≤ N (b − a)|B3c | sup

κ˜

(a,b)×B3c Z Bc L (Ω)

where N = N (N0 ).

Proof. The proof of this lemma is similar to that of Lemma 4.1 in [9]. In order to read easily, we give the outline of the proof. By H¨ older’s inequality and Assumption 2.4, E

Z bZ a

|Gg(t, x)|q dxdt Bc (q0 −q)/q0

≤ (b − a)

(q0 −q)/q0

|Bc |

(q0 −q)/q0

≤ N (b − a)

Z bZ E

(q0 −q)/q0

|Bc |

Z

a

q0

|Gg(t, x)| dxdt

Bc

T

0

Z

Z

q/q0

q/q0

kg(t, ·, z)kqL0q0 ν(dz)dt

.

Lκ˜ (Ω)

Since g vanishes on (a, b) × (B3c )c and (0, a) × Rd , the above term is equal to or less than (q0 −q)/q0

N (b − a)

(q0 −q)/q0

|Bc |

Z b Z

a

B3c

Z

q/q0

|g(t, x, z)| ν(dz)dxdt

κ˜ q0

Z

q/q0 Z

≤ N (b − a)|B3c | sup |g(·, ·, z)|q0 ν(dz) .

(a,b)×B3c Z

κ˜

L (Ω)

L (Ω)

The proof of lemma is complete.

Lemma 3.2 Let q ∈ [2, q0 ∧ κ], 0 ≤ a ≤ b ≤ T and Assumption 2.3 (i) hold. Suppose that g vanishes on (0, 3b−a 2 ) × B2c × Z. Then Z bZ

Z bZ

|Gg(t, x) − Gg(s, y)|q dxdtdsdy

Z q/2

≤ N (b − a)2 |Bc |2 [ϕ(bc−γ )]q/q0 kg(·, ·, z)k2L∞ (OT ) ν(dz)

κ

κ−q

Z L

Z

q

+ kg(·, ·, z)k ν(dz) , ∞

κ L (OT ) E

a

Bc

a

Bc

Z

where

∞ ∞

(3.1)

L κ−q

:= 1 and N = N (T, q).

Proof. Let (t, x) ∈ (a, b) × Bc and 0 ≤ r ≤ t. If |y| ≤ c, then (r, x − y) ∈ (0, 3b−a 2 ) × B2c and g(r, x − y, z) = 0 for all z ∈ Z. Hence, Assumption 2.3 (i), H¨ older inequality and Kunita’s first

7 inequality (2.1) implies q

E|Gg(t, x)|

≤ E

Z t Z 0

+E

Z

0

0

+E

K(t, r, y)g(r, x − y, z)dy| ν(dz)dr Rd

| |

Z

Z tZ 0

≤ T

2

Z

Z tZ

≤ E

|

Z tZ

Z

Z

Z

K(t, r, y)g(r, x − y, z)dy| ν(dz)dr Rd

|

(q0 −2)q/(2q0 )

Z

" Z Z t +E

!q/2

q

K(t, r, y)g(r, x − y, z)dy| ν(dz)dr |y|≥c



Z t Z E

×

0

Z

Z

|y|≥c

|y|≥c

q0 |K(t, r, y)|dy dr

!q/q0

!

q/2 #

kg(·, ·, z)k2L∞ (OT ) ν(dz)

# !Z q q kg(·, ·, z)kL∞ (OT ) ν(dz) |K(t, r, y)|dy dr Z

Z Z q0 q/q0

t

|K(t, r, y)|dy dr ≤ N (T )

κ/q

0 |y|≥c L 0

Z q/2

2 kg(·, ·, z)kL∞ (OT ) ν(dz) ×

κ

κ−q

Z L

Z Z q0 q/q0

t

|K(t, r, y)|dy dr +N (T )

κ/q

0 |y|≥c L 0

Z

q

× κ

kg(·, ·, z)kL∞ (OT ) ν(dz) κ−q Z L

Z q/2

−γ q/q0 2 ≤ N (T )[ϕ(bc )] kg(·, ·, z)kL∞ (OT ) ν(dz)

κ

κ−q

Z

Z

L

q

+ ,

kg(·, ·, z)kL∞ (OT ) ν(dz) κ Z

which implies that E

K(t, r, y)g(r, x − y, z)dy|2 ν(dz)dr |y|≥c

Z

0

q

q/2

Z bZ a

Bc

Z bZ a

L κ−q

|Gg(t, x) − Gg(s, y)|q dxdtdsdy

Bc

Z bZ

|Gg(t, x)|q dxdt a Bc

Z q/2

≤ N (T, q)(b − a)2 |Bc |2 [ϕ(bc−γ )]q/q0 kg(·, ·, z)k2L∞ (OT ) ν(dz)

κ−q κ

Z L

Z

q

. +

kg(·, ·, z)kL∞ (OT ) ν(dz) κ

≤ N (q)(b − a)|Bc |E

Z

L κ−q

The inequality (3.1) is obtained. The proof of lemma is complete.

8 Lemma 3.3 Let q ∈ [2, q0 ∧ κ], 0 ≤ a < b ≤ T such that 3a > b. Suppose that Assumption 2.3 3b−a holds and Gg is well-defined almost everywhere. Assume further that g vanishes on ( 3a−b 2 , 2 )× B2c × Z. Then Z bZ Z bZ |Gg(t, x) − Gg(s, y)|q dxdtdsdy E Bc a Bc a

Z q/2

2 2 2 kg(·, ·, z)kL∞ (OT ) ν(dz) ≤ N (b − a) |Bc | Φ(a, b, c)

κ

κ−q

Z L

Z

q

+ , (3.2)


L κ−q

where N = N (T, q, a, b, c) and

Φ(a, b, c) = [ϕ(2)]q/q0 + [ϕ((b − a)c−γ )]q/q0 + [ϕ(21+1/γ c(b − a)−1/γ )]q/q0 . Proof. Due to the Fubini’s Theorem, it suffices to prove that for all (t, x) ∈ (a, b) × Bc and (s, y) ∈ (a, b) × Bc , the following inequality holds:

Z q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz) E|Gg(t, x) − Gg(s, y)|q ≤ N Φ(a, b, c)

κ

κ−q

Z L

Z

q

+ , κ

kg(·, ·, z)kL∞ (OT ) ν(dz) κ−q Z

L

Obviously,

E|Gg(t, x) − Gg(s, y)|q ≤

N (E|Gg(t, x) − Gg(s, x)|q + E|Gg(s, x) − Gg(s, y)|q )

=: N (I1 + I2 ). Estimate of I1 . Without loss of generality we assume t ≥ s. Hence by Lemma 3.1 of [14] and

9 (2.1), we get I1

= =

≤

E|Gg(t, x) − Gg(s, x)|q Z t Z Z K(t, r, x − y)g(r, y, z)dy N˜ (dz, dr) E 0 Z Rd Z sZ Z q ˜ K(s, r, x − y)g(r, y, z)dy N (dz, dr) − d 0 Z R Z t Z Z K(t, r, x − y)g(r, y, z)dy N˜ (dz, dr) NE 0 Z Rd Z sZ Z q ˜ K(t, r, x − y)g(r, y, z)dy N (dz, dr) − d Z0 s ZZ ZR q (K(t, r, x − y) − K(s, r, x − y))g(r, y, z)dy N˜ (dz, dr) +N E d "Z 0 Z Z Z R q/2 # t

≤

NE

+N E

Z t Z s

+N E +N E

K(t, r, x − y)g(r, y, z)dy|2 ν(dz)dr

|

s

"Z

Z

0

Rd

Z

|

Z sZ

0 sZ

Z

Z

q

K(t, r, x − y)g(r, y, z)dy| ν(dz)dr Rd

|

Z

|

Z

Z

(K(t, r, x − y) − K(s, r, x − y))g(r, y, z)dy|2 ν(dz)dr Rd q

(K(t, r, x − y) − K(s, r, x − y))g(r, y, z)dy| ν(dz)dr Rd

=: N (I11 + I12 + I13 + I14 ). 3b−a Note that g vanishes on ( 3a−b 2 , 2 ) × B2c × Z and a > yields that "Z Z Z

3a−b 2 .

t

I11 + I12 = E

+E 

Z t Z s

2

Rd

Z

|

Z

q

K(t, r, y)g(r, x − y, z)dy| ν(dz)dr

Rd

Z

Z t Z  ≤ E

|y|≥c

s

"Z Z t +E s

Assumption 2.3 (i) with λ = s

K(t, r, y)g(r, x − y, z)dy| ν(dz)dr

|

s

|y|≥c

q/2 #

!q/2  2 Z  kg(·, ·, z)k2L∞ (OT ) ν(dz) |K(t, r, y)|dy dr Z

q Z kg(·, ·, z)kqL∞ (OT ) ν(dz)dt |K(t, r, y)|dy Z

#

Z q/2

≤ N [ϕ((b − a)c−γ )]q/q0 kg(·, ·, z)k2L∞ (OT ) ν(dz)

κ

Z

κ−q L

Z

q

. + ν(dz) kg(·, ·, z)k ∞

κ L (OT ) Z

q/2 #

L κ−q

3b−a 3a−b Similarly, due to g vanishes on ( 3a−b 2 , 2 ) × B2c × Z, we divide (0, s) into two parts (0, 2 ) and

10 ( 3a−b 2 , s). And thus we have  !q/2  Z s Z Z  (K(t, r, x − y) − K(s, r, x − y))g(r, y, z)dy|2 ν(dz)dr | I13 + I14 = E  +E

"Z 

+E  +E

"Z

3a−b 2

s

3a−b 2

Z

Rd

Z

Z

| Z

Z

3a−b 2

0

Z

3a−b 2

0

Z

q


Z

| Z

Z

|

Z

#

(K(t, r, x − y) − K(s, r, x − y))g(r, y, z)dy|2 ν(dz)dr Rd q


#

=: I131 + I141 + I132 + I142 . Using again Assumption 2.3 (i) with λ = 3a−b 2 , we get  !q/2  Z t Z Z  |K(t, r, x − y)g(r, y, z)|dy|2 ν(dz)dr | I131 + I141 ≤ E  3a−b 2



Z

+E  +E +E

"Z

"Z

Z

s

3a−b 2

t 3a−b 2

s 3a−b 2

Rd

Z

Z Z

|

Z

| Z

| Z

Z Z

Z

Rd

!q/2   |K(s, r, x − y)g(r, y, z)|dy|2 ν(dz)dr

|K(t, r, x − y)g(r, y, z)|dy|q ν(dz)dr Rd q

#

|K(s, r, x − y)g(r, y, z)|dy| ν(dz)dr Rd

#

Z q/2

2 ≤ N [ϕ(2(b − a)c )] kg(·, ·, z)k ν(dz)

L∞ (OT )

κ−q κ

Z L

Z

q

+ .

kg(·, ·, z)kL∞ (OT ) ν(dz) κ −γ

q/q0

Z

L κ−q

gives On the other hand, Assumption 2.3 (ii) with λ = 3a−b 2 " Z 3a−b Z 2 2 |K(t, r, x − y) − K(s, r, x − y)|dy dr I132 + I142 ≤ N E Rd

0

×

Z

Z

+E ×

Z

"Z

Z

q/2 #

kg(·, ·, z)k2L∞ (OT ) ν(dz) 3a−b 2

0

Z

Rd

q |K(t, r, x − y) − K(s, r, x − y)|dy

kg(·, ·, z)kqL∞ (OT ) ν(dz)dr

Z q/2

2 ≤ N [ϕ(2)] kg(·, ·, z)k ν(dz)

L∞ (OT ) κ

κ−q

Z L

Z

q

, +

kg(·, ·, z)kL∞ (OT ) ν(dz) κ q/q0

Z

where we used s −

3a−b 2

≥a−

3a−b 2

L κ−q

=

b−a 2

and (t − s)(s −

3a−b −1 2 )

≤ 2.

!q/2  

11 3b−a Estimate of I2 . By using the fact g = 0 on ( 3a−b 2 , 2 ) × B2c × Z again, we divide (0, s) into 3a−b 3a−b two parts (0, 2 ) and ( 2 , s). Direct calculations shows that

I2

≤

q/2 Z 2 K(s, r, w)(g(r, x − w, z) − g(r, y − w, z))dw| ν(dz)dr NE Rd 0 Z Z s Z Z q K(s, r, w)(g(r, x − w, z) − g(r, y − w, z))dw| ν(dz)dr | +E Z

sZ

0

≤

NE

Z

+N E

s 3a−b 2

Z

s

Z Z Z

0

+N E

Z

≤

NE

+N E +N E +N E

+N E

s 3a−b 2

Z

Z

Z

s

s 3a−b 2

s 3a−b 2

+N E

0

K(s, r, w)(g(r, x − w, z) − g(r, y − w, z))dw|q ν(dz)dr Rd

Z

|

Z

(K(s, r, x − w) − K(s, r, y − w))g(r, w, z)dw|q ν(dz)dr

Rd

|K(s, r, w)g(r, x − w, z)|dw|2 ν(dz)dr

Rd

Z

|

Z

Z

!q/2

|K(s, r, w)g(r, y − w, z)|dw|2 ν(dz)dr

Rd

Z

|

Z

Z Z

|K(s, r, w)g(r, x − w, z)|dw|q ν(dz)dr

Rd

|K(s, r, w)g(r, y − w, z)|dw|q ν(dz)dr Rd

!

!q/2

!

!

!q/2 Z Z (K(s, r, x − w) − K(s, r, y − w))g(r, w, z)dw|2 ν(dz)dr Rd

Z

3a−b 2

!

Rd

Z

Z Z

3a−b 2

!q/2

!q/2 Z Z (K(s, r, x − w) − K(s, r, y − w))g(r, w, z)dw|2 ν(dz)dr

Z

0

Z

Z

Z Z

3a−b 2

Z

K(s, r, w)(g(r, x − w, z) − g(r, y − w, z))dw| ν(dz)dr

Z

3a−b 2

0

Z

|

Z

3a−b 2

2

Rd

Z

3a−b 2

Z

+N E

Rd

Z

Z

=: I21 + · · · + I26 .

|

Z

Z

(K(s, r, x − w) − K(s, r, y − w))g(r, w, z)dw|q ν(dz)dr

Rd

Similar to I11 + I12 , the four terms I21 + · · · + I24 is less than or equal to

Z q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz) N [ϕ(2(b − a)c−γ )]q/q0

κ

κ−q

Z L

Z

q

ν(dz) + kg(·, ·, z)k . ∞

κ

L (OT ) Z

L κ−q

!

12 Using Assumption 2.3 (iii) with λ = Z

I25 + I26 ≤ N E

3a−b 2

Rd

0

×

Z

Z

+N E ×

Z

we get

2 |K(s, r, x − w) − K(s, r, y − w)|dw dr q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz)

Z

Z

Z

3a−b 2 ,

3a−b 2

0

Z

Rd

q |K(s, r, x − w) − K(s, r, y − w)|dw dr

kg(·, ·, z)kqL∞ (OT ) ν(dz)

Z q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz) ≤ N ϕ(21+1/γ c(b − a)−1/γ )q/q0

κ

κ−q

Z L

Z

q

ν(dz) kg(·, ·, z)k + . ∞

κ

L (OT ) Z

L κ−q

Combining the above discussion, (3.2) is obtained. The proof of this lemma is complete. Now, we are ready to prove the main result. The proof is similar to that of Theorem of 2.4 in [9]. Proof of Theorem 2.1. Let q ∈ [2, q0 ∧ κ]. It suffices to prove that for each Q = Qc (t0 , x0 ) := (t0 − cγ , t0 + cγ ) × Bc (x0 ) ⊂ OT , c > 0, t0 > 0, we have

Z Z 1 |Gg(t, x) − Gg(s, y)|q dtdxdsdy E Q2 Q Q

Z q/2

≤ N kg(·, ·, z)k2L∞ (OT ) ν(dz)

κ−q κ

Z L

q/q0

Z

q0

|g(·, ·, z)| ν(dz) +

κ˜

L∞ (OT ) Z L (Ω)

Z

q

+ kg(·, ·, z)kL∞ (OT ) ν(dz) κ , Z

L κ−q

where N = N (T, q, ϕ). Since the operator G is translation invariant with respect to x, i.e. Gg(·, ·)(t, x + x0 ) = Gg(·, x0 + ·)(t, x), we may assume that x0 = 0. We divide the left hand side of (3.3) into two parts. Indeed, Z Z 1 |Gg(t, x) − Gg(s, y)|q dtdxdsdy E Q2 Q Q Z 2 |Gg1 (t, x)|q dtdxdsdy E ≤ Q Q Z Z 1 + 2E |Gg2 (t, x) − Gg2 (s, y)|q dtdxdsdy Q Q Q =: J1 + J2 , where g1 (t, x, z) = I((t0 −2cγ )∨0,t0 +2cγ )×B2c ×Z (t, x, z)g(t, x, z), g2 = g − g1 .

(3.3)

13 Estimate of J1 . Since Q ⊂ OT , it holds that t0 − cγ ≥ 0 and thus (t0 − cγ , t0 + cγ ) ⊂ (t0 − 2cγ ) ∨ 0, t0 + 2cγ ) and g vanishes on h i[h i c ((t0 − 2cγ ) ∨ 0, t0 + 2cγ ) × B2c ×Z (0, (t0 − 2cγ ) ∨ 0) × Rd × Z .

It follows from Lemma 3.1 with a = (t0 − 2cγ ) ∨ 0 and b = t0 + 2cγ that

Z

q/q0

q 0

J1 ≤ N .

|g(·, ·, z)|L∞ (OT ) ν(dz) Z

(3.4)

Lκ˜ (Ω)

Estimate of J2 . If t0 ≤ 2cγ , we apply Lemma 3.2 with a = t0 − cγ and b = t0 + cγ . In this case, one can easily check that bc−γ ≤ 3 and h i g2 = 0 on (0, t0 + 2cγ ) × B2c × Z . (3.1) of Lemma 3.2 yields that J2

Z q/2

kg(·, ·, z)k2L∞ (OT ) ν(dz) ≤ N

κ

κ−q

Z

Z

L

q

+ ν(dz) kg(·, ·, z)k ∞

κ L (OT ) Z

(3.5)

L κ−q

On the other hand, if t0 > 2cγ , we apply Lemma 3.3 with a = t0 − cγ and b = t0 + cγ . In this case, one can easily check that 3a > b and h i g2 = 0 on (t0 − 2cγ , t0 + 2cγ ) × B2c × Z . Moreover, by using the nondecreasing of ϕ, we have sup t0 ∈R+ ,c>0

=

sup t0 ∈R+ ,c>0

Φ(t0 − cγ , t0 + cγ , c) n

< ∞.

o [ϕ(2)]q/q0 + [ϕ((b − a)c−γ )]q/q0 + [ϕ(21+1/γ c(b − a)−1/γ )]q/q0   

a = t0 − cγ b = t0 + cγ

(3.2) implies that J2

Z q/2

2 kg(·, ·, z)kL∞ (OT ) ν(dz) ≤ N

κ

κ−q

Z

Z L

q

+ .


(3.6)

L κ−q

Combining (3.4), (3.5) and (3.6), we obtain (3.3). The proof of Theorem 2.1 is complete. Remark 3.1 In this paper, we only consider the simply case. Actually, one can use the similar method and Kunita’s second inequality (see Page 268 in [1]) to deal with the following case Z tZ K(t, s, x − y)h(s, y)dydW (s) Gˆ g (t, x) = 0 Rd Z tZ Z K(t, s, x − y)g(s, y, z)dy N˜ (dz, ds), + 0

Z

Rd

˜ is a Wiener process and a compensated Poisson measure, respectively. Also see where W and N [14] for this case.

14

4

Applications

In this section, applying Theorem 2.1, we obtain the BMO estimate of the following stochastic singular integral operator Gg(t, x) =

∞ Z tZ X k=1

Rm

0

Z

Rd

˜k (dz, ds), K(t, s, x − y)gk (s, y)dyz N

(4.1)

where K(t, s, x) = ∇β p(t, s, x) and p(t, s, x) is the heat kernel of the equation α

∂t u = ∆ 2 u. The fractional derivative of spatial variable is understood in sense of Fourier transform. It is easy to see that Z ∞ Z tZ X ˜k (dz, ds) K(t, s, x − y)gk (s, y)dyz N 0

k=1

Rm

Rd

is the fundamental solution to the following equation α 2


∞ Z X

m k=1 R

gk (t, x)z N˜k (dz, dt),

u0 = 0, 0 ≤ t ≤ T,

(4.2)

R ˜k (t, dz) =: Y k are independent m-dimensional pure jump Lévy processes with Lévy where Rm z N t measure of ν k . Indeed, one can use the method of [9] (see the proof of Lemma 6.1) to prove the above result. On the other hand, Kim-Kim [12] considered the general case. We only recall the results concerned with this paper. In section 3 of [12], Kim-Kim studied the following linear equation (see Page 3935 of [12]): α 2

du = (a(ω, t)∆ u + f )dt +

∞ X

k

h

dWtk

+

∞ X m X

gk,j · dYtk,j , u(0) = u0 ,

(4.3)

k=1 j=1

i=1

R ˜k (t, dz). where h = (h1 , h2 , · · · ), Wtk is independent one-dimensional Wiener processes and Ytk := Rm z N Note that Ytk are independent m-dimensional pure jump Lévy processes with Lévy measure of ν k . For any q, k = 1, 2, · · · , denote cˆk,q :=

Z

1 q |z| ν (dz) . q k

Rm

Fix p ∈ [2, ∞) and set cˆk := cˆk,2 ∨ cˆk,p . Assume that cˆ := sup cˆk < ∞.

(4.4)

k≥1

Let P be the predictable σ-field generated by {Ft , t ≥ 0} and P¯ be the completion of P with ¯ Hpη ), that is, Hηp (T ) is the set of respect to dP × dt. For η ∈ R, define Hηp (T ) := Lp (Ω × [0, T ], P, ¯ all P-measurable processes u : Ω × [0, T ] 7→ Hpη so that kuk

Hηp (T )

Z := E

0

T

k|u(ω, t, ·)kpH η dt p

1/p

< ∞,

where Hpη (Rd ) := {u : D n u ∈ Lp (Rd ), |n| ≤ η} for η = 1, 2, . . . . And when η is not an integer, Hpη (Rd ) is defined by Fourier transform.

15 ¯ For ℓ2 -valued P-measurable processes g = (g 1 , g2 , · · · ), we write g ∈ Hηp (T, ℓ2 ) if kgkHηp (T,ℓ2 ) := =

Z E

T

2

p

0

Z E

kg(ω, t, ·)kpH η (T,ℓ ) dt

T

η/2

k|(1 − ∆)

0

1/p

g(ω, t, ·)|ℓ2 kpp dt

1/p

< ∞.

Lastly, we define kukHη+α (T ) := kukHη+α (T ) + kf kHη+α (T ) + khkHη+α/2 (T,ℓ p p p p

+

m X

2)

kg·,j kHη+α/2 (T,ℓ ) + ku(0)kU η+α−α/p , p

j=1

2

p

1/p where ku(0)kU η+α−α/p := E[ku0 kpH η ] . p

p

Proposition 4.1 [12, Theorem 3.6] Suppose (4.4) holds. Then for any f ∈ Hpη (T ), h ∈ η+α/2 η+α−α/p η+α−α/p Hp (T, ℓ2 ), g ·,j ∈ Hp (T, ℓ2 ), 1 ≤ j ≤ m and u0 ∈ Up , Eq. (4.3) has a unique η+α solution u in Hp , and for this solution kukHη+α ≤ N (p, T, a) kf kHηp (t) + khkHη+α/2 (t,ℓ ) (t) p p

+

m X j=1

2

kg·,j kHη+α−α/p (t,ℓ ) + ku(0)kU η+α−α/p p

2

p

for every t ≤ T . In order to investigate the BMO estimate of the solution, we recall some properties of kernel p(t, s, x) (see [2, 3, 5, 7] for more details). • for any t > 0, kp(t, ·)kL1 (Rd ) = 1 for all t > 0. • p(t, x, y) is C ∞ on (0, ∞) × Rd × Rd for each t > 0; • for t > 0, x, y ∈ Rd , x 6= y, the sharp estimate of pb(t, x) is t −d/α p(t, x, y) ≈ min ,t ; |x − y|d+α • for t > 0, x, y ∈ Rd , x 6= y, the estimate of the first order derivative of pb(t, x) is t − d+2 α ,t . |∇x p(t, x, y)| ≈ |y − x| min |y − x|d+2+α

(4.5)

The notation f (x) ≈ g(x) means that there is a number 0 < C < ∞ independent of x, i.e. a constant, such that for every x we have C −1 f (x) ≤ g(x) ≤ Cf (x). The estimate (4.5) for the first order derivative of p(t, x) was derived in [2, Lemma 5]. Xie et al. [17] the estimate of the m-th order derivative of p(t, x) by induction.

16 Proposition 4.2 [17, Lemma 2.1] For any m ≥ 0, we have n=⌊ m X2 ⌋ d+2(m−n) t m−2n − m α Cn |x| min ,t ∂x p(t, x) = , |x|d+α+2(m−n) n=0

where ⌊ m 2 ⌋ means the largest integer that is less than

(4.6)

m 2.

α

Next, we claim that the kernel ∇ q0 p(t, s, x), q0 ≥ 2, satisfies the Assumption 2.3 with γ = α and κ = ∞. Lemma 4.1 Let β = qα0 . The following estimates hold. (i) For any t > λ > 0 and c > 0, Z t Z q0 |∇β p(t, r, x)|dx dr ≤ N [(t − λ)c−α ]q0 +1 + [(t − λ)c−α ] ; λ

|x|≥c

(ii) For any t > s > λ > 0, q0 Z λ Z β β dr ≤ N [(t − s)(t ∧ s − λ)−1 ]q0 ; |∇ p(t, r, x) − ∇ p(s, r, x)|dx 0

Rd

(iii) For any s > λ ≥ 0 and h ∈ Rd , q0 Z λ Z β β dr ≤ N ϕ(|h|(s − λ)−1/α ). |∇ p(s, r, x + h) − ∇ p(s, r, x)|dx 0

Rd

Proof. Note that β =

1

α q0

< 2. By using Proposition 4.2, we have if c > (t − r) α , Z t Z q0 β |∇ p(t, r, x)|dx dr |x|≥c

λ

Z t Z ≤ N

q0 t−r |x| dx dr d+α+2β |x| |x|≥c λ Z t Z ∞ q0 t−r |x|β · |x|d−1 d+α+2β d|x| dr ≤ N |x| λ c Z t = N c−α(q0 +1) (t − r)q0 dr β

λ

≤ N [(t − λ)c−α ]q0 +1 . 1

When c ≤ (t − r) α , we have (t − r)−1 ≤ c−α Z t Z q0 |∇β p(t, r, x)|dx dr |x|≥c

λ

≤ N

Z

t

Z

∞

1 (t−r) α

λ

+

Z

|x|β · |x|d−1

t−r d|x| |x|d+α+2β

1

(t−r) α

− d+2β α

|x|β · |x|d−1 (t − r)

c

≤ N

Z t Z λ

∞

|x|β · |x|d−1

c

+

Z

t−r d|x| |x|d+α+2β

d|x| dr

1

(t−r) α

− d+2β α

|x|β · |x|d−1 (t − r)

0

≤ N c−α(q0 +1)

Z

t

(t − r)q0 dr + N c−α

λ −α q0 +1

≤ N [(t − λ)c

]

Z

t

λ

+ [(t − λ)c−α ].

q0

q0

d|x| dr

dr

17 Hence we obtain the first estimate. 0 ⌋ = 0. Using the fact that ∂t p = ∆α/2 p, βq0 = 1 and Proposition 4.2, When α + qα0 < 2, ⌊ α+α/q 2 we get q0 Z λ Z β β dr |∇ p(t, r, x) − ∇ p(s, r, x)|dx 0 Rd q0 Z λ Z α+β q0 dr |∇ p(ξ − r, x)|dx ≤ (t − s) 0 Rd  Z λ Z (ξ−r) α1 d+2α+2β  ≤ N (t − s)q0 |x|α+β |x|d−1 (ξ − r)− α d|x| 0

0

+ q0

≤ N (t − s)

Z

λ

Z

∞

1 (ξ−r) α

|x|α+β |x|d−1

ξ −r

|x|d+3α+2β

! q0

d|x|

dr

(ξ − r)−q0 −1 dr

0

≤ N [(t − s)(t ∧ s − λ)−1 ]q0 , where ξ = θt + (1 − θ)s, θ ∈ [0, 1]. Since q0 ≥ 2 and 0 ≤ α ≤ 2, we have α + Z

λ Z

α q0

< 4. When 2 ≤ α +

< 4, we have

q0

α q0

β

α q0

|∇ p(t, r, x) − ∇ p(s, r, x)|dx dr q0 Z λ Z dr ≤ (t − s)q0 |∇α+β p(ξ − r, x)|dx 0 Rd  Z λ Z (ξ−r) α1 d+2α+2β  |x|α+β |x|d−1 (ξ − r)− α d|x| ≤ N (t − s)q0 0

Rd

0

0

Z

1

(ξ−r) α

0

+ + q0

≤ N (t − s)

Z

λ

Z

|x|α+β−2 |x|d−1 (ξ − r)−

∞ 1 (ξ−r) α

Z

|x|α+β |x|d−1

∞

α+β−2

1

|x|

(ξ−r) α

ξ−r d|x| |x|d+3α+2β

d−1

|x|

d+2α+2β−2 α

ξ−r |x|d+3α+2β−2

(ξ − r)−q0 −1 dr

0

≤ N [(t − s)(t ∧ s − λ)−1 ]q0 , where ξ = θt + (1 − θ)s, θ ∈ [0, 1]. Thus we obtain the second estimate.

d|x|

! q0

d|x|

dr

18 For the last estimate (iii), noting that 1 + β ≤ 2, we have for 1 + β < 2 Z

λ Z

β

q0

β

dr |∇ p(s, r, x + h) − ∇ p(s, r, x)|dx q 0 Z Z λ 1+β q0 dr |∇ p(s, r, x + θh)|dx h ≤ N d 0  R Z λ Z (s−r) α1 d+2+2β hq0  ≤ N |x|1+β · |x|d−1 (s − r)− α d|x| 0

Rd

0

0

+

Z

∞

1

|x|1+β · |x|d−1

(s−r) α

s−r

|x|d+α+2+2β

!q 0

d|x|

dr

≤ N [h(s − λ)−1 ]q0 ,

where θ ∈ [0, 1]. When 1 + β = 2, similar the case (ii), one can get the same estimate. The proof of Lemma is complete. It follows from the Proposition 4.1 that ∇β p(t, s, x) satisfies the Assumption 2.4. By using Theorem 2.1, we have the following result. η+α−α/p

Theorem 4.1 Let q0 ≥ 2. Suppose (4.4) with p ≥ q0 holds. Then for any g ∈ Hp Eq. (4.2) has a unique solution u in Hpη+α (η ∈ R), and for this solution kukHη+α (t) ≤ N (p, T )kgkHη+α−α/p (t,ℓ p p

(T, ℓ2 ),

2)

for every t ≤ T . Moreover, we have for q ∈ [2, q0 ] q/q0 [∇β u]BMO(T,q) ≤ N cˆ E[k|g|ℓ2 kqL0∞ (OT ) ] ,

where β = α/q0 and cˆ is defined as in (4.4).

When the Lévy noise is replaced by Brownian motion in (4.2), i.e., α

dut (x) = ∆ 2 ut (x)dt +

∞ X

hk (t, x)dWtk ,

u0 = 0, 0 ≤ t ≤ T,

(4.7)

k=1

where Wtk are independent one-dimensional Wiener processes. Denote h = (h1 , h2 , · · · ). α Similar to Lemma 4.1, one can prove ∇ 2 p(t, s, x) satisfies the Assumption 2.1. From Propoα sition 4.1, we know that Assumption 2.2 holds for ∇ 2 p(t, s, x). Thus we can get the following result. Theorem 4.2 Suppose that h ∈ Lp (T, ℓ2 ), there exists a uniqueness solution u in Hpη+α (η ∈ R), and for this solution kukHη+α (t) ≤ N (p, T )khkHη+α/2 (t,ℓ p p

2)

for every t ≤ T . Moreover, we have for any q ∈ (0, p] 1/p α . [∇ 2 u]BMO(T,q) ≤ N E[k|h|ℓ2 kpL∞ (OT ) ]

19 Remark 4.1 1. In Lemma 4.1, the second part (ii) is essential. From the proof of Theorem 2.1, the bound of the BMO norm can be controlled by the function ϕ and some norm of g, where the bound of the function ϕ depends on the choice of scale of time and space. In second part (ii), we must prove that the left hand side of (ii) can be controlled by the function of (t − s)(t ∧ s − λ)−1 . Only in this form, the left hand side of (ii) can be controlled by a constant. α 2. Particularly, taking q0 = 2, we have Lemma 4.1 holds for ∇ 2 p(t, s, x). Hence we have Theorem 4.2. Noting that if α = 2, Theorem 4.2 becomes [9, Theorem 3.4]. Thus we generalize the result of [9].

5

Discussion

In this section, we give another proof of Theorem 2.1 under some assumptions on g. Similarly, one can give another proof of [9, Theorem 2.4] under the same assumptions on g. Firstly, let us recall the proofs of Theorem 2.1 and [9, Theorem 2.4]. The reason why we divide the interval (0, s) into 3a−b two parts (0, 3a−b 2 ) and ( 2 , s) in proof of Lemma 3.3 is the singularity of K at time t. In order to see it clearly, we look at the Section 4 and recall that for any t > λ > 0 and c > 0, Z t Z q0 β |∇ p(t, r, x)|dx dr ≤ N [(t − λ)c−α ]q0 +1 + [(t − λ)c−α ] . λ

|x|≥c

Note that if we choose c = 0, then the above integral will be infinity. Indeed, direct calculations show that Z t Z t Z q0 β (t − r)−1 dr = ∞. |∇ p(t, r, x)|dx dr ≈ N Rd

λ

λ

Obviously, the singularity of ∇β p appears at t. But p ∈ L1 (Rd ), thus a natural question appears: when the singularity of p does not appear at t, is there another proof ? Moreover, it is easy to see that the derivative of p deduces the singularity of ∇β p at t. In this section, we first give a similar theorem to Theorem 2.1 under different assumptions. Then as an application, we use the method of integration by part to deal with the derivative of p and obtain the BMO estimate by direct calculation. Theorem 5.1 Assume that the kernel function is a deterministic function and satisfies that for all t ≥ r ≥ 0, Z tZ |K(t, r, x)|dxdr ≤ N (T ). 0

Rd

Assume further that there exists a positive constant q0 > 2 such that E

Z

Z

q0

kg(·, ·, z)k̟ L∞ (OT ) ν(dz)

2

< ∞,

̟ = 2 or q0 .

Then for any q ∈ (0, q0 ], one has [Gg]BMO(T,q) ≤ N E +E

Z

Z Z Z

where N = N (N0 , d, q, q0 , T ).

q

kg(·, ·, z)k2L∞ (OT ) ν(dz)


2

,

20 Proof. It suffices to prove that for each Q = Qc (t0 , x0 ) := (t0 − cγ , t0 + cγ ) × Bc (x0 ) ⊂ OT , c > 0, t0 > 0, we have

Z Z 1 |Gg(t, x) − Gg(s, y)|q dtdxdsdy E Q2 Q Q q Z Z 2 q 2 kg(·, ·, z)kL∞ (OT ) ν(dz) kg(·, ·, z)kL∞ (OT ) ν(dz) , ≤ NE +E Z

(5.1)

Z

where N = N (T, q, ϕ). Since the operator G is translation invariant with respect to x, we may assume that x0 = 0. Kunita’s first inequality implies that q/2 Z t Z Z 2 q k(t − r, y)g(r, x − y, z)dy| ν(dz)dr | E|Gg(t, x)| ≤ E Rd 0 Z Z t Z Z q k(t − r, y)g(r, x − y, z)dy| ν(dz)dr | +E ≤ E

Z

+E

0

Z

Z

Rd

kg(·, ·, z)k2L∞ (OT ) ν(dz)

Z

Z

×


Z

Z

t

|

0

×

Z

2

k(t − r, y)dy| dr

Rd

t

|

0

q

kg(·, ·, z)k2L∞ (OT ) ν(dz)

≤ N (T )E Z Z q kg(·, ·, z)kL∞ (OT ) ν(dz) +E

Z

Z

q

2q

k(t − r, y)dy| dr Rd

2

Z

< ∞. Thus we have

Z Z 1 |Gg(t, x) − Gg(s, y)|q dtdxdsdy E Q2 Q Q Z 2 ≤ |Gg(t, x)|q dtdx E Q Q q Z 2 2 kg(·, ·, z)kL∞ (OT ) ν(dz) ≤ N (T )E Z Z q kg(·, ·, z)kL∞ (OT ) ν(dz) , +E Z

which implies (5.1) holds. The proof of Theorem 5.1 is complete. As an application, for simplicity, we consider the following stochastic evolution equation Z g(t, x, z)N˜ (dt, dz) u(0, x) = 0. (5.2) du = ∆udt + Z

It is easy to check that the solution of (5.2) is Z tZ Z K(t − r, y)g(r, y, z)dydN˜ (dr, dz). u(t, x) = 0

Z

Rd

It follows the properties of heat kernel that Z |K(t, r, x)|dx = 1 for all t > r > 0. Rd

Applying Theorem 5.1, we have

21 Theorem 5.2 Assume that there exists a positive constant q0 > 2 such that E

Z

Z

q0 2

kg(·, ·, z)k̟ L∞ (OT ) ν(dz)

< ∞,

̟ = 2 or q0 .

Then for any q ∈ (0, q0 ], one has [u]BMO(T,q) ≤ N E +E

Z

Z Z Z

q

kg(·, ·, z)k2L∞ (OT ) ν(dz)

2

kg(·, ·, z)kqL∞ (OT ) ν(dz) ,

where N = N (N0 , d, q, q0 , T ). Moreover, if we further assume that E

Z

Z

q0

k∇x g(·, ·, z)k̟ L∞ (OT ) ν(dz)

2

< ∞,

̟ = 2 or q0 .

Then for any q ∈ (0, q0 ], one has [∇u]BMO(T,q) ≤ N E +E

Z

Z Z Z

q

k∇x g(·, ·, z)k2L∞ (OT ) ν(dz)

k∇x g(·, ·, z)kqL∞ (OT ) ν(dz)

2

,

where N = N (N0 , d, q, q0 , T ) and ∇x g = ∇x g(t, ·, z). Proof. Denote u(t, x) = Gg(t, x). Noting that Z tZ Z ˜ (dr, dz). k(t − r, y)∇x g(r, x − y, z)dy N ∇x Gg(t, x) = 0

Z

Rd

Then similar to the proof of Theorem 5.1, one can get the desired result. Remark 5.1 Comparing with the proofs of Theorems 2.1 and 5.1, we find if we assume the function g has high regularity, then the proof of BMO estimate will be very simple. The proof of Theorem 4.1 will be also simple if we improve the regularity of g. If g ≡ 0, then u ≡ 0. That is to say, the noise has effect on the regularity of the solutions. Acknowledgment The first author was supported in part by NSFC of China grants 11301146, 11171064.

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