Abstract
This paper is dedicated to the study of a nonlinear SPDE on a bounded domain in , with zero initial conditions and Dirichlet boundary, driven by an -stable Lévy noise with , , and possibly nonsymmetric tails. To give a meaning to the concept of solution, we develop a theory of stochastic integration with respect to this noise. The idea is to first solve the equation with “truncated” noise (obtained by removing from the jumps which exceed a fixed value ), yielding a solution , and then show that the solutions coincide on the event , for some stopping times converging to infinity. A similar idea was used in the setting of Hilbert-space valued processes. A major step is to show that the stochastic integral with respect to satisfies a th moment inequality. This inequality plays the same role as the Burkholder-Davis-Gundy inequality in the theory of integration with respect to continuous martingales.
1. Introduction
Modeling phenomena which evolve in time or space-time and are subject to random perturbations are a fundamental problem in stochastic analysis. When these perturbations are known to exhibit an extreme behavior, as seen frequently in finance or environmental studies, a model relying on the Gaussian distribution is not appropriate. A suitable alternative could be a model based on a heavy-tailed distribution, like the stable distribution. In such a model, these perturbations are allowed to have extreme values with a probability which is significantly higher than in a Gaussian-based model.
In the present paper, we introduce precisely such a model, given rigorously by a stochastic partial differential equation (SPDE) driven by a noise term which has a stable distribution over any space-time region and has independent values over disjoint space-time regions (i.e., it is a Lévy noise). More precisely, we consider the SPDE: with zero initial conditions and Dirichlet boundary conditions, where is a Lipschitz function, is a second-order pseudo-differential operator on a bounded domain , and is the formal derivative of an -stable Lévy noise with , . The goal is to find sufficient conditions on the fundamental solution of the equation on , which will ensure the existence of a mild solution of (1). We say that a predictable process is a mild solution of (1) if for any , , We assume that is a function in , which excludes from our analysis the case of the wave equation with .
To explain the connections with other works, we describe briefly the construction of the noise (the details are given in Section 2). This construction is similar to that of a classical -stable Lévy process and is based on a Poisson random measure (PRM) on of intensity , where for some , with . More precisely, for any set , where is the compensated process and is a constant (specified by Lemma 3). Here, is the class of bounded Borel sets in and is the Lebesgue measure of .
As the term on the right-hand side of (2) is a stochastic integral with respect to , such an integral should be constructed first. Our construction of the integral is an extension to random fields of the construction provided by Giné and Marcus in [1] in the case of an -stable Lévy process . Unlike these authors, we do not assume that the measure is symmetric.
Since any Lévy noise is related to a PRM, in a broad sense, one could say that this problem originates in Itô’s papers [2, 3] regarding the stochastic integral with respect to a Poisson noise. SPDEs driven by a compensated PRM were considered for the first time in [4], using the approach based on Hilbert-space-valued solutions. This study was motivated by an application to neurophysiology leading to the cable equation. In the case of the heat equation, a similar problem was considered in [5–7] using the approach based on random-field solutions. One of the results of [6] shows that the heat equation: has a unique solution in the space of predictable processes satisfying , for any . In this equation, is the compensated process corresponding to a PRM on of intensity , for an arbitrary -finite measure space with measure satisfying . Because of this later condition, this result cannot be used in our case with and . For similar reasons, the results of [7] also do not cover the case of an -stable noise. However, in the case , we will be able to exploit successfully some ideas of [6] for treating the equation with “truncated” noise , obtained by removing from the jumps exceeding a value (see Section 5.2).
The heat equation with the same type of noise as in the present paper was examined in [8, 9] in the cases and , respectively, assuming that the noise has only positive jumps (i.e., ). The methods used by these authors are different from those presented here, since they investigate the more difficult case of a non-Lipschitz function with . In [8], Mueller removes the atoms of of mass smaller than and solves the equation driven by the noise obtained in this way; here we remove the atoms of of mass larger than and solve the resulting equation. In [9], Mytnik uses a martingale problem approach and gives the existence of a pair which satisfies the equation (the so-called “weak solution”), whereas in the present paper we obtain the existence of a solution for a given noise (the so-called “strong solution”). In particular, when and , the existence of a “weak solution” of the heat equation with -stable Lévy noise is obtained in [9] under the condition which we encounter here as well. It is interesting to note that (6) is the necessary and sufficient condition for the existence of the density of the super-Brownian motion with “”-stable branching (see [10]). Reference [11] examines the heat equation with multiplicative noise (i.e., ), driven by an -stable Lévy noise which does not depend on time.
To conclude the literature review, we should point out that there are many references related to stochastic differential equations with -stable Lévy noise, using the approach based on Hilbert-space valued solutions. We refer the reader to Section 12.5 of the monograph [12] and to [13–16] for a sample of relevant references. See also the survey article [17] for an approach based on the white noise theory for Lévy processes.
This paper is organized as follows.(i)In Section 2, we review the construction of the -stable Lévy noise , and we show that this can be viewed as an independently scattered random measure with jointly -stable distributions.(ii)In Section 3, we consider the linear equation (1) (with ) and we identify the necessary and sufficient condition for the existence of the solution. This condition is verified in the case of some examples.(iii)Section 4 contains the construction of the stochastic integral with respect to the -stable noise , for . The main effort is dedicated to proving a maximal inequality for the tail of the integral process, when the integrand is a simple process. This extends the construction of [1] to the case random fields and nonsymmetric measure .(iv)In Section 5, we introduce the process obtained by removing from the jumps exceeding a fixed value , and we develop a theory of integration with respect to this process. For this, we need to treat separately the cases and . In both cases, we obtain a th moment inequality for the integral process for if and if . This inequality plays the same role as the Burkholder-Davis-Gundy inequality in the theory of integration with respect to continuous martingales.(v)In Section 6 we prove the main result about the existence of the mild solution of (1). For this, we first solve the equation with “truncated” noise using a Picard iteration scheme, yielding a solution . We then introduce a sequence of stopping times with a.s. and we show that the solutions , coincide on the event . For the definition of the stopping times , we need again to consider separately the cases and .(vi)Appendix A contains some results about the tail of a nonsymmetric stable random variable and the tail of an infinite sum of random variables. Appendix B gives an estimate for the Green function associated with the fractional power of the Laplacian. Appendix C gives a local property of the stochastic integral with respect to (or ).
2. Definition of the Noise
In this section we review the construction of the -stable Lévy noise on and investigate some of its properties.
Let be a Poisson random measure on , defined on a probability space , with intensity measure , where is given by (3). Let be a sequence of positive real numbers such that as and . Let
For any set , we define
Remark 1. The variable is finite since the sum above contains finitely many terms. To see this, we note that , and hence .
For any , the variable has a compound Poisson distribution with jump intensity measure ; that is, It follows that and for any . Hence, for any and . If , then is finite. Define This sum converges a.s. by Kolmogorov’s criterion since are independent zero-mean random variables with .
From (9) and (10), it follows that is an infinitely divisible random variable with characteristic function: Hence, and .
Lemma 2. The family defined by (10) is an independently scattered random measure; that is,(a)for any disjoint sets in , are independent;(b)for any sequence of disjoint sets in such that is bounded, a.s.
Proof. (a) Note that for any function with compact support , we can define the random variable where . For any , we have
For any disjoint sets and for any , we have
using (12) with for the second equality and (9) for the last equality. This proves that are independent.
(b) Let and , where . By Lévy’s equivalence theorem, converges a.s. if and only if it converges in distribution. By (13), with for all , we have
This clearly converges to , and hence converges in distribution to .
Recall that a random variable has an -stable distribution with parameters , , , and if, for any , or (see Definition of [18]). We denote this distribution by .
Lemma 3. has a distribution with , and . If , then .
Proof. We first express the characteristic function (11) of in Feller’s canonical form (see Section XVII.2 of [19]): with and . Then the result follows from the calculations done in Example XVII.3.(g) of [19].
From Lemmas 2 and 3, it follows that is an -stable random measure, in the sense of Definition of [18], with control measure and constant skewness intensity . In particular, has a distribution.
We say that is an -stable Lévy noise. Coming back to the original construction (10) of and noticing that it follows that can be represented as Here is the compensated Poisson measure associated with ; that is, for any relatively compact set in .
In the case , we will assume that so that is symmetric around , for all , and admits the same representation as in the case .
3. The Linear Equation
As a preliminary investigation, we consider first equation (1) with : with zero initial conditions and Dirichlet boundary conditions. In this section is a bounded domain in or .
By definition, the process given by is a mild solution of (23), provided that the stochastic integral on the right-hand side of (24) is well defined.
We define now the stochastic integral of a deterministic function :
If , this can be defined by approximation with simple functions, as explained in Section 3.4 of [18]. The process has jointly -stable finite dimensional distributions. In particular, each has a -distribution with scale parameter:
More generally, a measurable function is integrable with respect to if there exists a sequence of simple functions such that a.e., and, for any , the sequence converges in probability (see [20]).
The next results show that condition is also necessary for the integrability of with respect to . Due to Lemma 2, this follows immediately from the general theory of stochastic integration with respect to independently scattered random measures developed in [20].
Lemma 4. A deterministic function is integrable with respect to if and only if .
Proof. We write the characteristic function of in the form used in [20]: with , if and if . By Theorem 2.7 of [20], is integrable with respect to if and only if where and . Direct calculations show that, in our case, if , if , and .
The following result follows immediately from (24) and Lemma 4.
Proposition 5. Equation (23) has a mild solution if and only if for any , In this case, has jointly -stable finite-dimensional distributions. In particular, has a distribution.
Condition (29) can be easily verified in the case of several examples.
Example 6 (heat equation). Let . Assume first that . Then , where and condition (29) is equivalent to (6). In this case, . If is a bounded domain in , then (see page 74 of [11]) and condition (29) is implied by (6).
Example 7 (parabolic equation). Let where is the generator of a Markov process with values in , without jumps (a diffusion). Assume that is a bounded domain in or . By Aronson estimate (see, e.g., Theorem 2.6 of [12]), under some assumptions on the coefficients , , there exist some constants such that for all and , . In this case, condition (29) is implied by (6).
Example 8 (heat equation with fractional power of the Laplacian). Let for some . Assume that is a bounded domain in or . Then (see, e.g., Appendix of [12])
where is the fundamental solution of on and is the density of the measure , being a convolution semigroup of measures on whose Laplace transform is given by
Note that if , is the density of , where is a -stable subordinator with Lévy measure .
Assume first that . Then , where
If , then is the density of , with being a symmetric -stable Lévy process with values in defined by , with a Brownian motion in with variance 2. By Lemma B.1 (Appendix B), if , then (29) holds if and only if
If is a bounded domain in , then (by Lemma 2.1 of [8]). In this case, if , then (29) is implied by (36).
Example 9 (cable equation in ). Let and . Then , where and condition (29) holds for any .
Example 10 (wave equation in with ). Let and with or . Then , where Condition (29) holds for any . In this case, if and if .
4. Stochastic Integration
In this section we construct a stochastic integral with respect to by generalizing the ideas of [1] to the case of random fields. Unlike these authors, we do not assume that has a symmetric distribution, unless .
Let where is the -field of negligible sets in and is the -field generated by for all , and for all Borel sets bounded away from . Note that where is the -field generated by , , and .
A process is called elementary if it is of the form where , , and is -measurable and bounded. A simple process is a linear combination of elementary processes. Note that any simple process can be written as with and , where are -measurable and are disjoint sets in . Without loss of generality, we assume that .
We denote by the predictable -field on , that is, the -field generated by all simple processes. We say that a process is predictable if the map is -measurable.
Remark 11. One can show that the predictable -field is the -field generated by the class of processes such that is left continuous for any , and is -measurable for any .
Let be the class of all predictable processes such that for all and . Note that is a linear space.
Let be an increasing sequence of sets in such that . We define
We identify two processes and for which ; that is, a.e., where . In particular, we identify two processes and if is a modification of ; that is, a.s. for all .
The space becomes a metric space endowed with the metric : This follows using Minkowski’s inequality if and the inequality if .
The following result can be proved similarly to Proposition 2.3 of [21].
Proposition 12. For any there exists a sequence of bounded simple processes such that as .
By Proposition 5.7 of [22], the -stable Lévy process has a càdlàg modification, for any . We work with these modifications. If is a simple process given by (40), we define Note that, for any , is -measurable for any , and is càdlàg. We write
The following result will be used for the construction of the integral. This result generalizes Lemma 3.3 of [1] to the case of random fields and nonsymmetric measures .
Theorem 13. If is a bounded simple process then for any and , where is a constant depending only on .
Proof. Suppose that is of the form (40). Since is càdlàg, it is separable. Without loss of generality, we assume that its separating set can be written as where is an increasing sequence of finite sets containing the points . Hence,
Fix . Denote by the points of the set . Say for some . Then each interval can be written as the union of some intervals of the form :
where . By (44), for any and ,
For any , let , and, for any , define , , and . With this notation, we have
Consequently, for any
Using (47) and (51), it is enough to prove that for any ,
First, note that
This follows from the definition (40) of and (48), since .
We now prove (52). Let . For the event on the left-hand side, we consider its intersection with the event and its complement. Hence, the probability of this event can be bounded by
We treat separately the two terms.
For the first term, we note that is -measurable and is independent of . By Fubini’s theorem
where and is the law of .
We examine the tail of for a fixed . By Lemma 3, has a distribution. Since the sets are disjoint, the variables are independent. Using elementary properties of the stable distribution (Properties 1.2.1 and 1.2.3 of [18]), it follows that has a distribution with parameters:
By Lemma A.1 (Appendix A), there exists a constant such that
for any . Hence,
We now treat . We consider three cases. For the first two cases we deviate from the original argument of [1] since we do not require that .
Case 1 . Note that
where is a submartingale. By the submartingale maximal inequality (Theorem 35.3 of [23]),
Using the independence between and it follows that
Let . Using (57) and Remark A.2 (Appendix A), we get
Hence,
From (59), (60), and (63), it follows that
Case 2 . We have
where and .
We first treat the term . Note that is a zero-mean square integrable martingale, and
Let . Using (57) and Remark A.2 (Appendix A), we get
As in Case 1, we obtain that
and hence
We now treat . Note that is a semimartingale and hence, by the submartingale inequality,
To evaluate , we note that, for almost all ,
due to the independence between and . We let with . Since , . Using (57) and Remark A.2, we obtain
Hence, and
Case 3 . In this case we assume that . Hence, has a symmetric distribution for any . Using (71), it follows that a.s. for all . Hence, is a zero-mean square integrable martingale. By the martingale maximal inequality,
The result follows using (68).
We now proceed to the construction of the stochastic integral. If is a jointly measurable random process, we define
Let be arbitrary. By Proposition 12, there exists a sequence of simple functions such that as . Let and be fixed. By linearity of the integral and Theorem 13, as . In particular, the sequence is Cauchy in probability in the space equipped with the sup-norm. Therefore, there exists a random element in such that, for any , Moreover, there exists a subsequence such that as . Hence, is -measurable for any . The process does not depend on the sequence and can be extended to a càdlàg process on , which is unique up to indistinguishability. We denote this extension by and we write If and are disjoint sets in , then
Lemma 14. Inequality (46) holds for any .
Proof. Let be a sequence of simple functions such that . For fixed , we denote . We let be the sup-norm on . For any , we have Multiplying by and using Theorem 13, we obtain Let . Using (76) one can prove that . We obtain that . The conclusion follows letting .
For an arbitrary Borel set (possibly ), we assume, in addition, that satisfies the condition: Then we can define as follows. Let where is an increasing sequence of sets in such that . By (80), Lemma 14, and (83), as . This shows that is a Cauchy sequence in probability in the space equipped with the sup-norm. We denote by its limit. As above, this process can be extended to and is -measurable for any . We denote Similarly, to Lemma 14, one can prove that, for any satisfying (83),
5. The Truncated Noise
For the study of nonlinear equations, we need to develop a theory of stochastic integration with respect to another process which is defined by removing from the jumps whose modulus exceeds a fixed value . More precisely, for any , we define
We treat separately the cases and .
5.1. The Case
Note that is an independently scattered random measure on with characteristic function given by
We first examine the tail of .
Lemma 15. For any set , where is a constant depending only on (given by Lemma A.3).
Proof. This follows from Example 3.7 of [1]. We denote by the restriction of to . Note that and hence . Next we observe that we do not need to assume that the measure is symmetric since we use a modified version of Lemma 2.1 of [24] given by Lemma A.3 (Appendix A).
In fact, since the tail of vanishes if , we can obtain another estimate for the tail of which, together with (90), will allow us to control its th moment for . This new estimate is given below.
Lemma 16. If , then If , then for all .
Proof. We use the same idea as in Example 3.7 of [1]. For each , let be a random variable with characteristic function:
Since converges in distribution to , it suffices to prove the lemma for . Let be the restriction of to . Since is finite, has a compound Poisson distribution with
where denotes the -fold convolution. Note that
where are i.i.d. random variables with law .
Assume first that . To compute we consider the intersection with the event and its complement. Note that for any . Using this fact and Markov’s inequality, we obtain that, for any ,
Note that if . Hence, for any
Combining all these facts, we get that for any
and the conclusion follows from (94).
Assume now that . In this case, since has a symmetric distribution. Using Chebyshev’s inequality this time, we obtain
The result follows as above using the fact that, for any ,
Lemma 17. If then where is a constant depending on and . If , then where is a constant depending on .
Proof. Note that We consider separately the integrals for and . For the first integral we use (90): For the second one we use Lemma 16: if then and if , then
We now proceed to the construction of the stochastic integral with respect to . For this, we use the same method as for . Note that , where is the -field generated by for all and . For any , we will work with a càdlàg modification of the Lévy process .
If is a simple process given by (40), we define by the same formula (44) with replaced by . The following result shows that has the same tail behavior as .
Proposition 18. If is a bounded simple process then for any and , where is a constant depending only on .
Proof. As in the proof of Theorem 13, it is enough to prove that where . This reduces to showing that satisfies an inequality similar to (57) for any ; that is, for any , for some . We first examine the tail of . By (90), where . Letting , we obtain that, for any , By Lemma A.3 (Appendix A), it follows that, for any , for any sequence of real numbers. Inequality (110) (with ) follows by applying this to .
In view of the previous result and Proposition 12, for any process , we can construct the integral in the same manner as , and this integral satisfies (108). If in addition the process satisfies (83), then we can define the integral for an arbitrary Borel set (possibly ). This integral will satisfy an inequality similar to (108) with replaced by .
The appealing feature of is that we can control its moments, as shown by the next result.
Theorem 19. If , then for any and for any , for any and , where is a constant depending on , . If is an arbitrary Borel set and we assume, in addition, that the process satisfies then inequality (115) holds with replaced by .
Proof. Consider the following steps.
Step 1. Suppose that is an elementary process of the form (39). Then where . Note that is independent of . Hence, is independent of . Let denote the law of . By Fubini’s theorem,
We evaluate the inner integral. We split this integral into two parts, for and , respectively. For the first integral, we use (90). For the second one, we use Lemma 16. Therefore, the inner integral is bounded by
Step 2. Suppose now that is a simple process of the form (40). Then where .
Using the linearity of the integral, the inequality , and the result obtained in Step 1 for the elementary processes , we get
Step 3. Let be arbitrary. By Proposition 12, there exists a sequence of bounded simple processes such that . Since , it follows that . By the definition of there exists a subsequence such that converges to a.s. Using Fatou’s lemma and the result obtained in Step 2 (for the simple processes ), we get
Step 4. Suppose that satisfies (116). Let where is an increasing sequence of sets in such that . By the definition of , there exists a subsequence such that converges to a.s. Using Fatou’s lemma, the result obtained in Step 3 (for ) and the monotone convergence theorem, we get
Remark 20. Finding a similar moment inequality for the cases and remains an open problem. The argument used in Step 2 above relies on the fact that . Unfortunately, we could not find another argument to cover the case .
5.2. The Case
In this case, the construction of the integral with respect to relies on an integral with respect to which exists in the literature. We recall briefly the definition of this integral. For more details, see Section 1.2.2 of [6], Section 24.2 of [25], or Section 8.7 of [12].
Let endowed with the measure and let be the class of bounded Borel sets in . For a simple process , the integral is defined in the usual way, for any , . The process is a (càdlàg) zero-mean square-integrable martingale with quadratic variation and predictable quadratic variation By approximation, this integral can be extended to the class of all -measurable processes such that for any and The integral is a martingale with the same quadratic variations as above and has the isometry property: . If, in addition, , then the integral can be extended to . By the Burkholder-Davis-Gundy inequality for discontinuous martingales, for any ,
The previous inequality is not suitable for our purposes. A more convenient inequality can be obtained for another stochastic integral, constructed for fixed, as suggested on page 293 of [6]. More precisely, one can show that, for any bounded simple process , where is the constant appearing in (125) (see Lemma 8.22 of [12]).
By the usual procedure, the integral can be extended to the class of all -measurable processes such that . The integral is defined as an element in the space and will be denoted by Its appealing feature is that it satisfies inequality (126).
From now on, we fix . Based on (88), for any , we let for any predictable process for which the rightmost integral is well defined. Letting , we see that this is equivalent to saying that and . By (126), where . If, in addition, the process satisfies (116) then (129) holds with replaced by , for an arbitrary Borel set .
Note that (129) is the counterpart of (115) for the case . Together, these two inequalities will play a crucial role in Section 6.
Table 1 summarizes all the conditions.
6. The Main Result
In this section, we state and prove the main result regarding the existence of a mild solution of (1). For this result, is a bounded domain in . For any , we denote
Theorem 21. Let , . Assume that for any for some if , or for some if . Then (1) has a mild solution. Moreover, there exists a sequence of stopping times with a.s. such that, for any and ,
Example 22 (heat equation). Let . Then where is the fundamental solution of on . Condition (133) holds if . If , this condition holds for any . If , this condition holds for any , as long as satisfies (6). Conditions (131) and (132) hold by the continuity of the function in and , by applying the dominated convergence theorem. To justify the application of this theorem, we use the trivial bound for both and , which introduces the extra condition . Unfortunately, we could not find another argument for proving these two conditions (In the case of the heat equation on , Lemmas A.2 and A.3 of [6] estimate the integrals appearing in (132) and (131), with in (131). These arguments rely on the structure of and cannot be used when is a bounded domain.).
Example 23 (parabolic equations). Let where is given by (31). Assuming (32), we see that (133) holds if . The same comments as for the heat equation apply here as well (Although in a different framework, a condition similar to (131) was probably used in the proof of Theorem 12.11 of [12] (page 217) for the claim . We could not see how to justify this claim, unless .).
Example 24 (heat equation with fractional power of the Laplacian). Let for some . By Lemma B.23 of [12], if , then condition (133) holds for any , provided that satisfies (36) (This condition is the same as in Theorem 12.19 of [12], which examines the same equation using the approach based on Hilbert-space valued solution.).
To verify conditions (131) and (132), we use the continuity of in and and apply the dominated convergence theorem. To justify the application of this theorem, we use the trivial bound for both and , which introduces the extra condition . This bound can be seen from (33), using the fact that where and are the fundamental solutions of on and , respectively. (In the case of the same equation on , elementary estimates for the time and space increments of can be obtained directly from (35), as on page 196 of [26]. These arguments cannot be used when is a bounded domain.)
The remaining part of this section is dedicated to the proof of Theorem 21. The idea is to solve first the equation with the truncated noise (yielding a mild solution ) and then identify a sequence of stopping times with a.s. such that, for any , , and , a.s. on the event . The final step is to show that process defined by on is a mild solution of (1). A similar method can be found in Section 9.7 of [12] using an approach based on stochastic integration of operator-valued processes, with respect to Hilbert-space-valued processes, which is different from our approach.
Since is a Lipschitz function, there exists a constant such that In particular, letting , we have
For the proof of Theorem 21, we need a specific construction of the Poisson random measure , taken from [13]. We review briefly this construction.
Let be a partition of with sets in and let be a partition of such that for all . We may take for all . Let be independent random variables defined on a probability space , such that where . Let for all . Then is a Poisson random measure on with intensity .
This section is organized as follows. In Section 6.1 we prove the existence of the solution of the equation with truncated noise . Sections 6.2 and 6.3 contain the proof of Theorem 21 when and , respectively.
6.1. The Equation with Truncated Noise
In this section, we fix and we consider the equation: with zero initial conditions and Dirichlet boundary conditions. A mild solution of (139) is a predictable process which satisfies (2) with replaced by . For the next result, can be a bounded domain in or (with no boundary conditions).
Theorem 25. Under the assumptions of Theorem 21, (139) has a unique mild solution . For any , and the map is continuous from into .
Proof. We use the same argument as in the proof of Theorem 13 of [27], based on a Picard iteration scheme. We define and
for any . We prove by induction on that (i) is well defined; (ii) for any ; (iii) is -measurable for any and ; (iv) the map is continuous from into for any .
The statement is trivial for . For the induction step, assume that the statement is true for . By an extension to random fields of Theorem 30, Chapter IV of [28], has a jointly measurable modification. Since this modification is -adapted (in the sense of (iii)), it has a predictable modification (using an extension of Proposition 3.21 of [12] to random fields). We work with this modification, that we call also .
We prove that (i)–(iv) hold for . To show (i), it suffices to prove that , where . By (136) and (133),
In addition, if , we have to prove that satisfies (83) if , or (116) if (see Table 1). If , this follows as above, since and hence ; the argument for is similar.
Combined with the moment inequality (115) (or (129)), this proves (ii), since
for any . Property (iii) follows by the construction of the integral .
To prove (iv), we first show the right continuity in . Let . Writing the interval as the union of and , we obtain that , where
Using again (136) and the moment inequality (115) (or (129)), we obtain
It follows that both and converge to as , using (131) for and the Dominated Convergence Theorem and (133) for , respectively. The left continuity in is similar, by writing the interval as the difference between and for . For the continuity in , similarly as above, we see that is bounded by
which converges to as due to (132). This finishes the proof of (iv).
We denote . Similarly to (143), we have
where . By applying Lemma 15 of Erratum to [27] with , , , and , we obtain that
We now prove that converges in , uniformly in . To see this, let for . Using the moment inequality (115) (or (129)) and (135), we have
where . By Lemma 15 of Erratum to [27], converges uniformly on (Note that this lemma is valid for all .).
We denote by the limit of in . One can show that satisfies properties (ii)–(iv) listed above. So has a predictable modification. This modification is a solution of (139). To prove uniqueness, let be another solution and denote . Then
Using (133), it follows that for all .
6.2. Proof of Theorem 21: Case
In this case, for any and , we have (see (21)) The characteristic function of is given by Note that is not a compound Poisson process since is infinite.
We introduce the stopping times , as on page 239 of [13]: where . Clearly, for all .
We first investigate the relationship between and and the properties of . Using construction (138) of and definition (87) of , we have
We observe that is a compound Poisson process with
Note that means that all the jumps of in are smaller than in modulus; that is, for all for which and . Hence, on , for any , , and with . Using an approximation argument and the construction of the integrals and , it follows that, for any and for any , a.s. on , we have
The next result gives the probability of the event .
Lemma 26. For any and , Consequently, and a.s.
Proof. Note that , where
Since and are independent, it is enough to prove that, for any ,
Note that for all for which and and are the jump times of a Poisson process with intensity . Hence,
which yields (160).
To prove the last statement, let . Then for any , and hence . Hence, with probability , for any , there exists some such that . Since is nondecreasing, this proves that with probability .
Remark 27. The construction of given above is due to [13] (in the case of a symmetric measure ). This construction relies on the fact that is a bounded set. Since (and consequently ) is not well defined, we could not see why this construction can also be used when , as it is claimed in [13]. To avoid this difficulty, one could try to use an increasing sequence of sets in with . Using (157) with and letting , we obtain that a.s. on , where . But for any , which means that a.s. Finding a suitable sequence of stopping times which could be used in the case remains an open problem.
In what follows, we denote . Let be the solution of (139), whose existence is guaranteed by Theorem 25.
Lemma 28. Under the assumptions of Theorem 21, for any , , and ,
Proof. By the definition of and (157),
a.s. on the event . Using the definition of and Proposition C.1 (Appendix C), we obtain that, with probability ,
Let . Using the moment inequality (115) and the Lipschitz condition (135), we get
where . Using (133), it follows that for all .
For any and , let , where and are positive integers. Let .
The next result concludes the proof of Theorem 21.
Proposition 29. Under the assumptions of Theorem 21, the process defined by is a mild solution of (1).
Proof. We first prove that is predictable. Note that
The process is clearly predictable, being in the class defined in Remark 11. By the definition of , since are predictable, it follows that is -measurable. Hence, is predictable.
We now prove that satisfies (2). Let and be arbitrary. Using (157) and Proposition C.1 (Appendix C), with probability , we have
For the second last equality, we used the fact that processes and are modifications of each other (i.e., a.s. for all , ), and, hence, and a.s. The conclusion follows letting , since a.s.
6.3. Proof of Theorem 21: Case
In this case, for any and , we have (see (22))
To introduce the stopping times we use the same idea as in Section 9.7 of [12].
Let and , where was defined in Section 2. Note that is a zero-mean square-integrable martingale and is a compound Poisson process with where . With this notation,
We let , where and . Recalling definition (88) of , it follows that
For any , we let where .
Lemma 26 holds again, but its proof is simpler than in the case , since is a compound Poisson process. By (138), Hence, on , for any , , and with ,
Let . Using (170) and (172), it follows that for any , , and with . Let be fixed. Using an approximation argument and the construction of the integrals and , it follows that, for any and for any , a.s. on , we have
We denote . We consider the following equation: with zero initial conditions and Dirichlet boundary conditions. A mild solution of (178) is a predictable process which satisfies for any , . The existence and uniqueness of a mild solution of (178) can be proved similarly to Theorem 25. We omit these details. We denote this solution by .
Lemma 30. Under the assumptions of Theorem 21, for any , , and ,
Proof. By the definition of and (177), a.s. on the event , is equal to
Using the definition of and Proposition C.1 (Appendix C), we obtain that, with probability ,
Letting , we see that where
We estimate separately the two terms. For the first term, we use the moment inequality (129) and the Lipschitz condition (135). We get
where . For the second term, we use Hölder’s inequality with and , where . Hence,
where (Since is a bounded set, where is a constant depending on and . Since , by (133). This shows that .). Therefore,
where . From (184) and (186), we obtain that
where . This implies that for all .
For any and , we let where and are positive integers, and . By Lemma 30, .
Proposition 31. Under the assumptions of Theorem 21, the process defined by is a mild solution of (1).
Proof. We proceed as in the proof of Proposition 29. In this case, with probability , we have The conclusion follows letting , since a.s. and .
Appendices
A. Some Auxiliary Results
The following result is used in the proof of Theorem 13.
Lemma A.1. If has a distribution then where is a constant depending only on .
Proof. Consider the following steps.
Step 1. We first prove the result for . We treat only the right tail, with the left tail being similar. We denote by to emphasize the dependence on . By Property 1.2.15 of [18], , where . We use the fact that, for any ,
for some (see Property of [18] or Section 1.5 of [29]). Since , there exists such that
It follows that for all and . Clearly, for all and , .
Step 2. We now consider the general case. Since has a distribution, by Step 1, it follows that for any . The conclusion follows multiplying by .
In the proof of Theorem 13 and Lemma A.3 below, we use the following remark, due to Adam Jakubowski (personal communication).
Remark A.2. Let be a random variable such that for all , for some and . Then, for any ,
The next result is a generalization of Lemma 2.1 of [24] to the case of nonsymmetric random variables. This result is used in the proof of Lemma 15 and Proposition 18.
Lemma A.3. Let be independent random variables such that for some and . If , we assume that for all , and, if , we assume that has a symmetric distribution for all . Then for any sequence of real numbers, we have where is a constant depending only on .
Proof. We consider the intersection of the event on the left-hand side of (A.6) with the event and its complement. Hence,
Using (A.5), we have . To treat , we consider 3 cases.
Case 1 . By Markov’s inequality and Remark A.2, we have
Case 2 . Let . Since ,
where we used Remark A.2 for the last inequality. From here, we infer that
where . By Chebyshev’s inequality, for any ,
using Remark A.2 for the last inequality. On the other hand, if ,
Case 3 . Since has a symmetric distribution, we can use the original argument of [24].
B. Fractional Power of the Laplacian
Let be the fundamental solution of on , .
Lemma B.1. For any , there exist some constants , depending on , , and such that
Proof. The upper bound is given by Lemma B.23 of [12]. For the lower bound, we use the scaling property of the functions . We have and hence
C. A Local Property of the Integral
The following result is the analogue of Proposition 8.11 of [12].
Proposition C.1. Let and be a Borel set. Let be a predictable process such that if , or for some if . If is unbounded, assume in addition that satisfies (83) if , or satisfies (116) for some , if . Suppose that there exists an event such that Then for any , a.s. on .
Proof. We only prove the result for , with the proof for being the same. Moreover, we include only the argument for ; the case is similar. The idea is to reduce the argument to the case when is a simple process, as in the proof Proposition of 8.11 of [12].
Step 1. We show that the proof can be reduced to the case of a bounded set . Let where and is an increasing sequence of sets in such that . Then satisfies (C.1). By the dominated convergence theorem,
By the construction of the integral, a.s. for a subsequence . It suffices to show that a.s. on for all . But and is bounded.
Step 2. We show that the proof can be reduced to the case of a bounded processes. For this, let . Clearly, is bounded and satisfies (C.1) for all . By the dominated convergence theorem, , and hence a.s. for a subsequence . It suffices to show that a.s. on for all .
Step 3. We show that the proof can be reduced to the case of bounded continuous processes. Assume that is bounded and satisfies (C.1). For any and , we define
where . Clearly, is bounded and satisfies (C.1). We prove that . Since is bounded, . To prove that is predictable, we consider
Since is predictable, it is progressively measurable; that is, for any , the map from to is -measurable. Hence, is -measurable for any . Since the map is left continuous for any , it follows that is predictable, being in the class defined in Remark 11. Hence, is predictable, being a sum of terms involving .
Since is continuous in , is continuous in . By Lebesgue differentiation theorem in , for any , and . By the bounded convergence theorem, . Hence, a.s. for a subsequence . It suffices to show that a.s. on for all .
Step 4. Assume that is bounded, continuous, and satisfies (C.1). Let be a partition of in Borel sets with Lebesgue measure smaller than . Let be arbitrary. Define
Since is continuous in , . By the bounded convergence theorem, , and hence a.s. for a subsequence . Since on the event ,
it follows that a.s. on .
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The author is grateful to Robert Dalang for suggesting this problem. This research is supported by a grant from the Natural Sciences and Engineering Research Council of Canada.