On comparison principle and strict positivity of solutions to the nonlinear stochastic fractional heat equations

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1 On comparison principle and strict positivity of solutions to the nonlinear stochastic fractional heat equations Le Chen and Kunwoo Kim University of Utah Abstract: In this paper, we prove a sample-path comparison principle for the nonlinear stochastic fractional heat equation on with measure-valued initial data. We give quantitative estimates about how close to zero the solution can be. These results extend Mueller s comparison principle on the stochastic heat equation to allow more general initial data such as the Dirac delta measure and measures with heavier tails than linear exponential growth at ±. These results generalize a recent work by Moreno Flores [25], who proves the strict positivity of the solution to the stochastic heat equation with the delta initial data. As one application, we establish the full intermittency for the equation. As an intermediate step, we prove the Hölder regularity of the solution starting from measure-valued initial data, which generalizes, in some sense, a recent work by Chen and Dalang [6]. MSC 2 subject classifications: Primary 6H5. Secondary 6G6, 356. Keywords: nonlinear stochastic fractional heat equation, parabolic Anderson model, comparison principle, measure-valued initial data, stable processes. Introduction The comparison principle for differential equations tells us whether two solutions starting from two distinct initial conditions can compare with each other when the initial conditions are comparable. The sample-path comparison principle for stochastic differential equations SDEs and also for stochastic partial differential equations SPDEs have been studied extensively; see e.g. [7, Chapter VI] and [3, Chapter V. 4] for SDEs, and [2, 2, 24, 26, 3] for SPDEs. A related problem is the stochastic comparison principle, which is of the form: E Φu t E Φv t, for all t >, esearch supported by a fellowship from Swiss National Science Foundation.

2 where {u t x} and {v t x} solve SDEs or SPDEs, with the same initial data but comparable drift and diffusion coefficients. One looks for as large a class of functions Φ as possible. See [2, 6, 8, 9]. In this paper, we will focus on the pathwise comparison principle for the following nonlinear stochastic fractional heat equation: t xdδ a ut, x = ρ ut, x Ẇ t, x, t + := ], + [, x. u, = µ where a ], 2] is the order of the fractional differential operator x Dδ a and δ δ 2 a is its skewness, Ẇ is the space-time white noise on +, µ denotes the initial data a measure, and the function ρ : is Lipschitz continuous. Throughout this paper, we assume that a and δ are fixed constants such that a ], 2] and δ 2 a,.2 unless we state otherwise see Corollary.2. When a = 2 and δ =, the fractional operator x Dδ a reduces to the Laplacian on, which is the infinitesimal operator for a Brownian motion. On the other hand, when a ], 2[ and δ 2 a, the operator x Dδ a is the infinitesimal generator of an a-stable process with skewness δ. In particular, x D a = a/2. This fractional Laplace operator has been paid many attentions for several decades because of its non-local property, and thus it is widely used in many areas such as physics, biology, and finance to model non-local anomalous diffusions. We refer to [23, 32, 34] for more details on these fractional operator and the related stable random variables. The existence and uniqueness of a random field solution to. have been studied in [7, 9,, 3, 4, 5]. In particular, the existence, uniqueness, and moment estimates under measure-valued initial data have been established recently in [7, 9, ]. We now specify the weak and strong comparison principles. Let u t, x and u 2 t, x be two solutions to. with initial measures µ and µ 2, respectively. We say that. satisfies the weak comparison principle if u t, x u 2 t, x for all t > and x, a.s., whenever µ µ 2 i.e., µ 2 µ is a nonnegative measure. And the equation. is said to satisfy the strong comparison principle if u t, x < u 2 t, x for all t > and x, a.s., whenever µ < µ 2 i.e., the measure µ 2 µ is nonnegative and nonvanishing. Note that the stochastic comparison principle for. with Φz = z k for k 2 so-called the moment comparison principle has been shown lately by Joseph, Khoshnevisan and Mueller [9]. When a = 2, the equation. reduces to the stochastic heat equation SHE. The special case when ρu = λu for some constant λ is called the parabolic Anderson model; see [4, 5, 5]. The weak comparison principle can be derived readily from the Feynman- Kac formula; see [4]. But the proof of the strong comparison principle requires some more efforts. This question is important because, e.g., the Hopf-Cole solution to the famous Kardar-Parisi-Zhang equation KPZ [2] is the logarithm of the solution to the SHE. 2

3 For general ρ which is Lipschitz continuous, we do not have the Feynman-Kac formula. The weak comparison principle is no longer obvious. In this case, Mueller [26] proves the strong comparison principle for the SHE on for the initial data being absolutely continuous with respect to the Lebesgue measure with a bounded density function. Mueller uses the discrete Laplacian and discretizes time to approximate the solution to the SHE, which results in the weak comparison principle. He then obtains the strong comparison principle by employing some large deviation estimates for the stochastic integral part of the solution. Using Mueller s large deviation estimates, Shiga [3] gives another proof of the strong comparison principle for the initial data being a so-called C tem function, that is, a continuous function with both tails growing no faster than e λ x for all λ >. He outlines a different approach for proving the weak comparison principle in the appendix of his paper: smooth both the Laplace operator and the white noise so that one can apply the comparison principle for SDEs. We will follow his approach in our proof for the weak comparison principle. In both Mueller [26] and Shiga [3], the initial data should be functions. One natural question is whether the solution remains strictly positive if we run the system. starting from a measure, such as the Dirac delta measure. Using the polymer model and following a convergence result by Alberts, Khanin and Questel [], Moreno Flores [25] recently proved the strict positivity result for the Anderson model i.e., the case where a = 2 and ρu = λu with the delta initial data. Our results below generalize their result to the stochastic fractional heat equation i.e., a ], 2], and moreover, we consider general measure-valued initial data and allow ρ to be any Lipschitz continuous function. ecently, Conus, Joseph and Khoshnevisan [] give a more precise estimate on the strong comparison principle for the SHE. When the initial data is the Lebesgue measure, they prove that for every t >, there exist two finite constants A > and B > such that for all ɛ ], [ and x, P ut, x < ɛ A exp B [ logɛ log logɛ ] 3/2..3 Clearly, this result implies the strong comparison principle. In [28], Mueller and Nualart prove that when a = 2 and the space domain is [, ] with the zero Dirichlet boundary condition, for some constants C and C, P ut, x < ɛ C exp C log ɛ 3/2 ɛ..4 We will generalize these results to the stochastic fractional heat equation. following []. This shows how close to zero the solution to. can be. In order to state our results, we need some notation. Let M be the set of signed regular Borel measures on. From the Jordan decomposition, µ = µ + µ such that µ ± are two non-negative Borel measures with disjoint support and denote µ = µ + + µ. As proved in [9], the admissible initial data for. is { M a := µ M : sup y µ dx + y x 3 } < + +a, for a ], 2].

4 Moreover, when a = 2, the admissible initial data can be more general than M 2 : It can be any measures from the following set { } M H := µ M : e cx2 µ dx < +, for all c > ; see [7]. Clearly, M a M H. In the following, a + sign in the subscript means the subset of nonnegative measures. An important example in M a,+ is the Dirac delta measure. For simplicity, denote { M M a if < a < 2, a := M H if a = 2. We will follow Shiga s arguments [3] to prove the following weak comparison principle. This result allows more general initial conditions than those in [26] and [3]. Theorem. Weak comparison principle. Let u t, x and u 2 t, x be two solutions to. with the initial data µ and µ 2 M a, respectively. If µ µ 2, then P u t, x u 2 t, x, for all t and x =..5 Here is one example. Let δ z be the Dirac delta function with unit mass at x = z. Suppose that µ = δ and µ 2 = 2δ. Then u t, x u 2 t, x for all t > and x, a.s. As a direct consequence of Theorem., one can turn weak intermittency statements in [9, 5] into the full intermittency. More precisely, define the upper and lower Lyapunov exponents of order p by m p x := lim sup t + t log E ut, x p, m p x := lim inf t + t log E ut, x p,.6 for all p 2 and x. According to Carmona and Molchanov [5, Definition III.., on p. 55], u is fully intermittent if inf x m 2 x > and m x for all x. Corollary.2. Suppose that a ], 2[, δ < 2 a strict inequality, µ M a,+, and ρ satisfies that for some constants l ρ > and ς, ρx 2 l 2 ρ ς 2 +x 2 for all x. If either µ or ς, then the solution to. is fully intermittent. Proof. By [9, Theorem 3.4], inf x m 2 x >. By Theorem., ut, x a.s. and so, E[ ut, x ] = E[ut, x] = J t, x see 2.3. Therefore, m x for all x. We adapt both Mueller and Shiga s arguments see [26, 3] to prove the strong comparison principle. In the proof, following the idea of [, Theorem 5.], we develop a large deviation result similar to [26] using the Kolmogorov continuity theorem. 4

5 Theorem.3 Strong comparison principle. Let u t, x and u 2 t, x be two solutions to. with the initial data µ and µ 2 M a, respectively. If µ < µ 2, then P u t, x < u 2 t, x for all t > and x =. The following theorem gives more precise information on the positivity of the solutions. Let supp f denote the support of function f, i.e., supp f := {x : fx }. Theorem.4 Strict positivity. Suppose ρ = and let ut, x be the solution to. with the initial data µ M a. Then we have the following two statements: If µ >, then for any compact set K +, there exist finite constants A > and B > which only depend on K such that for small enough ɛ >, P inf ut, x < ɛ t,x K A exp B logɛ /a log logɛ 2 /a..7 2 If µdx = fxdx with f C, fx for all x and supp f, then for any compact set D supp f and any T >, there exist finite constants A > and B > which only depend on D and T such that for all small enough ɛ >, P inf ut, x < ɛ t,x ],T ] D A exp B { logɛ log logɛ } 2 /a..8 Theorem.4 shows that for all t >, the function x ut, x does not have a compact support see [26] and [27, Section 6.3] for some other scenarios where the compact support property can be preserved. We also note that thanks to Theorem.4, one can regard the solution ut, x to. as the density at location x of a continuous particle system at time t, where particles move as independent a-stable processes but branch independently according to the noise term; see [9]. Furthermore, Theorem.4 implies that for all compact sets K + and all p >, the negative moments exists: E [ inf t,x K ut, x p] <. Note that part 2 of Theorem.4 gives essentially the same rate as those in.3 and.4 when a = 2. The next three theorems will be used in the proofs of the above theorems. Since they are interesting by themselves, we list them below. The first one, which is used in the proof of Theorem., says that we can approximate a solution to. starting from µ M a by a solution to. starting from smooth initial conditions. Define if x /ɛ, ψ ɛ x = + /ɛ x if /ɛ x + /ɛ,.9 if x + /ɛ. 5

6 Theorem.5. Suppose that µ M a. Let ut, x and u ɛ t, x be the solutions to. starting from µ and µ ψ ɛ δ G a ɛ, x, respectively. Then lim E [ ut, x u ɛ t, x 2] =, for all t > and x. ɛ The following theorem, which is used in the proof of Theorem.3, shows the samplepath regularity for the solutions to.. When the initial data has a bounded density, this has been proved in [4]. For general initial data, the case where a = 2 is proved in [6]. The theorem below covers the cases where < a < 2. We need some notation: Given a subset K + and positive constants β, β 2, denote by C β,β 2 K the set of functions v : + with the property that for each compact set D K, there is a finite constant C such that for all t, x and s, y in D, vt, x vs, y C [ t s β + x y β 2 ]. Denote C β,β 2 D := α ],β [ α2 ],β 2 [ C α,α 2 D. Theorem.6. Let ut, x be the solution to. starting from µ M a. Then we have u C a a, 2a 2 +, a.s. The last one, which is also used in the proof of Theorem.3, shows that the solution ut, x to. converges to the initial measure µ in the weak sense as t. The case when a = 2 is proved in [6, Proposition 3.4]. Let C c be the set of continuous functions with compact support and f, g be the inner product in L 2. Theorem.7. Let ut, x be the solution to. starting from µ M a. Then, lim ut,, φ L2 Ω = µ, φ for all φ C c. t In the following, we first list some notation and preliminary results in Section 2. Then we prove Theorem. in Section 3 with many technical lemmas proved in the Appendix. The proof of Theorem.3 is presented in Section 4, Theorem.4 is proved in Section 5. Finally, the three Theorems.5,.6 and.7 are proved in Sections 6, 7, and 8, respectively. 2 Notation and some preliminaries The Green function associated to the problem. is δg a t, x := F [exp { δ ψ a t}] x = dξ exp { iξx t ξ a e iδπ sgnξ/2}, 2. 2π 6

7 where F is the inverse Fourier transform and δψ a ξ = ξ a e iδπ sgnξ/2. Denote the solution to the homogeneous equation t xdδ a ut, x =, t +, x, u, = µ, 2.2 by J t, x := δ G a t, µ x = µdy δ G a t, x y, 2.3 where denotes the convolution in the space variable. Following notation in [7], let W = {W t A, A B b, t } be a space-time white noise defined on a probability space Ω, F, P, where B b is the collection of Borel sets with finite Lebesgue measure. Let F t, t be the natural filtration generated by W and augmented by the σ-field N generated by all P -null sets in F: F t = σ W s A : s t, A B b N, t. Define F t := Ft+ = s>t Fs for t. In the following, we fix this filtered probability space {Ω, F, {F t : t }, P }. We use p to denote the L p Ω-norm p. The rigorous meaning of the SPDE. is the integral mild form ut, x = J t, x + It, x, where It, x = δg a t s, x y ρ us, y W ds, dy, [,t] 2.4 where the stochastic integral is the Walsh integral [33]. Definition 2.. A process u = ut, x, t, x + is called a random field solution to. if: u is adapted, i.e., for all t, x +, ut, x is F t -measurable; 2 u is jointly measurable with respect to B + F; 3 δg 2 a ρu 2 2 t, x < + for all t, x +, where denotes the simultaneous convolution in both space and time variables. Moreover, the function t, x It, x mapping + into L 2 Ω is continuous; 4 u satisfies 2.4 a.s., for all t, x +. 7

8 Throughout the paper, we assume that the function ρ : is Lipschitz continuous with Lipschitz constant Lip ρ >, and moreover, for some constants L ρ > and ς, ρx 2 L 2 ρ ς 2 +x 2, for all x. 2.5 Note that the above growth condition 2.5 is a consequence of ρ being Lipschitz continuous. Let a be the dual of a, i.e., /a + /a =. The following constant is finite: Λ = δ Λ a := sup x δg a, x, 2.6 and in particular, Λ a = π Γ + /a; see [9, 3.]. In the following, we often omit the dependence of this constant on δ and a and simply write δ Λ a as Λ. This rule will also apply to other constants. For all t, x +, n N and λ, define L t, x; λ := λ 2 δg 2 at, x L n t, x; λ := L L, }{{} for n, 2.7 n + factors L, ;λ K t, x; λ := L n t, x; λ. 2.8 n= We apply the following conventions to Kt, x; λ: Kt, x := Kt, x; λ, Kt, x := K t, x; L ρ, Kp t, x := K t, x; 4 p L ρ, for p 2. The following theorem is from [9, Theorem 3.] for < a < 2 and [7, Theorem 2.4] for a = 2. Theorem 2.2 Existence,uniqueness and moments. Suppose that µ M a, and ρ is Lipschitz continuous and satisfies 2.5. Then the SPDE. has a unique in the sense of versions random field solution {ut, x: t, x + } starting from µ. Moreover, for all even integers p 2, all t > and x, { J 2 ut, x 2 p t, x + [ς 2 +J 2 ] K t, x, if p = 2, 2J 2 t, x + [ς 2 +2J 2 ] K p t, x, if p > In order to use the moment bounds in 2.9, we need some estimates on Kt, x. ecall that if the partial differential operator is the heat operator ν where ν >, then t 2 4ν Φ K heat t, x; λ = G ν t, x 2 λ 2 4πνt + λ4 2ν e λ 4 t t λ 2 2ν, 8

9 where Φx is the distribution function of the standard normal random variable and G ν t, x = 2πνt exp x 2 /2νt; see [7]. If the partial differential operator is the wave operator 2 t 2 κ 2 where κ >, then K wave t, x; λ = λ2 4 I λ2 κt 2 x 2 { x κt}, 2κ where I x is the modified Bessel function of the first kind of order ; see [8]. Except these two cases, there are no explicit formulas for Kt, x. The following upper bound on Kt, x from [9, Proposition 3.2] will be useful in this paper. Proposition 2.3. Let γ := λ 2 Λ Γ /a. For some finite constant C = Cλ >, Kt, x; λ C t /a δ G a t, x + t /a exp γ a t, for all t and x Proof of Theorem. Before proving Theorem., we need some preparation. operator, denoted by δ G a t for clarity, as follows: One may view δ G a t, x as an δg a tfx := δ G a t, f x. Let I be the identity operator: Ifx = δ fx = fx. Set Let ɛ δg a t = expt δd ɛ a = e t/ɛ ɛ δd a = δ G a ɛ I ɛ t/ɛ n n= n!. δg a nɛ = e t/ɛ I + ɛ δ a t, 3. where the operator ɛ δ at has a density, denoted by ɛ δ at, x, which is equal to ɛ δ a t, x = e t/ɛ t/ɛ n n= n! δg a nɛ, x. 3.2 One may also write the kernel of ɛ δ G a t as ɛ δg a t, x = e t/ɛ δ x + δ ɛ a t, x. 3.3 The two operators ɛ δ G a and δ G a are close in many senses; see Appendix for more details. 9

10 Proof of Theorem.. Denote φ ɛ x = 2πɛ /2 exp x 2 /2ɛ. For ɛ > and x, denote t Wxt ɛ := φ ɛ x yw ds, dy, for t. Clearly, t Wxt ɛ is a one-dimensional Brownian motion. Denote Ẇ xt ɛ = d W ɛ dt xt. Then the quadratic variation of dwxt ɛ is d Wxt ɛ = φ 2 ɛx ydydt = dt πɛ Consider the following stochastic partial differential equation t u ɛt, x = δd ɛ a u ɛ t, x + ρu ɛ t, xẇ xt, ɛ t >, x, u ɛ, x = µ δ G a ɛ, x, x. Since ρ is globally Lipschitz continuous, 3.5 has a unique strong solution u ɛ t, x = µ δ G a ɛ, x + t ds ɛ δd a u ɛ s, x + t ρu ɛ s, xdw ɛ xs. 3.5 Step. Let u ɛ,i t, x be the solutions to 3.5 with initial data µ i, i =, 2, respectively. Denote v ɛ t, x := u ɛ,2 t, x u ɛ, t, x. We will prove that P v ɛ t, x, for every t > and x =. 3.6 Let a n = 2n 2 + n + 2, n. Then a n as n and a n a n x 2 dx = n. Let ψ n x, n =, 2,..., be nonnegative continuous functions supported on ]a n, a n [ such that ψ n x 2 nx 2 and an a n ψ n xdx =. Define x y Ψ n x := dy ψ n zdz. Clearly, Ψ n x C 2 with Ψ nx = ψ n x, Ψ n x = for x, and Ψ nx = x ψ nzdz for all x. Let I denote the indicator function. Here are three important properties: For all x, as n +, Ψ n x x =: Ψx, Ψ nx Ix < and Ψ nx x Ψx, 3.7 Because for each x fixed, u ɛ t, x is a semi-martingale, by Itô s formula, Ψ n v ɛ t, x = t Ψ nv ɛ s, x [ρu ɛ,2 s, x ρu ɛ, s, x] dw ɛ xs

11 + 2 + t t By the Lipschitz condition on ρ, Ψ nv ɛ s, x [ρu ɛ,2 s, x ρu ɛ, s, x] 2 4πɛ ds Ψ nv ɛ s, x ɛ δd a v ɛ s, xds. Ψ nv ɛ s, x [ρu ɛ,2 s, x ρu ɛ, s, x] 2 Lip 2 ρ Ψ nv ɛ s, xv 2 ɛ s, x 2 Lip 2 ρ /n. Hence, [ E [Ψ n v ɛ t, x] Lip2 ρ t n 4πɛ + E ɛ t ] ds Ψ nv ɛ s, x dy δ G a ɛ, x y [v ɛ s, y v ɛ s, x]. Now let n go to +, by 3.7 and the monotone convergence theorem, Notice that ɛ E [Ψv ɛ t, x] ɛ t ds t ɛ E [Iv ɛ s, x < v ɛ s, x] ds ds dy δ G a ɛ, x ye [Iv ɛ s, x < v ɛ s, y]. t dy δ G a ɛ, x ye [Iv ɛ s, x < v ɛ s, y] t ds dy δ G a ɛ, x ye [Iv ɛ s, x <, v ɛ s, y < v ɛ s, y] ɛ = t ds dy δ G a ɛ, x ye [Iv ɛ s, x <, v ɛ s, y < v ɛ s, y ] ɛ t ds dy δ G a ɛ, x ye [Iv ɛ s, y < v ɛ s, y ] ɛ Then using the fact that x Ix < = Ψx, we have that E [Ψv ɛ t, x] t ds dy δ G a ɛ, x ye [Ψv ɛ s, y]. ɛ Therefore, by Gronwall s lemma applied to sup y E [Ψv ɛ s, y], one can conclude that E [Ψv ɛ s, y] = for every t > and x. This proves 3.6. Step 2. In this step, we assume that µdx = fxdx with f L and fx for all x. Denote f ɛ x := µ δ G a ɛ, x. We will prove that lim sup ɛ x u ɛ t, x ut, x 2 2 =, for all t >, 3.8

12 where u ɛ is a solution to 3.5 with u ɛ, x = f ɛ x and u is a solution to.. Fix T >. Notice that u ɛ t, x can be written in the following mild form using the kernel ɛ of δ G a t in 3.3: u ɛ t, x = f ɛ δg ɛ a t, x + e t s/ɛ ρ u ɛ s, x dwxs ɛ t ɛ + δ a t s, x yρ u ɛ s, y dwysdy, ɛ where the last term equals to t dz δ ɛ a t s, x zρu ɛ s, zφ ɛ y z W ds, dy. By 8.2 below, the boundedness of the initial data implies that for all t >, A t := sup sup ɛ ],] s [,t] By the linear growth condition 2.5, t sup u ɛ s, x 2 2 us, x 2 2 < x u ɛ t, x ut, x f ɛ δg ɛ a t, f δ G a t, 2 x t + 6 L 2 ρ ds e 2t s/ɛ ς 2 + u ɛ s, x 2 2 4πɛ t [ + 6 ds dy E dz δ ɛ a t s, x z [ρu ɛ s, z ρus, z] φ ɛ y z =: 6 t t t ds ds ds 6 I n t, x; ɛ. n= [ dy E [ dy E [ dy E dz δ ɛ a t s, x z [ρus, z ρus, y] φ ɛ y z dz [ δ ɛ a t s, x z δ G a t s, x z] ρus, yφ ɛ y z dz [ δ G a t s, x z δ G a t s, x y] ρus, yφ ɛ y z Denote C f := sup x fx sup x f ɛ x. Using the semigroup property, we see that I t, x; ɛ 2 2 ] 2 ] 2 ] 2 ]

13 [f ɛ δg ɛ a t, δ G a t, x + f δ G a t + ɛ, δ G a t, x] [f ɛ δg ɛ a t, x + f δ G a t, x] 2 Cf 2 e t/ɛ + 2 C 2 f dy ɛ δ a t, y δ G a t, y + 2e t/ɛ + C + C ɛ/t /2, dy δ G a t + ɛ, y δ G a t, y where the last step is due to Lemma 8.2, 8.7 and the fact that log + x x for all x, and C and C are the constants defined in 8.5 and 8.8. For simplicity, define C := C + C. As for I 2 t, x; ɛ, ɛ I 2 t, x; ɛ L 2 ρ ς 2 +A t e 2t/ɛ, 4π which implies lim sup ɛ t T sup I 2 t, x; ɛ =. x By the Hölder inequality and the Lipschitz continuity of the function ρ, t I 3 t, x; ɛ Lip 2 ρ ds dy dz δ ɛ at 2 s, x z u ɛ s, z us, z 2 2 φ ɛy z t Lip 2 ρ ds dy δ ɛ as, 2 x y u ɛ t s, y ut s, y 2 2, 3. and similarly, I 4 t, x; ɛ Lip 2 ρ t ds dy dz δ ɛ as, 2 z ut s, x z ut s, x y 2 2 φ ɛy z. By Lemma 8.4, for some constant C := CT, a, δ, µ, ut s, x z ut s, x y 2 2 Ct s /a y z + CA T y z a. Hence, integrating over dy first and then integrating over dz using 8.2 give that I 4 t, x; ɛ Lip 2 ρ C t Lip2 ρ C C a,δ π ds t dz δ ɛ as, 2 z [t s /a 2ɛ/π + A ] T 2 a /2 Γ a/2 ɛ a /2 π ds s /a [ t s /a 2ɛ + A T 2 a /2 Γ a/2 ɛ a /2 ], where C a,δ is defined in Lemma 8.3. Finally, integrating over ds using the Beta integral, we have that for some finite constant C := C T, a, δ, µ, A T >, I 4 t, x; ɛ C Lip 2 ρ t 2/a ɛ /2 + ɛ a /2. 3

14 By Hölder inequality, 2.5 and 3.9, I 5 t, x; ɛ L 2 ρ ς 2 +A t t ds dy dz φ ɛ y z [ δ ɛ a t s, x z δ G a t s, x z] 2. Integrate dy and enlarge the integral interval for ds from [, t] to [, T ], I 5 t, x; ɛ L 2 ρ and then apply 8. to obtain ς 2 +A T T lim sup ɛ t T Similarly to the case of I 5, we have that ds T I 6 t, x; ɛ L 2 ρ ς 2 +A T T =2 L 2 ρ ς 2 +A T ds dy [ dy ds dz [ δ ɛ a s, z δ G a s, z] 2, sup I 5 t, x; ɛ =. x dz φ ɛ y z [ δ G a s, x z δ G a s, x y] 2 δg 2 as, y δ G a s, y dz δ G a s, zφ ɛ y z Define F ɛ s, y := δ G 2 as, y δ G a s, y dz δg a s, zφ ɛ y z. Clearly, lim ɛ F ɛ s, y = for all s > and y. On the other hand, F ɛ s, y δ G 2 as, y + δ G a s, ys /a Λ, where the constant Λ is defined in 2.6. In fact, this upper bound is integrable: T ds dy δg 2 as, y + δ G a s, ys /a Λ T = ds [ δg a 2s, + Λs /a] = Hence, the dominated convergence theorem implies that lim sup ɛ t T sup I 6 t, x; ɛ =. x δg a, a + Λ 2 /a a T /a. Now set Mt; ɛ := sup y u ɛ t, y ut, y 2 2. Fix T >. Combining things together, we can get that for some constant C T >, t Mt; ɛ C T ds t s /a Ms; ɛ + HT ; ɛ + Ĥt; ɛ, ]. 4

15 where HT ; ɛ := 6 Ĥt; ɛ := 2 C 2 f sup n=2,5,6 t T sup I n t, x; ɛ, x 2e t/ɛ + C ɛ/t /2 + 6 C Lip 2 ρ t 2/a ɛ /2 + ɛ a /2. Then by Chandirov s lemma, which is a variation of Bellman s inequality see [3, Theorem.4, on p. 5], for < t T, t t Mt; ɛ HT ; ɛ + Ĥt; ɛ + ds HT ; ɛ + Ĥs; ɛ t s /a exp dτ t τ /a s t a = HT ; ɛ + Ĥt; ɛ + ds HT ; ɛ + Ĥs; ɛ t s /a exp t s /a a HT ; ɛ + a a T a T /a /a exp a t a t s /a + Ĥt; ɛ + ds Ĥs; ɛt s /a exp. a Clearly, as ɛ, the first two terms in the above upper bound go to zero. The integral also goes to zero by applying the dominated convergence theorem. This proves 3.8. Finally, suppose that µ i dx = f i xdx with f i L, i =, 2. If f x f 2 x for almost all x, then by Step we know that v ɛ t, x := u ɛ,2 t, x u ɛ, t, x for all t > and x, a.s. Then Step 2 implies v ɛ t, x converges to vt, x = u 2 t, x u t, x in L 2 Ω for all t > and x. Therefore, the nonnegativity of vt, x is inherited from that of v ɛ t, x, that is, P u t, x u 2 t, x, for all t > and x =. Step 3. Now we assume that µ i M a. ecall the definition of ψ ɛ in.9. Fix ɛ >. Let u ɛ,i, i =, 2, be the solutions to. starting from [µ i ψ ɛ ] δ G a ɛ, x. Denote vt, x = u 2 t, x u t, x and v ɛ t, x = u ɛ,2 t, x u ɛ, t, x. Because ψ ɛ is a continuous function with compact support on, the initial data for u ɛ,i t, x is bounded: sup [µ i ψ ɛ ] δ G a ɛ, x Λ ψ x ɛ /a ɛ y µ i dy < +, where Λ is defined in 2.6. Hence, by Step 2, we have that P v ɛ t, x, for all t > and x =, for all ɛ >. Applying Theorem.5, we obtain This completes the proof of Theorem.. P vt, x, for all t > and x =. 5

16 4 Proof of Theorem.3 We need several lemmas. Lemma 4. below plays a role to initialize the induction procedure. Lemma 4.. Let d >. For all t > and M >, there exist some constants < m = m t, M < and < γ /4 such that for all m m, all s [t/2m, t/m] and x, δg a s, ] d,d[ x γ ] d M/m,d+M/m[ x. 4. Proof. Let Z be a random variable with the stable density δ G a, x. Define γ := min{p Z, P Z }/2. Clearly, < γ /4. We first consider the case where d M/m x. Because t/2 ms t, we have δg a s, ] d,d[ x = d d δg a s, x ydy = x+d s /a x d s /a δg a, zdz P d 2m /a t /a Z M m a/a t /a. Similarly, when x d + M/m, we have δg a s, ] d,d[ x P M t /a m a/a Z d2m /a t /a. Therefore, when m is large enough, the above probabilities are bigger than γ. This completes the proof of Lemma 4.. Lemma 4.2. If µ M a and ρ satisfies 2.5, then for all p 2, there exists some finite constant C := Ca, δ, L ρ, ς, µ, p > such that, sup ut, x 2 p Ct 2+/a t [ 2/a + t /a + t exp γ a t ], x for all t >, where γ := 8p L ρ ΛΓ/a. 2 If µdx = c dx, c and ρ =, then for some constant Q := Qc, a, Lip ρ, Λ >, sup E ut, x p Q p exp Qp 2a a t, for all p 2 and t. x Proof. Part is from [9, Lemma 4.9 and 4.2]. As for part 2, notice that J t, x c. Then by 2.9 and 2., for p 2 and p N, ut, x 2 p 2c2 + C c 2 t a 2c 2 + C c 2 a t a a ds s /a + exp γ a p s + γ a p exp γ a p t, 6

17 where γ p = 6 p L 2 ρ ΛΓ/a and the constant C = CLip ρ is defined in Proposition 2.3. Notice that logx βx for all x whenever β e. So by choosing θ = a ae γ a 2, we have that exp θγp a t t a a for all t >. Hence, if c, then [ ] a ut, x 2 p 2c 2 + C c 2 a + γ a 2 exp θγp a t. Then, raise both sides by a power of p/2. This completes the proof of Lemma 4.2. The following lemma proves the inductive step. Lemma 4.3. Let d >, t > and M >. If ρ = and µdx = [ d,d] xdx, then there are some finite constants Q := Qβ, Lip ρ, Λ, t >, < β /8, and m > such that for all m m, t P us, x β ] d M/m,d+M/m[ x for all 2m s t m and x exp Q m /a [logm] 2 /a. Proof. Define S := S t,m,d,m := {s, y : t/2m s t/m, x d + M/m}. By Lemma 4., for some constant < β /8, µ δ G a s, x 2β ] d M/m,d+M/m[ x for all s [t/2m, t/m] and x. 4.2 Hence, t P us, x < β ] d M/m,d+M/m[ x for some 2m s t m and x P P Is, x < β for some s, x S sup Is, x > β s,x S β p E [ ] sup Is, x p, s,x S where we have applied Chebyshev s inequality in the last step. Denote τ = t/m and S := {s, y : s t/m, x d + M/m}. By the fact that I, x for all x, a.s., we see that for all < η < 2a+ pa, E [ sup s,x S Is, x p] E τ a 2a η sup s,x,s,x S Is, x Is, x x x a 2 + s s a 2a η p

18 Let us find the upper bound of 4.3. By the Burkholder-Davis-Gundy inequality and [9, Proposition 4.4], for some universal constant C >, p/2 E [ Is, x Is, x p ] C x x a 2 + s s a 2a sup ut, y p p, t,y S for all s, x and s, x S. Hence, by part of Lemma 4.2, sup ut, y p p Q p exp Qp 2a a τ =: C p,τ, t,y S for some constant Q := Qa, Lip ρ, Λ > and for all p [2, [. Notice that when m is sufficiently large, S [, ] 2. Hence, the right hand side of 4.3 is bounded by the same quantity with S replaced by [, ] 2. Then by Kolmogorov s continuity theorem see [22, Theorem.4.] and [6, Proposition 4.2], for some universal constant C >, the expectation on the right hand side of 4.3 is bounded above by C p C p,τ. We consider the case where p = O [m log m] /a as m see 4.4 below. In this case, we have p a/a τ = Olog m as m since τ = t/m. This implies that there exists some constant Q := Q β, Lip ρ, Λ, t such that β p E [ By denoting η = θ 2 p ] sup Is, x p s,x S Q τ a η 2a p exp Q p 2a a τ = Q exp Q p 2a p logτ. a η a τ + 2a a+ with θ ], [, the above exponent becomes a fp := Q τp 2a logτ θ p[a ] 2[a + ] a +. 2a It is easy to see that fp for p 2 is minimized at a 2 /a θ log/τ a 2 /a θ m logm/t p = =. 2a2a Q τ 2a2a Q t Thus, for some constants A := Aβ, Lip ρ, Λ, t and Q := Q β, Lip ρ, Λ, t, min p 2 fp fp = Q m /a [logm] 2 /a, with p = A [m logm] /a. 4.4 This completes the proof of Lemma 4.3. Proof of Theorem.3. Let ut, x := u 2 t, x u t, x and ρu := ρu + u ρu. Then ut, x is, in fact, a solution to. with the nonlinear function ρ and the initial data µ := µ 2 µ. We note that ρ = and ρ is a Lipschitz continuous function with the same Lipschitz constant as for ρ. For simplicity, we will use ρ instead of ρ. By the weak comparison principle Theorem., we only need to consider the case when µ has compact support and prove that ut, x > for all t > and x, a.s. 8

19 t d = m 2 2a+ [t/2,t] [ M/2,M/2] m =4 k =3 k =2 k = k = d m M m d M m d B 3 B 2 B B A 3 A 2 A A d d + M m 5t 2m 3t 2m t 2m d + m M m Figure : Induction schema for the strong comparison principle. t 2m t 2m 3t m 2t m t m x Case I. We first assume that µdx = fxdx with f C and fx for all x. Since µ >, there exists x such that fx >. By the weak comparison principle Theorem., we only need to consider the case where fx = [ d,d] x for some d > this is because, if fx = [a,b] x, then we can use fx a + b/2 as our initial function; in addition, if fx = c [ d,d] x, then we can consider ũt, x := cut, x which is the unique solution to. with the initial function [ d,d] x and with replacing ρz by cρz/c which is also Lipschitz continuous with the same Lipschitz constant as ρz. Let γ ], /4] be the constant defined in Lemma 4. and let β := γ/2. For any M > and k =,,, m, define the events { [ ] } 2k + t A k := us, x β k+ k + t S m k x for all s, and x, where B k := { us, x β k+ S m k x for all s S m k := 2m [ kt 2k + t, m 2m ] d Mk m, d + Mk [ m. m ] } and x, See Figure for an illustration of the schema. By Lemma 4.3, there are constants Q > and m > such that for all m m, P A cm, 9

20 where cm := exp Q m /a [logm] 2 /a. 4.5 By definition, on the event A k, k, kt u m, x β k S m k x, for all x. Let w k s, x be the solution to the following SPDE: t xdδ a w k s, x = ρ k w k s, x Ẇks, x, s + := ], + [, x, w k, x = S m k x, where ρ k x := β k ρβ k x and {Ẇks, x := Ẇ s + kt/m, x} k is the time-shifted white noise. Note that ρ k x is also a Lipschitz continuous function with the same Lipschitz constant as for ρ and ρ k =. Thus, by Lemma 4.3, we see that by the same constants Q and m as in 4.5, for all m m, [ t P w k s, x β S m k x for all s 2m, t ] and x cm. 4.6 m Let vs, x be a solution to. with the initial data µdx := β k S m k xdx, subject to the above time-shifted noise Ẇk with the same ρ. Then vs, x = β k w k s, x a.s. for all s and x. Since us + kt/m, x vs, x for all x and s by the Markov property and the weak comparison principle Theorem., 4.6 implies that Hence, P A k F kt/m cm, a.s. on Ak for k m. P A k A k A cm, for all k m. Furthermore, because A B, on the event A, we see that Similarly, one can prove that Then, P B P A cm. P B k B k B cm, for all k m. P k m [A k B k ] P k m A k P k m B k Therefore, for all t > and M >, cm m P A + cm m P B 2 cm m. 4.7 P us, x > for all t/2 s t and x M/2 lim P k m [A k B k ] m lim 2 m cmm =. Since t and M are arbitrary, this completes the proof for Case I. 2

21 Case II. Now we assume that µ M a,+. We only need to prove that for each ɛ >, P ut, x > for t ɛ and x =. 4.8 Fix ɛ >. Denote V t, x := ut + ɛ, x. By the Markov property, V t, x solves. with the time-shifted noise Ẇɛt, x := Ẇ t + ɛ, x starting from V, x = uɛ, x, i.e., V t, x = uɛ, δ G a t, x + ρv s, y δ G a t s, x yw ɛ ds, dy [,t] 4.9 =: J t, x + Ĩt, x. We first claim that P uɛ, x =, for all x =. 4. Notice that by Theorem.6, the function x ut, x is Hölder continuous over a.s. The weak comparison principle Theorem. shows that ut, x a.s. Hence, if 4. is not true, then by the Markov property and the strong comparison principle in Case I, at all times η [, ɛ], with some strict positive probability, uη, x = for all x, which contradicts Theorem.7 as η goes to zero. Therefore, there exists a sample space Ω with P Ω = such that for each ω Ω, there exists x such that uɛ, x, ω >. Since uɛ, x, ω is continuous at x, one can find two nonnegative constants c and β such that uɛ, y, ω β [x c,x+c] y for all y. Then Case I implies that P V ω t, x > for all t and x =, where V ω is the solution to 4.9 starting from uɛ, x, ω. Therefore, 4.8 is true. This completes the whole proof of Theorem.3. 5 Proof of Theorem.4 We first prove part. For any compact sets K +, one can find η >, T > and N > such that K [η, T ] [ N, N]. Then choose M = 2NT/η. If µdx = fxdx with f C and fx for all x, then following the proof of Theorem.3, from 4.7, we see that P inf us, x < βm P k m [A k B k ] s,x K 2 [ cm m ], where cm is defined in 4.5. Because log x 2x for < x /2, when m is sufficiently large, so that m exp Q m /a [logm] 2 /a /2, 5. 2

22 we have that cm m exp 2 m exp Q m /a [logm] 2 /a. Since e x x for x, if m is sufficiently large such that 5. holds, then [ cm m ] 2 m exp Q m /a [logm] 2 /a. If µ M a,+, then we follow the notation of Case II in the proof of Theorem.3 with ɛ = η/2. Using the Markov property, for each initial data uɛ, x, ω, we apply the previous case to get.7 with ut, x replaced by V ω t ɛ, x. Because the upper bound of which does not depend on ω and uɛ, x is independent of V t, x,.7 holds for V t ɛ, x = ut, x. This completes the proof of part of Theorem.4. Now we prove part 2. Since f is a continuous function, there exists finite constant c > such that fx c D x. Without loss of generality, we assume that c =. Let vt, x be the solution to. with the initial data D xdx. By Theorem., ut, x vt, x for all t > and x, a.s. Hence, it suffices to prove that for all n, P inf x D inf vt, x e n t ],T ] A exp Bn logn 2a /a. We define a set of {F t } t -stopping times as follows: T :=, and { } T k+ := inf s > T k : inf vs, x e k, x D where we use the convention that inf φ =. Similar to the proof of Theorem.3, let {Ẇkt, x : k N} be time-shifted space-time white noises and let v k t, x be the unique solution to. subject to the noise Ẇk, starting from v k, x = e k D x. Then w k t, x := e k v k t, x solves t xdδ a w k t, x = ρ k w k t, x Ẇkt, x, t + := ], + [, x, w k, x = D x, where ρ k x := e k ρe k x. From the definitions of the stopping times, we see that e k vt k, x D x, for all x, a.s. on {T k < }, for all k. 22

23 Therefore, by the strong Markov property and the weak comparison principle in Theorem., we obtain that on {T k < }, P T k T k 2t F Tk P sup w k t, x w k, x /e. n t,x ],2T/n] D Since ρ k is Lipschitz continuous with the same Lipschitz constant as ρ, a suitable form of the Kolmogorov continuity theorem see the arguments in the proof of Lemma 4.3 implies that for all η ], 2a + /pa [, there exists a finite constant Q >, not depending on p, n and τ, such that for all p 2, n, and τ ], [, [ E ] sup w k s, x w k, x p Q τ pηa /2a exp Q τp 2a /a. 5.2 s,x ],τ] D Letting τ := 2t/n for < t < T and minimizing the right hand side of 5.2 over p, we obtain that for some finite constant Q >, not depending on n, P T k T k 2t F Tk Q exp { Q n a /a log n 2a /a}. n Therefore, we obtain the following: P inf inf vt, x e n P {T n t} x D t ],T ] P at least n/2 -many distinct values k {, 2,..., n} such that T k T k 2t/n n n/2 c n/2 exp { c 2 n/2 n a /a log n 2a /a}. This completes the proof of Theorem.4. 6 Proof of Theorem.5 Proof of Theorem.5. Fix ɛ >. By Theorems 2.2, both ut, x and u ɛ t, x are well-defined solutions to.. By Lipschitz continuity of ρ and the moment formulas 2.9, ut, x u ɛ t, x 2 2 [µ ψ ɛ δ G a ɛ, δ G a t, x µ δ G a t, x] 2 t + Lip 2 ρ ds dy us, y u ɛ s, y 2 2 δg 2 at s, x y. Denote the first part on the above upper bound as I ɛ t, x. Let Kt, x := Kt, x; Lip ρ and denote f ɛ t, x := ut, x u ɛ t, x 2 2. Then formally, f ɛ Kt, x I ɛ K t, x + Lip 2 ρ f ɛ δ G 2 a K t, x. 23

24 Using the fact that Lip 2 ρ δg 2 a K t, x = Kt, x Lip 2 ρ δg 2 at, x, one has that fɛ δ Ga 2 t, x Lip 2 ρ I ɛ K t, x. Hence, it reduces to show that lim I ɛ K t, x =, ɛ for all t > and x. 6. We first assume that a ], 2[. Notice that I ɛ t, x = [µ ψ ɛ [ δ G a t + ɛ, δ G a t, ] x + [µψ ɛ µ] δ G a t, x] 2. By [9, 4.3], for < t T and x, µψ ɛ δ G a t, x µ δ G a t, x C T t /a, 6.2 with C T := A a K a, T +/a, where A a is defined as µ dz A a := sup. 6.3 y + y z +a Hence, if < t + ɛ T, then I ɛ t, x 4C T t /a [ µ ψ ɛ δ G a t + ɛ, δ G a t, x + µψ ɛ µ δ G a t, x]. Now use the upper bound on Kt, x in 2., I ɛ K t, x 2C T C C T t ds s /a t s /a [g t, s, ɛ, x + g 2 t, x, ɛ, x] 6.4 where C := C a, δ, Lip ρ is defined in Proposition 2.3, C T = +T /a exp Lip 2 ρ ΛΓ/a a T, /a + /a =, and g t, s, ɛ, x = µ ψ ɛ δ G a s + ɛ, δ G a s, δ G a t s, x, g 2 t, s, ɛ, x = µψ ɛ µ δ G a s, δ G a t s, x. By the semigroup property and the dominated convergence theorem, g 2 t, s, ɛ, x = µψ ɛ µ δ G a t, x, as ɛ. Clearly, g 2 t, s, ɛ, x 2µ δ G a t, x. Again, by the dominated convergence theorem and by bounding δ G a t + ɛ, using [9, 4.3], one can show that lim ɛ g t, s, ɛ, x =. Then by the semigroup property and 6.2, for < t + ɛ T, g t, s, ɛ, x µ δ G a s + ɛ, + δ G a s, δ G a t s, x 24

25 = µ δ G a t + ɛ, x + µ δ G a t, x C T t + ɛ /a + t /a 2C T t /a. Hence, both upper bounds on g and g 2 are integrable over ds in 6.4. Therefore, by another application of the dominated convergence theorem, we have proved 6.. Since both functions f ɛ t, x and δ G a t, x are nonnegative and the support of δ G a t, x is over, we can conclude that lim ɛ f ɛ t, x = for almost all t > and x. When a = 2, one can apply the dominated convergence theorem to show that I ɛ t, x as ɛ. Another application of the dominated convergence theorem shows that 6. is true. The rest is same as the previous case. We leave the details for interested readers. This completes the proof of Theorem.5. 7 Proof of Theorem.6 Without loss of generality, we assume that µ. Let ut, x be the solution to. starting from µ M a. Fix T > and ɛ ], T/2 ]. Denote V t, x := ut + ɛ, x. By the Markov property, V t, x solves. with the time-shifted noise Ẇɛt, x := Ẇ t + ɛ, x starting from V, x = uɛ, x. ecall the integral form V t, x = J t, x + Ĩt, x in 4.9. Time increments ecall that ut, x = J t, x + It, x. Let < ɛ t t T ɛ. So It + ɛ, x It + ɛ, x 2 p 2 ut + ɛ, x ut + ɛ, x 2 p + 2 J t + ɛ, x J t + ɛ, x 2, with ut + ɛ, x ut + ɛ, x 2 p = V t, x V t, x 2 p 2 Ĩt, x Ĩt, x J t, x J t, x p Notice that for all p 2, by the Burkholder-Davis-Gundy inequality see [7, Lemma 3.3], Ĩt, x Ĩ t, x 2 2zp 2 L 2 ρ I t, t, x + 2zp 2 L 2 ρ I 2 t, t, x, p where z p 2 p and z 2 =, and I t, t, x = dsdy δ G a t s, x y δ G a t s, x y 2 ς 2 + V s, y 2 p, [,t] I 2 t, t, x = dsdy δ G 2 a t s, x y ς 2 + V s, y 2 p. [t,t ] 2 p. 25

26 By part 2 of Lemma 4.2, for some finite constant Q := Qa, δ, L ρ, ς, µ, p, ɛ, T >, sup V s, y 2 p = sup us, y 2 p Q. s,y [,t] s,y [ɛ,t+ɛ] Then apply [9, Proposition 4.4] to see that for some finite constant C = C a, δ >, Ĩt, x Ĩ t, x 2 C zp 2 L 2 ρ Q t t /a. 7. p By Minkowski s integral inequality and 8.7 below, for some finite constant C 2 := C 2 a >, J t, x J t, x 2 sup uɛ, y 2 p dy δ G a t, y δ G a t, y p y C 2 Q [logt /t] 2 C 2 t 2 Q t t 2, where in the last step, we have applied the inequality log + x x for x >. Because t t T a+ 2a t t a 2a and t ɛ, we have that J t, x J t, x 2 p C 2 ɛ 2 T a+ a 2 Q t t a a. 7.2 Similarly, 2 J t + ɛ, x J t + ɛ, x 2 sup J 2 ɛ, y dy δ G a t, x δ G a t, x y C 2 ɛ 2 T a+ a Q t t a a. Space increments Fix t ɛ. Let x, x [ T, T ]. Then It + ɛ, x It + ɛ, x 2 p 2 ut + ɛ, x ut + ɛ, x 2 p + 2 J t + ɛ, x J t + ɛ, x 2, with ut + ɛ, x ut + ɛ, x 2 p 2 Ĩt, x Ĩt, x J t, x J t, x p 2 p. For p 2, by the Burkholder-Davis-Gundy inequality and [9, Proposition 4.4], Ĩt, x Ĩ t, x 2 2zp 2 L 2 ρ dsdy δ G a t s, x y δ G a t s, x y 2 p [,t] ς 2 + V s, y 2 p 26

27 2z 2 p L 2 ρ Q x x a. By the Minkowski s integral inequality and 8.2, for some finite constant C 3 := C 3 a >, J t, x J t, x 2 sup uɛ, y 2 p dy δ G a t, x y δ G a t, x y p y C 3 Q t /a x x C 3 Q ɛ /a 2T 2 a x x a. 7.3 Similarly, 2 J t + ɛ, x J t + ɛ, x 2 sup J 2 ɛ, y dy δ G a t, x δ G a t, x y C 3 Q ɛ /a 2T 2 a x x a. Finally, combining the two cases, we see that for all compact sets D +, one can find T >, ɛ ], T/2 ], such that D Kɛ, T := [2ɛ, T ] [ T, T ]. There is some finite constant Q := Q a, δ, L ρ, ς, µ, p, ɛ, T > such that for all t, x and t, x D, It, x It, x 2 p Q t t /a + x x a. Then the Hölder continuity follows from Kolmogorov s continuity theorem see [22, Theorem.4.] and [6, Proposition 4.2]. Note that J t, x belongs to C + see [9, Lemma 4.9]. This completes the proof of Theorem.6. 8 Proof of Theorem.7 The case when a = 2 is proved in [6, Proposition 3.4]. Assume that < a < 2. Fix φ C c. For simplicity, we only prove the case where ρu = λu and µ. As in the proof [6, Proposition 3.4], we only need to prove that 2 lim dx It, xφx = t + in L 2 Ω. Denote Lt := It, xφxdx. By the stochastic Fubini theorem see [33, Theorem 2.6, p. 296], whose assumptions are easily checked, t Lt = dx δ G a t s, x yφx ρus, yw ds, dy. Hence, by Itô s isometry, E [ Lt 2] = λ 2 t 2 ds dy dx δ G a t s, x yφx us, y

28 Assume that t. Since for some constant C >, φx C δ G a, x for all x, we can apply the semigroup property to get E [ Lt 2] t C 2 λ 2 Λ ds dy t + s /a δ G a t + s, y us, y 2 2, where the constant Λ is defined in 2.6. Apply the moment formula 2.9, with and L t := L 2 t := t t E [ Lt 2] C 2 λ 2 Λ [L t + L 2 t], ds t + s /a ds t + s /a dy J 2 s, y δ G a t + s, y, dy J 2 K s, y δ G a t + s, y. We first consider L t. By [9, 4.2], for some constant C := C a, δ, µ >, J t, x C t /a. Thus, t L t C ds dy J t + s /a s /a s, y δ G a t + s, y =C J t +, C J t +, t t ds t + s /a s /a ds, as t. s /a s/a The case for L 2 t can be proved in a similar way, where one needs to apply 2.. We leave the details for interested readers. This completes the proof of Theorem.7. Appendix ecall the kernel function δ ɛ at, x defined in 3.2. The following two lemmas 8.2 and 8.3 ɛ below show that δ a t, x is an approximation of δg a t, x. The proofs of both Lemmas depend on Lemma 8. below. Lemma 8.. For all b, If b, then lim z e z z b+ C b := sup e z z b+ z 28 z k =. 8. k! kb z k < k! kb

29 Note that when b N, the series in 8. converges to the generalized hypergeometric function see [29, Chapter 6]: z k k! k = bf b b,..., }{{} b +, 2,..., 2 ; z }{{} b +, for b N and z C. Proof of Lemma 8.. Clearly, the series converges on z C and it defines an entire function. We first assume that b N. We will prove 8. by induction. Clearly, the case b = is true. Suppose that 8. is true for b. Now let us consider the case b + : Applying l Hôpital s rule and the induction assumption, we obtain lim z e z z b+2 z k = lim k! kb+ z = lim z e z z b+ z k k! k b+ e z z b+ = lim z z k k! k b =. z k k! k b e z z b+ b+z b z 2b+ 8.3 This proves Lemma 8. for b N. Now assume that /2 b < 3/2. Because the function fx = x b for x is either concave or convex, we have that for all k and z, z b k b b z b k b k z b z b + k b k z. Hence, for all z and k, k z b b z b = k b b k z z b k b z k + b k z b z b k z k /2 z /2 k Thus, for z, by Cauchy-Schwartz inequality and 8.8, z k k!k b z b ez z = z k k! k b z b k=2 k=2 b z k k z + b z k k z z k! k=2 k z k! k k=2 b /2 z k z k k z 2 z k! k! k k= + b /2 z k z k= = b e z z 2 z 3/2 29 z k k z 2 k! /2 z k + b k! k z 3/2 /2 k! k 2 e z z 3 /2 /2 z k k! k 2

30 Hence, by the previous proof for the case b N, and because b < 3/2, lim z zb e z z k k!k b z b ez z lim 2b =. z z3/2 b Therefore, lim z zb e z k=2 k=2 z k k! k b = lim z zb e z z b ez z =. Now assume that b < /2. Let c [/2, 3/2[ and n N such that b + n = c. Then apply l Hôpital s rule n times as in 8.3, = lim z e z z c+ z k k! k c = lim z e z z c z k = = lim k! kc z e z z c n+ z k k! k c n. Similarly, if b 3/2, then let c [/2, 3/2[ and n N such that b = c + n. Then apply l Hôpital s rule n times as in 8.3, lim z e z z b+ z k k! k b = lim z e z z b z k = = lim k! kb z e z z b n+ This proves 8. for all b. Finally, 8.2 follows from the fact that the function fz = z b+ e z z k =. k! kb n z k k! k b is continuous over + {+ } with f = and f < if b actually, f = if b > and f = if b =. This completes the proof of Lemma 8.. Lemma 8.2. There exists a finite constant C > such that ɛ /2 dx δ ɛ a t, x δ G a t, x e t/ɛ + C, for all ɛ > and t >, t where the constant C can be chosen as C = a [ a a K a, Γ Γ sup e z z4z 2 + 7z + a + 2 a + 2 z with the constant K a, defined in [9, 4.3]. Proof. From [9, 4.3], t δ G a t, x = δg a t, x + x δg a t, x at x δg a t, x + at x δg a t, x x 3 ] z k 2, 8.5 k! k 2

31 at Thus, for < t t, dx δ G a t, x δ G a t, x where C := a t t [ δg a t, x + t 2 a t dx t ds as ds t δ G a s, x + ] K a, x t /a x 2+a dy K a, y + y 2+a t C log t, dy K a, y = a a K + y 2+a a, Γ Γ, 8.8 a a + 2 a + 2 and the integral in 8.8 is evaluated by Lemma 8.5. Notice that ɛ δ a t, x δ G a t, x e t/ɛ δg a t, x + e t/ɛ By the above inequality, we have that dx ɛ δ a t, x δ G a t, x e t/ɛ + C e t/ɛ k t ɛ k! δg a t, x δ G a kɛ, x. k t ɛ k! logkɛ/t. Denote the summation over k in above upper bound by It/ɛ. Because the function x logx is concave, logt /t t t t t t t + t t. So, by letting z = t/ɛ, z k Iz k z k! k + z k = k 2 z 2. z k! k Then by Cauchy-Schwartz inequality, Notice that z k k! Iz /2 z k /2 z k [ k 2 z 2] 2. k! k 2 k! [ k 2 z 2] 2 = e z 4z 2 + 7z + z 3 e z 4z 2 + 7z To prove the equality in 8.9, one can write k 2 = Pk 2+P k and k4 = Pk 4+6P k 3+7P k 2+P k, where Pk n := kk k n +. Hence, k2 z 2 2 = Pk 4 + 6P k z2 Pk 2 + 2z2 Pk + z4. Then use the fact that for n, z k P n k! k = ez z n. 3

32 Therefore, Ize z e z z4z 2 + 7z + /2 z k z. k! k 2 By Lemma 8., the function fz = e z z4z 2 + 7z + z k is continuous over k! k 2 + {+ } with f = and f = 4. Thus, sup z fz < +. This completes the proof of Lemma 8.2. Lemma 8.3. We have that t lim ɛ lim ɛ ɛ δ a t, x = δ G a t, {x=}, 8. ds dx [ δ ɛ a s, x δ G a s, x] 2 =, for all t >, 8. and there is a nonnegative constant C a,δ < + such that dx δ ɛ at, 2 x C a,δ t a, for all t >. 8.2 Proof. Fix t >. Denote A := δ G a,. Clearly, where t ds dx [ δ ɛ a s, x δ G a s, x] 2 = I t, ɛ 2I 2 t, ɛ + I 3 t, I t, ɛ = I 2 t, ɛ = I 3 t = [,t] [,t] [,t] dsdx ɛ δ 2 as, x, 8.3 dsdx ɛ δ a s, x δ G a s, x, 8.4 dsdx δ G 2 as, x. 8.5 By the semigroup property and scaling property [9, 4.], we have that I 3 t = t ds δ G a 2s, = A t 2s a ds = aa 2 /a a t a. 8.6 Step. We first calculate I. Use the semigroup property and scaling property [9, 4.]: I t, ɛ = t ds e 2s/ɛ s ɛ n= m= n+m n!m! δ G a n + mɛ, 32

33 = A t ds e 2s/ɛ s ɛ n= m= n+m n!m!n + m /a ɛ /a. Then by change of variables u = s/ɛ and let z = t/ɛ, we have that I t, ɛ = At a z a z du e 2u n,m= u n+m n!m!n + m /a. By l Hôpital s rule, I t := lim ɛ I t, ɛ = Because k n= = n!k n! k! 2k 2, Hence, n,m= aa a t z n+m n!m!n + m = /a k=2 a z k k /a lim z /a e 2z z k n= n,m= n!k n! = z n+m n!m!n + m /a. k=2 z k k!k /a 2k 2. I t = aa a t a lim z /a e 2z z k=2 z k k!k /a 2k 2 = aa 2 /a a t a, 8.7 where the last equality is due to Lemma 8. with b = /a [/2, ]. Step 2. In this step, we calculate I 2 t := lim ɛ I 2 t, ɛ. Similarly to the Step, use the semigroup property and scaling property [9, 4.], and then change the variables u = s/ɛ and z = t/ɛ, I 2 t, ɛ = t ds e s/ɛ n= s n ɛ n! δ G a s + nɛ, = At a z a z du e u n= u n n!u + n /a. By l Hôpital s rule, I 2 t = aa a t a lim z /a e z z n= z n n!z + n /a. Apply the inequality 8.4 below with b = /a [/2, ], z + n /a 2z /a a n z 2z n + z + n + z 2z + 2 2a n z z 3/2, 33

34 for z and n. Notice that see the proof of 8.9, n= z n n! n z 2 = e z z. 8.8 By Cauchy-Schwartz inequality and 8.8, for z, z n n!z + n /a 2z /a ez = z n n! z + n /a 2z /a n= n= + 2 z n n z 2az 3/2 n! Therefore, + 2 2az 3/2 n= = + 2 2az ez. n= z n n z 2 n! /2 n= /2 z n n! I 2 t = aa a t a lim z /a e z z 2z /a ez = aa 2 /a a t a. 8.9 Finally, 8. is proved by combining 8.6, 8.7 and 8.9. Step 3. Now we prove 8.. Clearly, if x, then lim ɛ Lemma 8. with b = /a, lim ɛ ɛ δ a t, = A t /a lim z z/a e z ɛ δ a t, x =. Otherwise, by z k k!k /a = δg a t,. Step 4. As for 8.2, denote It; ɛ = dx ɛ δ 2 at, x. Following the arguments in Step, It, ɛ A sup z /a e z 2t /a z + k=2 z k. 8.2 k!k/a Clearly, the function fz = z k k=2 is an entire function over C. By Lemma 8. and k!k /a lim z z /a e z fz =, we know that the supremum in 8.2, which depends only on a, is finite. This completes the proof of Lemma 8.3 Lemma 8.4. If µdx = fxdx with f L, then for all < t T and x, y, ut, x ut, y 2 2 Ct /a x y + A T C x y a, 34

35 where K a, is defined in [9, 4.3], C := C a, δ is defined in [9, Proposition 4.4], and a + a + 3 C := 8K a, Γ Γ sup[fx] 2, A T := sup sup ρus, x 2 2 a + 2 a + 2. x s [,T ] x Proof. By Itô s isometry, ut, x ut, y 2 2 = [J t, x J t, y] 2 + ds dz ρus, z 2 2 [ δ G a t s, x z δ G a t s, y z] 2. Denote C f := sup x fx and fix < t < T. Then ut, x ut, y C2 f dz δ G a t, x z δ G a t, y z t + A T ds dz δ G a s, x z δ G a s, y z 2. By [9, Proposition 4.4], for some constant C := C a, δ, the second part of the above upper bound is bounded by A T C x y a. As for the first part, notice that by [9, 4.3], for all t > and x, y, y δ G a t, x δ G a t, y = y x δ G a t, zdz x δ G a t, z dz x t K a, t 2/a y x x dz + t /a z 2+a K a, t 2/a x y + t /a x t /a y 2+a. Thus, dz δ G a t, x z δ G a t, y z K a, t 2/a x y dz + t /a x z t /a y z 2+a [ ] K a, t 2/a x y dz + t /a x z + 2+a + t /a y z 2+a = 2K a, t 2/a dz x y + t /a z = 2K a,t /a dz x y 2+a + z. 2+a Hence, by letting dz a + a + 3 C a := = 2Γ Γ, + z 2+a a + 2 a + 2 where the integral is evaluated by Lemma 8.5, we have that dz δ G a t, x z δ G a t, y z 2K a, C a t /a x y, for all x, y. 8.2 This completes the proof of Lemma