This is the accepted version of the journal article: Bardina i Simorra, Xavier; Márquez, Juan Pablo; Quer i Sardanyons, Lluís. «Weak approximation of the complex Brownian sheet from a Lévy sheet and applications to SPDEs». Stochastic Processes and their Applications, Vol. 130, Issue 9 (September 2020), p. 5735-5767. DOI 10.1016/j.spa.2020.04.006 This version is available at https://ddd.uab.cat/record/274609 under the terms of the license Weak approximation of the complex Brownian sheet from a Lévy sheet and applications to SPDEs ∗ ∗ ∗† Xavier Bardina Juan Pablo Márquez Lluı́s Quer-Sardanyons April 11, 2020 Abstract We consider a Lévy process in the plane and we use it to construct a family of complex-valued random fields that we show to converge in law, in the space of continuous functions, to a complex Brownian sheet. We apply this result to obtain weak approximations of the random field solution to a semilinear one-dimensional stochastic heat equation driven by the space-time white noise. 2000 Mathematics Subject Classification: 60F17; 60G15; 60H15. Key words and phrases: Brownian sheet; Lévy sheet; stochastic heat equation; weak approximation. 1 Introduction Let {N (x, y); x, y ≥ 0} be a Poisson process in the plane and S, T > 0. For any ε > 0, define the following random field: ∫ εt ∫ εs √ xε (s, t) := ε xy (−1)N (x,y) dxdy, (s, t) ∈ [0, S] × [0, T ]. (1) 0 0 Then, in [2] (see Theorem 1.1 therein) the authors proved that, as ε tends to zero, xε converges in law, in the Banach space C([0, S] × [0, T ]) of continuous functions, to the Brownian sheet on [0, S] × [0, T ]. It is worth mentioning that this result was motivated by its one-dimensional counterpart, which was proved by Stroock in [12] and says the following: the family of processes ∫ t ε2 yε (t) := ε (−1)N (s) ds, ε > 0, 0 where N denotes a standard Poisson process, converges in law, in the space of continuous functions, to a standard Brownian motion. Note that this kind of processes had already been used by Kac in [9] in order to express the solution of the telegrapher’s equation in terms of a Poisson process. In the present paper, we aim to extend the above result of [2] to the case where the Poisson process is replaced by a Lévy sheet {L(x, y); x, y ≥ 0} (see Section 2 for the precise definition). Indeed, note that expression (−1)N (x,y) can be written in terms of the complex exponential as eiπN (x,y) . Hence, when ∗ Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra (Barcelona), Catalonia. E-mail addresses: bardina@mat.uab.cat, juanpablo@mat.uab.cat, quer@mat.uab.cat. Research supported by the grant PGC2018- 097848-B-I00 of the Ministerio de Economı́a y Competitividad. J.P. Márquez was supported by a fellowship of CONACYT- México † Corresponding author. 1 replacing N by L, we will use the form eiπL(x,y) = cos(πL(x, y)) + i sin(πL(x, y)) since the expression (−1)L(x,y) may not be well-defined in R. On the other hand, we will replace π by an arbitrary angle θ ∈ (0, 2π). The main result of the paper is the following: Theorem 1.1. Let {L(x, y); x, y ≥ 0} be a Lévy sheet and Ψ(ξ) := a(ξ)+ib(ξ), ξ ∈ R, its Lévy exponent. Let θ ∈ (0, 2π) and S, T > 0, and define, for any ε > 0 and (s, t) ∈ [0, S] × [0, T ], ∫ t ∫ s ε ε √ Xε (s, t) := εK xy {cos(θL(x, y)) + i sin(θL(x, y))}dxdy, (2) 0 0 where the constant K is given by 1 a(θ)2 + b(θ)2 K=√ . (3) 2 a(θ) Assume that a(θ)a(2θ) ̸= 0. Then, as ε tends to zero, Xε converges in law, in the space of complex-valued continuous functions C([0, S] × [0, T ]; C), to a complex Brownian sheet. We point out that the processes Xε provide a natural extension of the family (1). Nevertheless, the proof of Theorem 1.1 above involves new, and indeed significantly more, technical difficulties compared to that of [2, Thm. 1]. This is obviously due to the presence of the Lévy sheet L, which is a rather general random field that includes, for instance, the Brownian sheet and the isotropic stable sheet (see, e.g., [10, Sec. 2.1]). We recall that, by definition, a complex Brownian sheet is a complex random field whose real and imaginary parts are independent Brownian sheets. Hence, in view of the above theorem, we observe that the real and imaginary parts of Xε are clearly not independent, for any ε > 0, while in the limit they are. This phenomenon is not new, for it already appeared in the study of analogous problems in the one-parameter setting (see, e.g., [1, 4, 11]). Indeed, in [1], a family of processes that converges in law to a complex Brownian motion was constructed from a unique Poisson process. This result was generalized in [4], where the Poisson process was replaced by processes with independent increments whose characteristic functions satisfy some properties. Lévy processes are one of the examples where the latter results may be applied. Finally, the authors of [11] use Poisson and Lévy processes in order to obtain approximations in law of a complex fractional Brownian motion. The main strategy in order to prove the kind of weak convergence stated in Theorem 1.1 consists in proving that the underlying family of laws is relatively compact in the space of continuous functions (with the usual topology). By Prohorov’s theorem, this is equivalent to proving the tightness property of this family of laws. Next, we will check that every weakly convergent partial sequence converges to the limit law that we want to obtain. In the last part of the paper (see Section 5), we consider the following semilinear stochastic heat equation driven by the space-time white noise: ∂U ∂2U (t, x) − (t, x) = b(U (t, x)) + Ẇ (t, x), (t, x) ∈ [0, T ] × [0, 1], (4) ∂t ∂x2 where T > 0 and b is a globally Lipschitz function. We impose some initial datum and Dirichlet bound- ary conditions. In Theorem 5.1 below, we will prove that the random field solution U of (4) can be approximated in law, in the space of continuous functions, by a sequence of random fields {Uε }ε , where Uε is the mild solution to a stochastic heat equation like (4) but driven by either the real or imaginary part of the noise Xε . This result provides an example of a kind of weak continuity phenomenon in the path space, where convergence in law of the noisy inputs implies convergence in law of the corresponding solutions. Another example of this fact was provided by Walsh in [14], where a parabolic stochastic partial differential equation was used to model a discontinuous neurophysiological phenomenon. The proof of Theorem 5.1 will follow from [3, Thm. 1.4]. More precisely, Theorem 1.4 of [3] establishes sufficient conditions on a family of random fields that approximate the Brownian sheet (in some sense) 2 under which the solutions of (4) driven by this family converges in law, in the space of continuous functions, to the random field U . We refer to Section 5 for the precise statement of the above-mentioned conditions. In [3], the authors apply their main result to two important families of random fields that approximate the Brownian sheet: the Donsker kernels in the plane and the Kac-Stroock processes, where the latter are defined by √ √ √ θn (t, x) := n tx (−1)N ( n t, n x) , where N denotes a standard Poisson process in the plane (indeed, this case corresponds to (1)). As it will be exhibited in Section 5, the proof of Theorem 5.1 is strongly based on the treatment of the Kac-Stroock processes in [3] (see Section 4 therein), and also on some technical estimates contained in the proof of the tightness result given in Proposition 3.1 of the present paper. Eventually, we note that the kind of convergence results that are obtained in the present paper assure that the limit processes, which in our case correspond to the complex Brownian sheet and the solution to the stochastic heat equation, are robust when used as models in practical situations. Moreover, the obtained results provide expressions that can be useful to study simulations of these limit processes. The paper is organized as follows. Section 2 contains some preliminaries on two-parameter random fields and the definition of Lévy sheet. Section 3 is devoted to prove that the family of laws of (Xε )ε>0 is tight in the space of complex-valued continuous functions. The limit identification is addressed in Section 4. Finally, the result on weak convergence for the stochastic heat equation is obtained in Section 5. 2 Preliminaries Let (Ω, F, P) be a complete probability space. We will use some notation introduced by Cairoli and Walsh in [6]. Namely, let {Fs,t ; (s, t) ∈ [0, S] × [0, T ]} be a family of sub-σ-algebras of F satisfying: (i) Fs,t ⊂ Fs′ ,t′ , for all s ≤ s′ and t ≤ t′ . (ii) All zero sets of F are contained in F0,0 . (iii) For any z ∈ [0, S] × [0, T ], Fz = ∩z 0 whenever σ > 0 and/or η is nontrivial. 3 Tightness This section is devoted to prove that the family of probability laws of {Xε }ε>0 is tight in C([0, S]×[0, T ]; C). This will be a consequence of the next result and the tightness criterion [5, Thm. 3] (see also [7]), taking in the account that our processes Xε vanish on the axes. Proposition 3.1. Let {Xε }ε>0 be the family of random fields defined by (2). There exists a positive constant C such that, for all (0, 0) ≤ (s, t) < (s′ , t′ ) ≤ (S, T ), [ ] sup E |∆s,t Xε (s′ , t′ )| ≤ C(s′ − s)2 (t′ − t)2 . 4 ε>0 This implies that the the family of probability laws of (Xε )ε>0 is tight in C([0, S] × [0, T ]; C). 4 Proof. By definition of Xε and the properties of the modulus | · |, we have [ ] E |∆s,t Xε (s′ , t′ )| 4   ∫ tε′ ∫ sε′ 4 √ = ε4 K 4 E  xy {cos(θL(x, y)) + i sin(θL(x, y))}dxdy  t s ε ε   ∫ t′ ∫ s′ 4 ε ε √ = ε4 K 4 E  xy eiθL(x,y) dxdy  t s ε ε ( )2  ∫ t′ ∫ s′ ∫ t′ ∫ s′ ε ε √ ε ε √ = ε4 K 4 E  x1 y1 eiθL(x1 ,y1 ) dx1 dy1 x2 y2 e−iθL(x2 ,y2 ) dx2 dy2  t s t s ε ε ε ε ( )2  ∫ t′ ∫ s′ ∫ t′ ∫ s′ ε ε ε ε √ = ε4 K 4 E  x1 x2 y1 y2 eiθ(L(x2 ,y2 )−L(x1 ,y1 )) dx1 dx2 dy1 dy2  t s t s ε ε ε ε [∫ t′ ∫ s′ ∫ t′ ∫ s′ ε ε ε ε √ √ = ε4 K 4 E ... x1 x2 x3 x4 y1 y2 y3 y4 t s t s ] ε ε ε ε ×e iθ(L(x4 ,y4 )−L(x3 ,y3 )+L(x2 ,y2 )−L(x1 ,y1 )) dx1 . . . dx4 dy1 . . . dy4 . ∑4 j ∑4 j Taking into account that we can write eiθ j=1 (−1) L(xj ,yj ) = eiθ j=1 (−1) ∆0,0 L(xj ,yj ) , we obtain [ ] ∫ t′ ∫ s′ ∫ t′ ∫ s′ ε ε ε ε √ √ E |∆s,t Xε (s′ , t′ )| = ε4 K 4 4 ... x1 x2 x3 x4 y1 y2 y3 y4 t s t s [ ∑4 j ε ε ] ε ε × E eiθ j=1 (−1) ∆0,0 L(xj ,yj ) dx1 . . . dx4 dy1 . . . dy4 . In order to estimate the expectation inside the above term, we need to consider all 24 possible orders of the x-variables and all 24 possible orders of the y-variables. Altogether, this amounts to take into account 576 possibilities. Let P4 be the group of permutations of degree 4. Then, [ ] E |∆s,t Xε (s′ , t′ )| 4 ∑ ∑ ∫ t′ ∫ s′ ∫ t′ ∫ s′ [ ∑4 ] ε ε ε ε √ √ j = 4 ε K 4 ··· x1 x2 x3 x4 y1 y2 y3 y4 E eiθ j=1 (−1) ∆0,0 L(xj ,yj ) t s t s σ∈P4 β∈P4 ε ε ε ε × I{xσ(1) 0 be the family of probability laws in C([0, S] × [0, T ]; C) corresponding to {Xε }ε>0 . By Proposition 3.1, there exists a subsequence {Pεn }n≥1 of {Pε }ε>0 converging, in the weak sense in the space C([0, S] × [0, T ]; C), to some probability measure P. This section is devoted to prove that P is the law of a complex Brownian sheet, that is a random field whose real and imaginary parts are independent Brownian sheets. We will use the following characterization of the complex Brownian sheet. It is an adaptation of the characterization of the (real-valued) Brownian sheet given in [2, Thm. 4.1] (other characterizations of Brownian sheet can be found, e.g., in [8, 13]). Theorem 4.1. Let X = {X(s, t); (s, t) ∈ [0, S] × [0, T ]} be a complex-valued and continuous process such that X(s, 0) = X(0, t) = 0 for all s ∈ [0, S] and t ∈ [0, T ]. We write X = X 1 + iX 2 . Let {Fs,t ; (s, t) ∈ [0, S] × [0, T ]} be the natural filtration associated to X. Then, the following statements are equivalent: (i) X is a complex Brownian sheet. (ii) X 1 and X 2 are strong martingales and, for all (0, 0) < (s, t) ≤ (s′ , t′ ) ≤ (S, T ) and i = 1, 2, [( )2 ] E ∆s,t X i (s′ , t′ ) |Fs,T = (s′ − s)(t′ − t), (10) [( )( ) ] E ∆s,t X 1 (s′ , t′ ) ∆s,t X 2 (s′ , t′ ) |Fs,T = 0. (11) Proof. It is obvious that (i) implies (ii) because, for all (0, 0) < (s, t) ≤ (s′ , t′ ) ≤ (S, T ) and i = 1, 2, it holds E[∆s,t X i (s′ , t′ )|FS,t ∨ Fs,T ] = E[∆s,t X i (s′ , t′ )] = 0 and [( )2 ] [ [( )2 ] ] E ∆s,t X i (s′ , t′ ) |Fs,T = E E ∆s,t X i (s′ , t′ ) |FS,t ∨ Fs,T |Fs,T = (s′ − s)(t′ − t). We will prove now that (ii) implies (i). First, we check that X 1 and X 2 define (real-valued) Brownian sheets. We will only write the proof for X 1 , because for X 2 it is exactly the same. Fix (0, 0) < (s, t) < (s′ , t′ ) ≤ (S, T ) and define the process Y = {Yu , u ∈ [s, S]} by Yu := ∆s,t X 1 (u, t′ ) (note that, indeed, it does not depend on s′ ). This process is a martingale with respect to the filtration {Fu,T , u ∈ [s, S]}, since it is adapted to it and, for any s ≤ u < u′ ≤ S, E[Yu′ − Yu |Fu,T ] = E[∆u,t X 1 (u′ , t′ )|Fu,T ] [ ] = E E[∆u,t X 1 (u′ , t′ )|Fu,T ∨ FS,t ]|Fu,T = 0, 11 because X 1 is a strong martingale, by hypothesis. On the other hand, note that, by (10), [ ] [ ] E (Yu′ − Yu )2 |Fu,T = E (∆u,t X 1 (u′ , t′ ))2 |Fu,T = (u′ − u)(t′ − t), which implies that the quadratic variation of the process Y is ⟨Y ⟩u = (u − s)(t′ − t). Hence, by Lévy’s characterization theorem, we obtain that Y = {Yu , u ∈ [s, S]} defines a Brownian motion with respect to the filtration {Fu,T , u ∈ [s, S]} and with variance function u 7→ (u − s)(t′ − t). Thus, the increments ∆s,t X 1 (s′ , t′ ) = Ys′ − Ys are normally distributed random variables with mean zero and variance (s′ −s)(t′ −t). Note that, so far, we are assuming that s, t > 0. In the case t = 0 (the case s = 0 can be argued in the same way), we consider the increments ∆s,ε X 1 (s′ , t′ ), with ε > 0. We know that they are Gaussian random variables, and they converge, as ε tends to zero, to ∆s,0 X 1 (s′ , t′ ), which will thus also be a centered Gaussian random variable with variance equal to the corresponding limit of variances, that is (s′ − s)t′ . Finally, observe that the (rectangular) increments of X 1 are independent since those of Y are. Therefore, X 1 defines a Brownian sheet. As we mentioned before, the same argument proves that X 2 also defines a Brownian sheet. Hence, in order to conclude the proof, it remains to check that X 1 and X 2 are independent. By (11), it suffices to verify that any linear combination of X 1 and X 2 defines a Gaussian random variable. For this, we use a similar argument as above, as follows. Let a, b ∈ R (not both equal to zero) and (0, 0) < (s, t) < (s′ , t′ ) ≤ (S, T ), and define Zu := a∆s,t X 1 (u, t′ ) + b∆s,t X 2 (u, t′ ), u ∈ [s, S]. Then, one verifies that the process Z = {Zu , u ∈ [s, S]} is a martingale with respect to the filtration {Fu,T , u ∈ [s, S]} and, owing to (11), it has quadratic variation ⟨Z⟩u = (u − s)(t′ − t)(a2 + b2 ). Lévy’s characterization theorem implies that Z is a Brownian motion with variance function u 7→ (u − s)(t′ − t)(a2 + b2 ). In particular, Zu defines a Gaussian random variable, for all u ∈ [s, S]. As we did above, this fact can be extended to the case where s = 0 and/or t = 0. This concludes the proof. Owing to Theorem 4.1 and Proposition 3.1, the following two propositions will guarantee the validity of (almost all) the statement of Theorem 1.1. Proposition 4.2. Recall that P denotes the weak limit in C([0, S] × [0, T ]; C) of a converging subsequence of the family {Pε }ε>0 . Let X = {X(s, t); (s, t) ∈ [0, S] × [0, T ]} be the corresponding (complex-valued) canonical process and {Fs,t ; (s, t) ∈ [0, S] × [0, T ]} its associated natural filtration. Then, the real and imaginary parts of X define strong martingales under the probability P. Proposition 4.3. Let X be the canonical process defined in the previous proposition. Then, for all (0, 0) < (s, t) ≤ (s′ , t′ ) ≤ (S, T ), it holds: [( )2 ] EP ∆s,t Re(X)(s′ , t′ ) |Fs,T = (s′ − s)(t′ − t), and [( )2 ] EP ∆s,t Im(X)(s′ , t′ ) |Fs,T = (s′ − s)(t′ − t), In the sequel, we will need to compute some limits, and we will use l’Hôpital’s rule in its usual form. However, sometimes it may be long and tedious to check whether we are under the hypotheses of l’Hôpital’s theorem. The following lemma is a version of this result which makes things easier. Its proof is an easy application of the mean value theorem. Lemma 4.4. Suppose that f : [M, ∞) → R is a derivable function, such that f ′ is continuous on [M, ∞), M ≥ 0, and assume that lim f ′ (u) = a < ∞. u→∞ Then, f (u) lim = a. u→∞ u 12 The proof of Proposition 4.2 is based on the following lemma. Lemma 4.5. Let Xε = {Xε (s, t); (s, t) ∈ [0, S] × [0, T ]} be the (complex-valued) random field defined in (2) and {Fs,t ε ; (s, t) ∈ [0, S] × [0, T ]} its natural filtration. Then, for all (0, 0) < (s, t) ≤ (s′ , t′ ) ≤ (S, T ), lim E[∆s,t Xε (s′ , t′ )|FS,t ε ∨ Fs,T ε ] = 0, (12) ε→0 where the limit is understood in L2 (Ω). Proof. We will use the notation Yε := E[∆s,t Xε (s′ , t′ )|FS,t ∨ Fs,T ]. First, note that we can write ( ) ∫ t′ ∫ s′ ε ε √ iθ L( εs ,y)+L(x, εt )−L( εs , εt )+∆ s , t L(x,y) ∆s,t Xε (s′ , t′ ) = εK xy e ε ε dxdy. t s ε ε Thus, we have ∫ t′ ∫ s′ [ iθ∆ L(x,y) ] ε ε √ xy eiθ(L( ε ,y)+L(x, ε )−L( ε , ε )) E e ε , ε s t s t s t Yε = εK dxdy t s ε ε ∫ t′ ∫ s′ ε ε √ xy eiθ(L( ε ,y)+L(x, ε )−L( ε , ε )) e−Ψ(θ)(x− ε )(y− ε ) dxdy s t s t s t = εK t s ε ε and, recalling that Ψ(ξ) = a(ξ) + ib(ξ) is the Lévy exponent, ∫ t′ ∫ t′ ∫ s′ ∫ s′ [ ] [ ] ε ε ε ε √ √ x1 x2 y1 y2 E eiθ(L( ε ,y1 )+L(x1 , ε )−L( ε ,y2 )−L(x2 , ε )) s t s t E |Yε |2 = ε2 K 2 t t s s ε ε ε ε × e−a(θ)( (x1 − εs )(y1 − εt )+(x2 − εs )(y2 − εt ) ) e−ib(θ)((x1 − εs )(y1 − εt )−(x2 − εs )(y2 − εt )) dx dx dy dy . 1 2 1 2 In order to compute the expectation inside the integral, we take into account the possible orders of x1 , x2 and y1 , y2 , respectively, which amounts to consider 4 possibilities. Then, in each case we express the exponent of the complex exponential in the above expectation as a suitable combination of rectangular increments of L, so that we can compute the corresponding expectation thanks to (5). In the resulting four terms, we get rid of the complex exponentials simply by applying the modulus’ triangle inequality and putting the modulus inside the integrals. Using this procedure, we end up with [ ] E |Yε |2 ≤ 2(I1 + I2 ), (13) where ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ √ x1 x2 y1 y2 e−a(θ)((x2 −x1 ) ε +(y2 −y1 ) ε ) t s 2 2 I1 = ε K t t s s ε ε ε ε × e−a(θ)((x1 − ε )(y1 − ε )+(x2 − ε )(y2 − ε )) dx1 dx2 dy1 dy2 s t s t and ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x1 √ √ x1 x2 y1 y2 e−a(θ)((x1 −x2 ) ε +(y2 −y1 ) ε ) t s 2 2 I2 = ε K t t s s ε ε ε ε × e−a(θ)( (x1 − εs )(y1 − εt )+(x2 − εs )(y2 − εt ) ) dx dx dy dy . 2 1 1 2 Applying Fubini theorem, one easily verifies that both I1 and I2 can be bounded by the term ∫ t′ ∫ t′ ∫ s′ ∫ ε ε ε x2 √ √ x1 x2 y1 y2 e−a(θ)((x2 −x1 ) ε +(y2 −y1 ) ε ) e−a(θ)(x2 − ε )(y1 − ε ) dx1 dx2 dy2 dy1 . t s s t 2 2 I=ε K t s s ε y1 ε ε 13 [ ] √ This implies that E |Yε |2 ≤ 4I. Let us check that I converges to zero as ε → 0. Indeed, bounding x1 √ √ √ t′ by x2 and y2 by ε , then integrating with respect to x1 and y2 , and finally applying a change of variables, we can infer that ∫ t′ ∫ t′ ∫ s′ ∫ √ ε ε ε x2 √ x2 y1 e−a(θ)((x2 −x1 ) ε +(y2 −y1 ) ε ) e−a(θ)(x2 − ε )(y1 − ε ) dx1 dx2 dy2 dy1 t s s t I ≤ Cε ε t s s ε y1 ε ε ∫ t′ ∫ s′ √ ε ε √ x2 y1 e−a(θ)(x2 − ε )(y1 − ε ) dx2 dy1 s t ≤ Cε 3 ε t s ε ε ∫ t′ −t ∫ s′ −s ( √ 3 √ ε ε s) t −a(θ)x2 y1 = Cε ε x2 + y1 + e dx2 dy1 0 0 ε ε ∫ t′ −t ε ∫ s′ −s ε ≤ Cε2 e−a(θ)x2 y1 dx2 dy1 , 0 0 where C is some positive constant whose value may change from line to line. The latter expression converges to zero as ε → 0. In order to see this, e.g., one splits the integral with respect to x2 on the ′ intervals (0, ε) and (ε, s ε−s ). In (0, ε) one bounds the exponential by 1 and then integrate, while in ′ (ε, s ε−s ) one first integrates with respect to y1 and then with respect to x2 . This concludes the proof. We are now in position to prove Proposition 4.2: Proof of Proposition 4.2. First, we check that X(s, 0) = X(0, t) = 0, for all (s, t) ∈ [0, S] × [0, T ]. By Skorohod’s representation theorem, there exist a probability space (Ω̃, F̃, P̃), a sequence {Yε }ε>0 and a random variable Y , all of them taking values in C([0, S] × [0, T ]), such that (a) For all ε > 0, Yε and Xε have the same law, (b) Y and X have the same law, (c) Yε converges to Y , as ε → 0, P̃-almost surely. Condition (c) means that lim sup |Yε (s, t) − Y (s, t)| = 0, P̃-a.s. ε→0 (s,t)∈[0,S]×[0,T ] In particular, for any t ∈ [0, T ], limε→0 Yε (0, t) = Y (0, t), P̃-a.s. By (a), it holds P̃(Yε (0, t) = 0) = Pε (Xε (0, t) = 0) = 1. Hence, letting ε → 0, we get P̃(Y (0, t) = 0) = 1. By (b), we end up with P(X(0, t) = 0) = 1. The same argument let us conclude that P(X(s, 0) = 0) = 1, for all s ∈ [0, S]. The remaining of the proof is similar to that of [2, Prop. 4.2]. Let (0, 0) < (s, t) < (s′ , t′ ) ≤ (S, T ). It suffices to prove that, for any n ≥ 1 and (s1 , t1 ), . . . , (sn , tn ) such that either si ≤ S and ti ≤ t, or si ≤ s and ti ≤ T , i = 1, . . . , n, and for any continuous and bounded function φ : Cn → R, it holds that |EP [φ(X(s1 , t1 ), . . . , X(sn , tn ))∆s,t X(s′ , t′ )]| = 0. We recall that the notation |z| stands for the modulus of z ∈ C. Without any loss of generality, the converging subsequence of probability measures to P will be simply denoted by {Pε }ε>0 . Thus, by Proposition 3.1, it suffices to check that lim |EPε [φ(X(s1 , t1 ), . . . , X(sn , tn ))∆s,t X(s′ , t′ )]| = 0. ε→0 14 For this, we recall that, as in the statement of Lemma 4.5, {Fs,t ε ; (s, t) ∈ [0, S] × [0, T ]} is the natural ε filtration associated to the (complex-valued) random field X introduced in (2). Then, we can argue as follows: |EPε [φ(X(s1 , t1 ), . . . , X(sn , tn ))∆s,t X(s′ , t′ )]| = |E[φ(Xε (s1 , t1 ), . . . , Xε (sn , tn ))∆s,t Xε (s′ , t′ )]| [ ] ≤ E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) E[∆s,t Xε (s′ , t′ )|FS,t ε ∨ Fs,T ε ] ( [ ]) 12 2 ≤ C E E[∆s,t Xε (s′ , t′ )|FS,t ε ∨ Fs,T ε ] . The latter term converges to zero as ε → 0, by Lemma 4.5. In order to prove Proposition 4.3, we need two auxiliary results. The first one is the following. Lemma 4.6. For any (0, 0) ≤ (s, t) ≤ (s′ , t′ ) ≤ (S, T ), it holds: [ ] lim E |∆s,t Xε (s′ , t′ )|2 = 2(s′ − s)(t′ − t). ε→0 Proof. We split the proof in three steps. Step 1. Owing to the definition of Xε (see (2)) and applying Fubini theorem, we have [ ] E |∆s,t Xε (s′ , t′ )|2 ∫ tε′ ∫ tε′ ∫ sε′ ∫ sε′ [ ( )] √ √ = ε2 K 2 x1 x2 y1 y2 E eiθ ∆0,0 L(x2 ,y2 )−∆0,0 L(x1 ,y1 ) dx1 dx2 dy1 dy2 . t t s s ε ε ε ε As in the proof of Lemma 4.5, we need to take into account the possible orders of x1 , x2 and y1 , y2 , respectively. Then, applying also some suitable changes of variables, we have [ ] E |∆s,t Xε (s′ , t′ )|2 ∫ tε′ ∫ y2 ∫ sε′ ∫ x2 √ 2 2 =ε K x1 x2 y1 y2 e−Ψ(θ)((y2 −y1 )x1 +(x2 −x1 )y2 ) dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ y2 ∫ s′ ∫ ε ε x2 √ +ε K2 2 x1 x2 y1 y2 e−Ψ(θ)(x2 −x1 )y1 e−Ψ(−θ)(y2 −y1 )x1 dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ +ε K2 2 x1 x2 y1 y2 e−Ψ(−θ)(x2 −x1 )y1 e−Ψ(θ)(y2 −y1 )x1 dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ +ε K2 2 x1 x2 y1 y2 e−Ψ(−θ)((y2 −y1 )x1 +(x2 −x1 )y2 ) dx1 dx2 dy1 dy2 . (14) t t s s ε ε ε ε Recalling that Ψ(θ) = a(θ) + ib(θ), where a(θ) = a(−θ) and b(θ) = −b(−θ), we observe that e−Ψ(θ)((y2 −y1 )x1 +(x2 −x1 )y2 ) + e−Ψ(θ)(x2 −x1 )y1 e−Ψ(−θ)(y2 −y1 )x1 + e−Ψ(−θ)(x2 −x1 )y1 e−Ψ(θ)(y2 −y1 )x1 + e−Ψ(−θ)((y2 −y1 )x1 +(x2 −x1 )y2 ) = e−a(θ)(x2 y2 −x1 y1 ) 2 cos (b(θ)(x2 y2 − x1 y1 )) + e−a(θ)((y2 −y1 )x1 +(x2 −x1 )y1 ) 2 cos (b(θ)((y2 − y1 )x1 −(x2 − x1 )y1 )) . As a consequence, we can infer that [ ] E |∆s,t Xε (s′ , t′ )|2 = 2(I1ε + I2ε ), (15) 15 where ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ I1ε 2 =ε K 2 x1 x2 y1 y2 e−a(θ)(x2 y2 −x1 y1 ) cos (b(θ)(x2 y2 − x1 y1 )) dx1 dx2 dy1 dy2 (16) t t s s ε ε ε ε and ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ I2ε =ε K 2 2 x1 x2 y1 y2 e−a(θ)((y2 −y1 )x1 +(x2 −x1 )y1 ) t t s s ε ε ε ε × cos (b(θ)((y2 − y1 )x1 + (x2 − x1 )y1 )) dx1 dx2 dy1 dy2 . (17) Step 2. Let us consider the case s = t = 0. In order to deal with I1ε , we make the changes of variables ′ ′ zi := xi yi and vi := sε′ xi , i = 1, 2, and we define u := sε2t . Thus, by l’Hôpital’s rule, we have ∫ 1 ∫ v2 ∫ uv1 √ z1 uv2 −a(θ)uv2 +a(θ)z1 lim I1ε = lim s′ t′ K 2 e cos (b(θ)(uv2 − z1 )) dz1 dv1 dv2 . ε→0 u→∞ 0 0 0 v1 Applying now the changes of variables v2′ := uv2 and v1′ := uv1 , and again l’Hôpital’s rule, we obtain that the latter limit equals to ∫ u ∫ v1′ √ ′ ′ z1 u −a(θ)u+a(θ)z1 lim s t K 2 e cos (b(θ)(u − z1 )) dz1 dv1′ . (18) u→∞ 0 0 v1′ In order to compute the above limit, we use the formula cos(θ) = 12 (eiθ + e−iθ ). Hence, the expression inside the limit (18) can be written as the sum 12 (Au + Bu ), where these terms are given by ∫ u∫ v √ zu −a(θ)u+a(θ)z ib(θ)(u−z) Au := s′ t′ K 2 e e dzdv, 0 0 v ∫ u∫ v √ ′ ′ 2 zu −a(θ)u+a(θ)z −ib(θ)(u−z) Bu := s t K e e dzdv. 0 0 v We will only deal with limu→∞ Bu , because limu→∞ Au can be treated in a similar way. Indeed, rewriting Bu as ∫ u ∫ v √z (a(θ)+ib(θ))z ′ ′ 2 0 0 v e dzdv Bu = s t K − 1 u e 2 (a(θ)+ib(θ))u and applying l’Hôpital’s rule twice, one easily proves that s′ t′ K 2 lim Bu = . u→∞ (a(θ) + ib(θ))2 Similarly, one gets s′ t′ K 2 lim Au = . u→∞ (a(θ) − ib(θ))2 Thus, a(θ)2 − b(θ)2 lim I1ε = s′ t′ K 2 . ε→0 (a(θ)2 + b(θ)2 )2 Now, we are going to compute limε→0 I2ε . Recall that the latter term is given in (17). The strategy that we have followed to deal with I1ε cannot be applied here. More precisely, we have not been able to compute the limit of I2ε directly, but we will introduce an auxiliary term which will converge to some quantity, and we will prove that the remainder converges to zero. 16 s′ t′ To start with, we apply the same changes of variables that we performed for I1ε , we set u := ε and apply l’Hôpital’s rule, so limε→0 I2ε equals to ∫ ∫ ∫ √ ( ( z1 v2 )) ( ) 1 v2 uv1 √ z1 a(θ) 2z1 −uv1 − z1vv2 lim K 2 s′ t′ uv2 e 1 cos b(θ) uv1 − dz1 dv1 dv2 . u→∞ 0 0 0 v1 v1 Next, we make the changes of variables v̄1 := uv1 and v̄2 := uv2 and we apply again l’Hôpital’s rule. Hence, the latter limit becomes ∫ u ∫ v̄1 √ ( ( z1 u )) ( ) √ z1 a(θ) 2z1 −v̄1 − zv̄1 u lim K 2 s′ t′ u e 1 cos b(θ) v̄1 − dz1 dv̄1 . u→∞ 0 0 v̄1 v̄1 z1 v̄1 Finally, performing the changes x := v̄1 and y := u, we end up with lim I2ε = lim Cu , ε→0 u→∞ with ∫ ∫ 2 ′ ′ 2 1 1 √ Cu = K s t u xy ea(θ)(2xy−y−x)u cos(b(θ)(y − x)u)dxdy. (19) 0 0 At this point, we introduce the auxiliary term mentioned above: ∫ 1∫ 1 2 ′ ′ 2 √ a(θ)(2xy−y−x)u C̃u := K s t u ye cos(b(θ)(y − x)u)dxdy, (20) 0 0 √ √ where we note that, compared to the right hand-side of (19), we have only replaced xy by y. For the moment, assume that limu→∞ (Cu − C̃u ) = 0. Let us compute the limit of C̃u , recalling that this term has been defined in (20). As in the analysis of the term I1ε , we use the formula cos(θ) = 12 (eiθ + e−iθ ), so we split C̃u as the sum of two terms (multiplied by 12 ), one of which is given by ∫ ∫ 2 ′ ′ 2 1 1 √ u(2xya(θ)−y(a(θ)−ib(θ))−x(a(θ)+ib(θ)) K stu ye dxdy, 0 0 and the other one is the same with a(θ) + ib(θ) instead of a(θ) − ib(θ). Integrating first with respect to x, applying l’Hôpital’s rule, and doing a change of variables, one gets that the limit of the above term equals to ∫ u √ (a(θ)+ib(θ))y 2 ′ ′( ye dy lim K s t 0 ) . (a(θ)+ib(θ))2 √ u→∞ (a(θ) + ib(θ)) − 2a(θ) u e(a(θ)+ib(θ))u K 2 s′ t′ It is straightforward to check that the latter limit is a(θ)2 +b(θ)2 . The limit of the term involving a(θ)+ib(θ) K 2 s′ t′ will also be given by a(θ)2 +b(θ)2 . Therefore, we have that 1 lim I2ε = lim C̃u = s′ t′ K 2 . ε→0 u→∞ a(θ)2 + b(θ)2 In conclusion, owing to (15) and the expression of K given in (3), the lemma’s statement holds in the case s = t = 0. In order to conclude the present step, we need to check that limu→∞ (Cu − C̃u ) = 0, that is ∫ ∫ 1 1 √ √ lim u2 ( xy − y) ea(θ)(2xy−y−x)u cos(b(θ)(2xy − y − x)u)dxdy = 0. u→∞ 0 0 17 Let us introduce the notation ∫ ∫ 1 1 √ √ Du := u2 ( xy − y) ea(θ)(2xy−y−x)u dxdy. 0 0 Then, it clearly holds that −Du ≤ Cu − C̃u ≤ Du . In order to apply a sandwich type argument, we will prove that both −Du and Du converge to zero as u tends to infinity. We will only tackle the term Du , since the analysis of −Du is analogous. Note that Du = Du1 − Du2 , where ∫ 1∫ 1 ∫ 1∫ 1 √ √ a(θ)(2xy−y−x)u Du1 = u2 xy ea(θ)(2xy−y−x)u dxdy and Du2 = u2 ye dxdy. 0 0 0 0 Regarding Du2 , observe that the integral in x can be computed explicitly and we can argue as follows: ∫ 1 √ ( ) y lim Du2 = lim u ea(θ)(y−1)u − e−a(θ)yu dy u→∞ u→∞ 0 a(θ)(2y − 1) ∫ 1 √y ( a(θ)(y− 1 )u ) − e−a(θ)(y− 2 )u dy 1 1 0 (2y−1) e 2 = lim a(θ) u→∞ a(θ) eu 2 ∫ 1 √ ( a(θ)(y− 1 )u ) u −a(θ)(y− 21 )u 0 y e 2 + e dy = lim a(θ) a(θ) . u→∞ a(θ)eu 2 2eu 2 u − u 2 In the last equality, we have applied l’Hôpital’s rule. By performing a change of variables, the latter expression equals to { ∫ u √ ya(θ) ∫ u √ −ya(θ) } ye dy ye dy lim √ 0 + 0 √ . u→∞ ua(θ)eua(θ) − √u2e ua(θ) u − √u 1 The second term in the above sum clearly converges to zero as u → ∞, while the limit of the first one equals to, thanks to l’Hôpital’s rule, √ a(θ)u 1 ue 1 lim √ √ = . u→∞ a(θ) ea(θ)u (a(θ) u + o( u)) a(θ)2 1 Thus, we have proved that limu→∞ Du2 = a(θ) 1 2 . On the other hand, in order to deal with Du we will use again a sandwich type argument, as follows. First, note that we trivially have Du1 ≤ Du2 . Next, applying the changes of variables v = uy and z = xv, we end up with ∫ u∫ v √ √ z −a(θ)(v+ zu −2z) lim Du1 = lim u e v dzdv u→∞ u→∞ 0 0 v ∫ u∫ v √ √ z −a(θ)(u−z) ≥ lim u e dzdv. u→∞ 0 0 v 1 Observe that the latter limit equals to a(θ)2 because it corresponds to the limit of Bu defined above in ′ ′ the particular case of s = t = K = 1 and b = 0. Hence, we obtain that 1 lim Du1 = u→∞ a(θ)2 and therefore limu→∞ Du = 0. 18 Step 3. Assume that either s ̸= 0 or t ̸= 0. By step 1, recall that we have [ ] E |∆s,t Xε (s′ , t′ )|2 = 2(I1ε + I2ε ), where the terms on the right hand-side have been defined in (16) and (17), respectively. Set ∫ εt ∫ εs ∫ εt ∫ εs ε 2 2 F (s, t) := ε K f (x1 , x2 , y1 , y2 )1{x1 ≤x2 ,y1 ≤y2 } dx1 dy1 dx2 dy2 , 0 0 0 0 √ where f (x1 , x2 , y1 , y2 ) := x1 x2 y1 y2 e−a(θ)(x2 y2 −x1 y1 ) cos(b(θ)(x2 y2 − x1 y1 )), and ∫ t ε ∫ s ε ∫ t ε ∫ s ε ε 2 2 G (s, t) := ε K g(x1 , x2 , y1 , y2 )1{x1 ≤x2 ,y1 ≤y2 } dx1 dy1 dx2 dy2 , 0 0 0 0 √ where g(x1 , x2 , y1 , y2 ) := x1 x2 y1 y2 e−a(θ)((y2 −y1 )x1 +(x2 −x1 )y1 ) cos(b(θ)((y2 − y1 )x1 − (x2 − x1 )y1 )). Ob- serve that I1ε and I2ε can be written as follows: ∫ t′ ε ∫ s′ ε ∫ t′ ε ∫ s ε ′ ′ I1ε = ∆s,t F (s , t ) − ε K ε 2 2 f (x1 , x2 , y1 , y2 )1{y1 ≤y2 } dx1 dy1 dx2 dy2 t s t ε ε ε 0 ∫ t′ ε ∫ s′ ε ∫ t ε ∫ s ε −ε K 2 2 f (x1 , x2 , y1 , y2 )dx1 dy1 dx2 dy2 t s ε ε 0 0 ∫ t′ ε ∫ s′ ε ∫ t ε ∫ s′ ε −ε K 2 2 f (x1 , x2 , y1 , y2 )1{x1 ≤x2 } dx1 dy1 dx2 dy2 t s s ε ε 0 ε =: ∆s,t F ε (s′ , t′ ) − I11 ε − I12 ε − I13 ε , and I2ε = ∆s,t Gε (s′ , t′ ) − I21 ε − I22 ε − I23 ε , ε where I2i , i = 1, 2, 3, are defined analogously by using the function g. By step 2, one verifies that 1 ′ lim ∆s,t F ε (s′ , t′ ) = lim ∆s,t Gε (s′ , t′ ) = (s − s)(t′ − t). ε→0 ε→0 2 In order to conclude the proof, it suffices to check that Iji ε converges to zero as ε → 0, for all j = 1, 2 and i = 1, 2, 3. For this, we estimate any Iji ε by I˜ji ε , where the latter are defined by simply bounding the cosinus by 1. Next, we note that I˜1,i ε ≤ I˜2,i ε , for all i = 1, 2, 3, and that any of the I˜2,i ε can be bounded by ∫ t′ ∫ s ∫ ∫ s′ ε ε y2 ε √ 2 ε K 2 x1 x2 y1 y2 e−a(θ)((y2 −y1 )x1 +(x2 −x1 )y1 ) dx2 dy1 dx1 dy2 . (21) s 0 0 0 ε In this integral, we perform the changes of variables x̄i := εxi and ȳi := εyi , i = 1, 2, we set u := ε12 , we use that x̄2 ≤ s′ and we integrate with respect to x̄2 . Thus, (21) can be bounded, up to some positive constant, by (using again the notation xi and yi for the variables) ∫ t′ ∫ s ∫ y2 √ x1 y2 −a(θ)((y2 −y1 )x1 +(s−x1 )y1 )u u √ e dy1 dx1 dy2 . 0 0 0 y1 Estimating now y2 by t′ inside the square root and integrating in y2 , the above expression can be bounded by (up to some constant) ∫ s ∫ t′ 1 √ e−a(θ)u(s−x1 )y1 dy1 dx1 . 0 0 x 1 y1 This expression converges to zero as u → ∞, by the Monotone convergence theorem. 19 Here is the second auxiliary result needed to prove Proposition 4.3. Lemma 4.7. Let (0, 0) ≤ (s, t) ≤ (s′ , t′ ) ≤ (S, T ). Then, there exists a sequence {Cε }ε>0 such that limε→0 Cε = 4(s′ − s)2 (t′ − t)2 and [( [ ])2 ] E E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε ≤ Cε . Proof. We split the proof in four steps. Step 1. By definition of the random field X ε , we first observe that [ ] E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε ∫ tε′ ∫ sε′ ∫ tε′ ∫ sε′ [ ] √ =K ε 2 2 x1 x2 y1 y2 E eiθ(L(x2 ,y2 )−L(x1 ,y1 )) |Fs,T ε dx1 dx2 dy1 dy2 . t s t s ε ε ε ε In order to compute the above conditional expectation, we have to consider all possible orders of x1 , x2 and y1 , y2 , respectively, which corresponds to a total of 4 possibilities. Hence, ∫ t′ ∫ s′ ∫ ∫ [ ] ε ε y2 x2 √ E |∆s,t X ε (s′ , t′ )|2 |Fs,T x1 x2 y1 y2 eiθ(L( ε ,y2 )−L( ε ,y1 )) s s ε = K 2 ε2 t s t s ε ε ε ε × e−Ψ(θ)((x2 −x1 )y2 +(y2 −y1 )(x1 − ε )) dx1 dy1 dx2 dy2 s ∫ tε′ ∫ sε′ ∫ y2 ∫ x2 √ x1 x2 y1 y2 e−iθ(L( ε ,y2 )−L( ε ,y1 )) s s 2 2 +K ε t s t s ε ε ε ε × e−Ψ(θ)(x2 −x1 )y1 e−Ψ(−θ)(y2 −y1 )(x1 − ε ) dx1 dy1 dx2 dy2 s ∫ tε′ ∫ sε′ ∫ y2 ∫ x2 √ x1 x2 y1 y2 eiθ(L( ε ,y2 )−L( ε ,y1 )) s s 2 2 +K ε t s t s ε ε ε ε −Ψ(−θ)(x2 −x1 )y1 −Ψ(θ)(y2 −y1 )(x1 − εs ) ×e e dx1 dy1 dx2 dy2 ∫ tε′ ∫ s′ ∫ ∫ ε y2 x2 √ x1 x2 y1 y2 e−iθ(L( ε ,y2 )−L( ε ,y1 )) s s + K 2 ε2 t s t s ε ε ε ε × e−Ψ(−θ)((x2 −x1 )y2 +(y2 −y1 )(x1 − ε )) dx1 dy1 dx2 dy2 . s We have also applied changes of variables in order to have x1 ≤ x2 and y1 ≤ y2 in all terms. We denote by Aεi , i = 1, 2, 3, 4, the above four terms, respectively. Thus, we have [( [ ])2 ] ∑ 4 [ ] E E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε = E Aεi Aεj . i,j=1 For the sake of clarity, we will only analyze one of the terms in the above sum, [ since ] the other ones can be treated exactly in the same way. So, we proceed to tackle the term E (Aε1 )2 . In fact, by Fubini theorem, we have that ∫ t′ ∫ s′ ∫ ∫ ∫ t′ ∫ s′ ∫ ∫ [ ] ε ε y4 x4 ε ε y2 x2 √ √ E (Aε1 )2 = K 4 ε4 x1 x2 x3 x4 y1 y2 y3 y4 t s t s t s t s [ ε ε ε ε ] ε ε ε ε × E eiθ(L( ε ,y2 )−L( ε ,y1 )+L( ε ,y4 )−L( ε ,y3 )) s s s s × e−Ψ(θ)((x2 −x1 )y2 +(y2 −y1 )(x1 − ε )+(x4 −x3 )y4 +(y4 −y3 )(x3 − ε )) s s × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 . (22) 20 Note that in the above integral we have y1 ≤ y2 and y3 ≤ y4 . However, in order to compute the expectation in (22), we need to consider all possible orders of the variables y1 , y2 , y3 , y4 , with the restrictions y1 ≤ y2 and y3 ≤ y4 . This amounts to take into account 6 different possibilities, which we split in two groups: (i) y1 ≤ y2 ≤ y3 ≤ y4 and y3 ≤ y4 ≤ y1 ≤ y2 , (ii) y1 ≤ y3 ≤ y2 ≤ y4 , y1 ≤ y3 ≤ y4 ≤ y2 , y3 ≤ y1 ≤ y4 ≤ y2 and y3 ≤ y1 ≤ y2 ≤ y4 . Then, we have that [ ] ∑6 E (Aε1 )2 = Bkε (1, 1), (23) k=1 where B1ε (1, 1), B2ε (1, 1) correspond to (22) with the orders of (i), respectively, while Bkε (1, 1), k = 3, 4, 5, 6, correspond to (22) with [ the ]orders of (ii), respectively. It turns out that we have a similar decomposition of any of the terms E Aεi Aεj , which we denote by [ ] ∑6 E Aεi Aεj = Bkε (i, j). k=1 Hence [( [ ])2 ] 4 ∑ ∑ 6 E E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε = Bkε (i, j). (24) i,j=1 k=1 In[the next ] two steps, we will focus on the analysis of (some[ of) the ] terms in the decomposition (23) of E (Aε1 )2 . As already mentioned, the terms arising from E Aεi Aεj can be treated analogously. We will come back to expansion (24) later in step 4. Step 2. We claim that, for any k = 3, 4, 5, 6, it holds ∫ √ √ |Bkε (1, 1)| ≤ K 4 ε4 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y2 ≤y3 ≤y4 } D × e−a(θ)((x4 −x3 ) ε +(x2 −x1 ) ε +(y4 −y3 ) ε +(y2 −y1 ) ε +(y3 −y2 )(x1 − ε )) t t s s s × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 , (25) ′ ′ where D := [ sε , sε ]4 × [ εt , tε ]4 , and we recall that a(θ) is the real part of Ψ(θ). We prove this estimate for B3ε (1, 1). For the remaining terms the argument is completely analogous. So, let us assume that in (22) we have the order y1 ≤ y3 ≤ y2 ≤ y4 . In this case, the expectation in (22) equals to e−Ψ(θ)((y4 −y2 ) ε +(y3 −y1 ) ε +2(y2 −y3 ) ε ) . s s s Plugging this term in (22) and shifting the modulus inside the integral, we can infer that ∫ √ √ |B3 (1, 1)| ≤ K ε ε 4 4 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y3 ≤y2 ≤y4 } D −a(θ)((x4 −x3 ) εt +(x2 −x1 ) εt +(y4 −y2 ) εs +(y3 −y1 ) εs +(y2 −y1 )(x1 − εs )) ×e × e−a(θ)(2(y2 −y3 ) ε +(y4 −y3 )(x3 − ε )) dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 s s ∫ √ √ ≤ K 4 ε4 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y3 ≤y2 ≤y4 } D −a(θ)((x4 −x3 ) εt +(x2 −x1 ) εt +(y4 −y2 ) εs +(y3 −y1 ) εs +(y2 −y1 )(x1 − εs )) ×e × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 . 21 Performing a change of variable, we obtain that the latter term equals to ∫ √ √ K 4 ε4 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y2 ≤y3 ≤y4 } D × e−a(θ)((x4 −x3 ) ε +(x2 −x1 ) ε +(y4 −y3 ) ε +(y2 −y1 ) ε +(y3 −y1 )(x1 − ε )) t t s s s × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 . In order to obtain (25), it suffices to observe that, in the domain of integration, it holds that (y3 −y1 )(x1 − ε ) ≥ (y3 − y2 )(x1 − ε ). s s Step 3. Here, we prove that the right hand-side of (25) converges to zero as ε → 0. Let us introduce the following notation: ∫ ε 4 √ √ β := ε x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y2 ≤y3 ≤y4 } D × e−a(θ)((x4 −x3 ) ε +(x2 −x1 ) ε +(y4 −y3 ) ε +(y2 −y1 ) ε +(y3 −y2 )(x1 − ε )) t t s s s × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 , so we want to check that limε→0 β ε = 0. To start with, in the expression of β ε we bound the two square roots by using the upper limit of any xi and yi . Next, we integrate with respect to x4 , x3 and x2 . We also use the fact that, according to the statement of Proposition 4.3, we may assume that t > 0. Thus, ∫ s′ ε ∫ t′ ε ∫ y3 ∫ t′ ε ∫ y2 e−a(θ)((y4 −y3 ) ε +(y2 −y1 ) ε +(y3 −y2 )(x1 − ε )) dy1 dy4 dy2 dy3 dx1 . s s s β ≤ Cε ε s t t t ε ε ε y3 ε At this point, we integrate with respect to y1 and y4 , thus ∫ s′ ε ∫ t′ ε ∫ y3 e−a(θ)(y3 −y2 )(x1 − ε ) dy2 dy3 dx1 s β ≤ Cε ε 3 s t t ε ε ε ∫ s′ −s ε ∫ t′ ε ∫ y3 = Cε 3 e−a(θ)(y3 −y2 )x dy2 dy3 dx t t ε ε ε ∫ ε ∫ t′ ε ∫ y3 + Cε 3 e−a(θ)(y3 −y2 )x dy2 dy3 dx. t t 0 ε ε Note that the second term in the latter sum may be bounded, up to some positive constant, by ε2 , which converges to zero. Regarding the first term, it can be bounded by ∫ s′ −s ε 1 Cε 2 dx = Cε2 (ln(s′ − s) − 2 ln(ε)) , ε x which also converges to zero as ε → 0. Step 4. By (24) in step 1 and steps 2 and 3, we have that [( [ ])2 ] ∑ 4 ∑ 6 E E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε = Bkε (i, j) i,j=1 k=1 ∑ 4 ∑ 2 = Bkε (i, j) + ρε , (26) i,j=1 k=1 22 [ ] where we recall that B1ε (i, j) and B2ε (i, j) are the terms in the decomposition of E Aεi Aεj with the orders of (i), respectively, and limε→0 ρε = 0. Focusing again (only) on the case i = j = 1, one easily verifies that ∑ 2 ∫ √ √ Bkε (1, 1) = K 4 ε2 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{{y1 ≤y2 ≤y3 ≤y4 }∪{y3 ≤y4 ≤y1 ≤y2 }} k=1 D −Ψ(θ)((x2 −x1 )y2 +(x4 −x3 )y4 +(y2 −y1 )x1 +(y4 −y3 )x3 ) ×e × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 , ′ ′ where we recall that D := [ sε , sε ]4 × [ εt , tε ]4 . Observing that 1{{y1 ≤y2 ≤y3 ≤y4 }∪{y3 ≤y4 ≤y1 ≤y2 }} ≤ 1{y1 ≤y2 } 1{y3 ≤y4 } , we end up with ∑ 2 ∫ √ √ Bkε (1, 1) ≤K ε 4 2 x1 x2 x3 x4 y1 y2 y3 y4 1{x1 ≤x2 } 1{x3 ≤x4 } 1{y1 ≤y2 } 1{y3 ≤y4 } k=1 D −Ψ(θ)((x2 −x1 )y2 +(x4 −x3 )y4 +(y2 −y1 )x1 +(y4 −y3 )x3 ) ×e × dx1 dy1 dx2 dy2 dx3 dy3 dx4 dy4 . One can get similar estimates for B1ε (i, j) + B2ε (i, j) with i, j ̸= 1. Gathering all the resulting bounds together, it can be verified that [( [ ])2 ] E E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε ≤ Θ2ε + ρε , (27) where ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ Θε = K ε 2 2 x1 x2 y1 y2 e−Ψ(θ)((x2 −x1 )y2 +(y2 −y1 )x1 ) dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ +K ε 2 2 x1 x2 y1 y2 e−Ψ(θ)(x2 −x1 )y1 e−Ψ(−θ)(y2 −y1 )x1 dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ +K ε 2 2 x1 x2 y1 y2 e−Ψ(−θ)(x2 −x1 )y1 e−Ψ(θ)(y2 −y1 )x1 dx1 dx2 dy1 dy2 t t s s ε ε ε ε ∫ t′ ∫ ∫ s′ ∫ ε y2 ε x2 √ +K ε 2 2 x1 x2 y1 y2 e−Ψ(−θ)((x2 −x1 )y2 +(y2 −y1 )x1 ) dx1 dx2 dy1 dy2 . t t s s ε ε ε ε Note that Θε coincides with the right hand-side of equality (14) in the proof of Lemma 4.6, where in the latter it was precisely proved that lim Θε = 2(t′ − t)(s′ − s). ε→0 Therefore, by (27) and recalling that limε→0 ρε = 0, we conclude the proof by taking Cε := Θ2ε + ρε . We can now provide the proof of Proposition 4.3. Proof of Proposition 4.3. We prove that, for all 0 ≤ s1 < · · · < sn ≤ s and 0 ≤ t1 < · · · < tn ≤ T , and any continuous and bounded function φ : C → R, we have [ (( )2 )] EP φ(X(s1 , t1 ), . . . , X(sn , tn )) ∆s,t Re(X)(s′ , t′ ) − (s′ − s)(t′ − t) = 0 23 and [ (( )2 )] EP φ(X(s1 , t1 ), . . . , X(sn , tn )) ∆s,t Im(X)(s′ , t′ ) − (s′ − s)(t′ − t) = 0. Since Pε converges to P weakly in C([0, S] × [0, T ]; C), it suffices to check that lim Aε = 0 and lim Bε = 0, (28) ε→0 ε→0 where [ (( )2 )] Aε := E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) ∆s,t Re(Xε )(s′ , t′ ) − (s′ − s)(t′ − t) and [ (( )2 )] Bε := E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) ∆s,t Im(Xε )(s′ , t′ ) − (s′ − s)(t′ − t) . Indeed, in order to check the validity of the limits in (28), we will prove that lim (Aε + Bε ) = 0 and lim (Aε − Bε ) = 0. ε→0 ε→0 We will first deal with the limit of Aε + Bε . More precisely, we have that [ ( )] Aε + Bε = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) |∆s,t X ε (s′ , t′ )|2 − 2(s − s′ )(t − t′ ) [ ( [ ] )] = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε − 2(s − s′ )(t − t′ ) . [ ] Hence, to prove that limε→0 (Aε + Bε ) = 0, it is enough to check that E |∆s,t X ε (s′ , t′ )|2 |Fs,T ε converges in L2 (Ω) to 2(s − s′ )(t − t′ ), as ε → 0. Indeed, by Lemma 4.7, we have: [( [ ] )2 ] E E |∆s,t X ε (s′ , t′ )|2 |Fs,Tε − 2(s − s′ )(t − t′ ) [ ] ≤ Cε − 4(s − s′ )(t − t′ )E |∆s,t X ε (s′ , t′ )|2 + 4(s − s′ )2 (t − t′ )2 , (29) where limε→0 Cε = 4(s′ − s)2 (t′ − t)2 . So, by Lemma 4.6, the right hand-side of (29) converges to zero as ε → 0. Let us now deal with the limit of Aε − Bε . To start with, note that [ 1 Aε − Bε = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) 2 ( )2 ( )2    ∫ tε′ ∫ sε′ ∫ tε′ ∫ sε′  √ √ . × Kε xy eiθL(x,y) dxdy + Kε xy e−iθL(x,y) dxdy  t s t s  ε ε ε ε We are going to prove that limε→0 Λε = 0, where  ( )2  ∫ t′ ∫ s′ ε ε √ Λε := E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) Kε xy eiθL(x,y) dxdy . t s ε ε The limit involving the complex conjugate e−iθL(x,y) can be tackled using analogous arguments. Expand- ing the squared integral of Λε , we end up with [ Λε = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) ∫ t′ ∫ s′ ∫ t′ ∫ s′ ] ε ε ε ε √ ×K ε 2 2 x1 x2 y1 y2 e iθ(L(x1 ,y1 )+L(x2 ,y2 )) dx1 dx2 dy1 dy2 . t s t s ε ε ε ε 24 As we have already done several times throughout the paper, we consider the four possible orders of x1 , x2 and y1 , y2 and, in each of these terms, we apply a change of variables so that we have x1 ≤ x2 and y1 ≤ y2 . Thus, [ Λε = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) ( ∫ t′ ∫ s′ ∫ ∫ ε ε y2 x2 √ × 2K ε 2 2 x1 x2 y1 y2 eiθ(L(x1 ,y1 )+L(x2 ,y2 )) dx1 dy1 dx2 dy2 t s t s ε ε ε ε ∫ t′ ∫ s′ ∫ ∫ )] 2 2 ε ε y2 x2 √ + 2K ε x1 x2 y1 y2 eiθ(L(x2 ,y1 )+L(x1 ,y2 )) dx1 dy1 dx2 dy2 . t s t s ε ε ε ε At this point, the idea is to write L(x1 , y1 ) + L(x2 , y2 ) and L(x2 , y1 ) + L(x1 , y2 ) as sums of suitable rectangular increments of L (which will be clearly specified in the next equation), and use the property of independent (rectangular) increments of L (see Definition2.1). Proceeding in this way, one obtains that [ ] Λε = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) ei2θL( ε , ε ) s t ( ∫ tε′ ∫ sε′ ∫ y2 ∫ x2 √ [ ] × 2K ε 2 2 x x y y E eiθ(∆x1 ,y1 L(x2 ,y2 )+∆0,y1 L(x1 ,y2 )+∆x1 ,0 L(x2 ,y1 )) 1 2 1 2 t s t s [ )] ε ε ε ε ( i2θ ∆ s , t L(x1 ,y1 )+∆0, t L( εs ,y1 )+∆ s ,0 L(x1 , εt ) ×E e ε ε ε ε dx1 dy1 dx2 dy2 ∫ t′ ∫ s′ ∫ ∫ [ ] ε ε y2 x2 √ + 2K ε 2 2 x1 x2 y1 y2 E eiθ(∆0,y1 L(x1 ,y2 )+∆x1 ,0 L(x2 ,y1 )) t s t s [ε ( ε ε ε )] ) i2θ ∆ s , t L(x1 ,y1 )+∆0, t L( εs ,y1 )+∆ s ,0 L(x1 , εt ) ×E e ε ε ε ε dx1 dy1 dx2 dy2 . Recalling that φ is a bounded function and computing the expectations of complex exponentials in terms of the Lévy exponent Ψ(ξ) = a(ξ) + ib(ξ), one can easily obtain that |Λε | ≤ C Λ̃ε , where C is a positive constant and ∫ t′ ∫ s′ ∫ ∫ ε ε y2 x2 √ Λ̃ε := K ε 2 2 x1 x2 y1 y2 e−a(θ)((y2 −y1 )x1 +(x2 −x1 )y1 ) t s t s ε ε ε ε × e−a(2θ)((y1 − ε ) ε +(x1 − ε )y1 ) dx1 dy1 dx2 dy2 . t s s We finally prove that limε→0 Λ̃ε = 0. Indeed, taking into account the integration limits of all variables and applying Fubini theorem, we have ∫ t′ ε ∫ s′ ε ∫ t′ ε ∫ s′ ε e− min(a(θ),a(2θ))((y2 − ε ) ε +(x2 − ε ) ε ) dx2 dy2 dx1 dy1 . t s s t Λ̃ε ≤ C t s ε ε y1 x1 We integrate with respect to x2 and y2 , thus ∫ t′ ε ∫ s′ ε e− min(a(θ),a(2θ))((y1 − ε ) ε +(x1 − ε ) ε ) dx1 dy1 t s s t Λ̃ε ≤ C ε 2 t s ε ε ≤ C ε4 . 25 Hence, we have limε→0 Λ̃ε = 0, which implies that limε→0 Λε = 0, and so limε→0 Aε − Bε = 0. The proof is complete. We have all needed ingredients to prove the main result of the paper: Proof of Theorem 1.1. Taking into account the tightness result Proposition 3.1 and Propositions 4.2 and 4.3, in order to apply Theorem 4.1 we only need to prove the validity of condition (11) in our case. Note that Re(X) and Im(X) satisfy (11) if, for any (0, 0) ≤ (s, t) ≤ (s′ , t′ ) ≤ (S, T ), 0 ≤ s1 < · · · < sn ≤ s and 0 ≤ t1 < · · · < tn ≤ t, and any continuous bounded function φ : Cn → R, we have lim E [φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) (∆s,t Re(Xε )(s′ , t′ )) (∆s,t Im(Xε )(s′ , t′ ))] = 0. ε→0 { } Using the equality αβ = i 4 (α − iβ)2 − (α + iβ)2 , we obtain that E [φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) (∆s,t Re(Xε )(s′ , t′ )) (∆s,t Im(Xε )(s′ , t′ ))] [ ( )2  ∫ tε′ ∫ sε′ i √ = E φ(Xε (s1 , t1 ), . . . , Xε (sn , tn )) Kε xy e−iθL(x,y) dxdy 4  t s ε ε ( ∫ tε′ ∫ sε′ )2  √  − Kε xy eiθL(x,y) dxdy . (30) t s  ε ε We observe that, in the analysis of the term Aε −Bε in the proof of Proposition 4.3, we indeed proved that the two terms arising from the difference in (30) converge to zero as ε → 0. So the proof is complete. 5 Weak convergence for the stochastic heat equation We consider the following one-dimensional quasi-linear stochastic heat equation: ∂U ∂2U (t, x) − (t, x) = b(U (t, x)) + Ẇ (t, x), (t, x) ∈ [0, T ] × [0, 1], (31) ∂t ∂x2 where T > 0 stands for a fixed time horizon, b : R → R is a globally Lipschitz function and Ẇ (t, x) denotes the space-time white noise. We impose the initial condition U (0, x) = u0 (x), x ∈ [0, 1], where u0 : [0, 1] → R is a continuous function, and boundary conditions of Dirichlet type: U (t, 0) = U (t, 1) = 0, t ∈ [0, T ]. For simplicity’s sake, throughout the section we will assume that T = 1. All results presented here can be easily extended to a general T > 0. The solution to equation (31) is interpreted in the mild sense, as follows. Let {W (t, x); (t, x) ∈ [0, 1]2 } be a Brownian sheet defined on some probability space (Ω, F, P) and {Ft ; t ∈ [0, 1]} its natural filtration. A jointly measurable and adapted random field U = {U (t, x); (t, x) ∈ [0, 1]2 } is a solution of (31) if it holds that ∫ 1 ∫ t∫ 1 U (t, x) = Gt (x, y)u0 (y)dy + Gt−s (x, y) b(U (s, y))dyds 0 0 0 ∫ t∫ 1 + Gt−s (x, y)W (ds, dy), a.s. (32) 0 0 for all (t, x) ∈ (0, 1] × (0, 1), where G denotes the Green function associated to the heat equation in [0, 1] with Dirichlet boundary conditions. Existence, uniqueness and pathwise continuity of the solution to 26 (32) are a consequence of [15, Thm 3.5]. For the reader’s convenience, we recall that the Green function G is given by ∞ ∑ sin(nπx) sin(nπy)e−n π t . 2 2 Gt (x, y) = 2 n=1 In this section, we aim to apply [3, Thm. 1.4] in order to deduce that the above solution U can be approximated in law, in the space C([0, 1]2 ) of continuous functions, by the family of mild solutions {Un }n≥0 , where Un solves a stochastic heat equation perturbed by (the formal derivative of) either the real or imaginary part of the family introduced in (2): ∫ t ∫ x ε ε √ Xε (t, x) = εK sy {cos(θL(s, y)) + i sin(θL(s, y))} dyds, 0 0 where we recall that {L(s, y); s, y ≥ 0} denotes a Lévy sheet and its Lévy exponent is given by Ψ(ξ) = a(ξ) + ib(ξ). The constant K is given in (3) and θ ∈ (0, 2π), where we assume that a(θ)a(2θ) ̸= 0. Note that, compared to (2), in the above expression of Xε we have modified the variables’ notation in order to properly match with the framework of stochastic partial differential equations. Let us be more precise about the above statement. First, we rewrite Xε in the following way: ∫ t∫ x √ { ( (√ √ )) ( (√ √ ))} Xε (t, x) = nK sy cos θL n s, n y + i sin θL n s, n y dyds, 0 0 −2 with n = ε . Set √ ( (√ √ )) √ ( (√ √ )) θn1 (s, y) := nK sy cos θL n s, n y and θn2 (s, y) := nK sy sin θL n s, n y . Let i ∈ {1, 2} and consider the stochastic heat equation ∂Uni ∂ 2 Uni (t, x) − (t, x) = b(Uni (t, x)) + θni (t, x), (t, x) ∈ [0, 1]2 , ∂t ∂x2 with initial condition u0 and Dirichlet boundary conditions. The mild form of this equation is given by ∫ 1 ∫ t∫ 1 Uni (t, x) = Gt (x, y)u0 (y)dy + Gt−s (x, y) b(Uni (s, y))dyds 0 0 0 ∫ t∫ 1 + Gt−s (x, y)θni (s, y)dyds. (33) 0 0 Owing to [3, Sec. 3], equation (33) admits a unique solution Uni whose paths are continuous almost surely. Here is the main result of the section: Theorem 5.1. For any i ∈ {1, 2}, the sequence {Uni }n≥1 converges in law, as n → ∞ and in the space C([0, 1]2 ), to the solution U of (32). The proof of this theorem is based on [3, Thm. 1.4], where sufficient conditions on a family of random fields {θn }n≥1 have been established such that the sequence of solutions to the stochastic heat equation driven by θn converges in law to U , in the space of continuous functions. The first requirement is that θn ∈ L2 ([0, 1]2 ) a.s., and then there are the following conditions (see hypotheses 1.1, 1.2 and 1.3 in [3]): (i) The finite dimensional distributions of the processes ∫ t∫ x ζn (t, x) := θn (s, y)dyds, (t, x) ∈ [0, 1]2 , 0 0 converge in law to those of the Brownian sheet. 27 (ii) For some q ∈ [2, 3), there exists a positive constant Cq such that, for any f ∈ Lq ([0, 1]2 ), it holds: [(∫ ∫ )2 ] (∫ ∫ ) q2 1 1 1 1 sup E f (t, x)θn (t, x) dxdt ≤ Cq |f (t, x)| dxdt q . n≥1 0 0 0 0 (iii) There exist m > 8 and a positive constant C such that the following is satisfied: for all s0 , s′0 ∈ [0, 1] and x0 , x′0 ∈ [0, 1] satisfying 0 < s0 < s′0 < 2s0 and 0 < x0 < x′0 < 2x0 , and for any f ∈ L2 ([0, 1]2 ), it holds: [∫ ′∫ ′ m] (∫ ′ ∫ ′ ) m2 s0 x0 s0 x0 sup E f (s, y) θn (s, y)dyds ≤C f (s, y)2 dyds . n≥1 s0 x0 s0 x0 Hence, in the proof of Theorem 5.1 we will prove the validity of all above conditions in the case where θn is given by θni , for any i ∈ {1, 2}. Indeed, as it will be explained below, we will use similar arguments as those used in one of the applications tackled in [3], namely the case where θn are given by the Kac-Stroock processes on the plane: √ θn (t, x) = n tx (−1)Nn (t,x) , √ √ where Nn (t, x) := N ( nt, nx), and {N (t, x); (t, x) ∈ [0, 1]2 } is a standard Poisson process in the plane. We start with the following technical lemma, which is the analogous of [3, Lem. 4.2]: Lemma 5.2. Let f ∈ L2 ([0, 1]2 ) and α ≥ 1. Then, for any u, u′ ∈ (0, 1) satisfying that 0 < u < u′ ≤ 2α u, it holds ( )2  ∫ 1 ∫ u′ ∫ 1 ∫ u′ 3 sup E  f (t, x)θni (t, x) dxdt  ≤ 2 (2 α+1 − 1)K 2 f (t, x)2 dxdt, n≥1 0 u a(θ) 0 u for any i ∈ {1, 2}. Proof. We will only deal with the case involving θn1 , since the result for θn2 follows exactly in the same way. Note that we clearly have ( )2    ∫ 1 ∫ u′ ∫ 1 ∫ u′ √ √ √ 2 E f (t, x)θni (t, x) dxdt  ≤ E  nK tx eiθL( n t, n x) f (t, x) dxdt  , 0 u 0 u and the latter term equals to ∫ ∫ u′ ∫ ∫ u′ [ ] 1 1 √ √ √ √ √ n K 2 2 t1 x1 t2 x2 E eiθ(∆0,0 L( n t1 , n x1 )−∆0,0 L( n t2 , n x2 )) 0 u 0 u × f (t1 , x1 )f (t2 , x2 )dx1 dt1 dx2 dt2 . Observe that this expression is completely analogous as that at the beginning of the first step in the proof of Lemma 4.6. Thus, the same arguments used therein yield ( )2  ∫ 1 ∫ u′ E f (t, x)θni (t, x) dxdt  ≤ 2 (I1n + I2n ) , (34) 0 u 28 where ∫ ∫ ∫ u′ ∫ 1 t2 x2 √ I1n =n K 2 2 f (t1 , x1 )f (t2 , x2 ) t1 x1 t2 x2 e−a(θ)n(x2 t2 −x1 t1 ) 0 0 u u × cos(b(θ)n(x2 t2 − x1 t1 ))dx1 dx2 dt1 dt2 and ∫ ∫ ∫ u′ ∫ 1 t2 x2 √ I2n 2 =n K 2 f (t1 , x1 )f (t2 , x2 ) t1 x1 t2 x2 e−a(θ)n((t2 −t1 )x1 +(x2 −x1 )t1 ) 0 0 u u × cos (b(θ)n ((t2 − t1 )x1 + (x2 − x1 )t1 )) dx1 dx2 dt1 dt2 . At this point, we apply the inequality zw ≤ 12 (z 2 + w2 ) in such a way that √ 1( ) f (t1 , x1 )f (t2 , x2 ) t1 x1 t2 x2 ≤ t1 x1 f (t1 , x1 )2 + t2 x2 f (t2 , x2 )2 . 2 This makes that both I1n and I2n can be bounded by the sum of two terms of the form Ij,1 n n + Ij,2 , j = 1, 2, n n respectively, where Ij,1 involves f (t1 , x1 ) and Ij,2 involves f (t2 , x2 ). Then, once all cosinus are simply bounded by 1, one observe that the resulting four terms are completely analogous as those appearing in the proof of Lemma 4.2 in [3], and can be treated using the same kind of arguments. Thus, we obtain that ∫ 1 ∫ u′ 1 1 I1,1 ≤ n K 2 f (t1 , x1 )2 dx1 dt1 , 2 a(θ)2 0 u ∫ 1 ∫ u′ 1 2α n I1,2 ≤ K 2 f (t2 , x2 )2 dx2 dt2 , 2 a(θ)2 0 u ∫ 1 ∫ u′ 1 2α I2,1 ≤ n K 2 f (t1 , x1 )2 dx1 dt1 , 2 a(θ)2 0 u ∫ ∫ ′ 1 4(2α − 1) 2 1 u n I2,2 ≤ K f (t2 , x2 )2 dx2 dt2 . 2 a(θ)2 0 u Plugging everything together and using (34), we conclude the proof. The above lemma allows us to prove the following proposition. In fact, its proof follows exactly the same lines as that of Proposition 4.1 in [3] and therefore will be omitted. Proposition 5.3. Let p > 1 and f ∈ L2p ([0, 1]2 ). Then, there exists a positive constant Cp which does not depend on f such that [(∫ ∫ )2 ] (∫ 1 ∫ 1 ) p1 1 1 sup E i f (t, x)θn (t, x) dxdt ≤ Cp |f (t, x)| dxdt 2p , n≥1 0 0 0 0 for any i ∈ {1, 2}. The last needed ingredient for the proof of Theorem 5.1 is the following result, which is the analogous of [3, Prop. 4.4] in our setting. Proposition 5.4. Let m ∈ N be an even number and f ∈ L2 ([0, 1]2 ). Then, there exists a positive constant Cm which does not depend on f such that, for all s0 , s′0 , x0 , x′0 ∈ [0, 1] satisfying 0 < s0 < s′0 < 2s0 and 0 < x0 < x′0 < 2x0 , we have that [(∫ ′ ∫ ′ )m ] (∫ ′ ∫ ′ ) m2 s0 x0 s0 x0 sup E f (s, y) θni (s, y)dyds ≤ Cm f (s, y)2 dyds , n≥1 s0 x0 s0 x0 for any i ∈ {1, 2}. 29 Proof. Let i ∈ {1, 2}. For any (s0 , x0 ) ∈ [0, 1]2 , we define ∫ s0 ∫ x0 i Zn (s0 , x0 ) := f (s, y)θni (s, y)dyds. 0 0 Observe that, for all (0, 0) ≤ (s0 , x0 ) < (s′0 , x′0 ) ≤ (1, 1), we have [( )m ] [ m] E ∆s0 ,x0 Zni (s′0 , x′0 ) ≤ E ∆s0 ,x0 Z̄n (s′0 , x′0 ) , (35) where the random field Z̄n , which does not depend on i, is complex-valued and given by ∫ s0 ∫ x0 ( ) Z̄n (s0 , x0 ) := f (s, y) θn1 (s, y) + iθn2 (s, y) dyds 0 0 √ (here i = −1). In order to bound the right hand-side of (35), we can proceed as in the first part of the proof of the tightness result Proposition 3.1, obtaining [( ∫ ∫ i ′ ′ )m ] ∏ m √ E ∆s0 ,x0 Zn (s0 , x0 ) ≤n K m m f (sj , yj ) sj yj [s0 ,s′0 ]m [x0 ,x′0 ]m j=1 [ ∑m j √ √ ] × E eiθ j=1 (−1) ∆0,0 L( n sj , n yj ) dy1 · · · dym ds1 · · · dsm . At this point, we apply that yj < x′0 < 2x0 and sj < s′0 < 2s0 , and we compute the modulus of the expectation as it has been done in the proof of Proposition 3.1; more precisely, using the method set up therein in order to end up with the estimate (8). Thus, we can infer that [( )m ] E ∆s0 ,x0 Zni (s′0 , x′0 ) ∫ ∫ ∏m f (sj , yj ) e−a(θ)ns0 ((y(m) −y(m−1) )+···+(y(2) −y(1) )) m ≤ 2m (s0 x0 ) 2 nm K m [s0 ,s′0 ]m [x0 ,x′0 ]m j=1 × e−a(θ)nx0 ((s(m) −s(m−1) )+···+(s(2) −s(1) )) dy1 · · · dym ds1 · · · dsm ∫ ∫ ∏ m f (sj , yj ) e−a(θ)ns0 ((y(m) −y(m−1) )+···+(y(2) −y(1) )) m = 2m (s0 x0 ) 2 m! nm K m [s0 ,s′0 ]m [x0 ,x′0 ]m j=1 × e−a(θ)nx0 ((sm −sm−1 )+···+(s2 −s1 )) 1{s1 ≤···≤sm } dy1 · · · dym ds1 · · · dsm . Note that the latter expression is almost equal to that in the right hand-side of equation (31) in the proof of [3, Prop. 4.4]. Hence, we can conclude the proof exactly in the same way as in that result. Proof of Theorem 5.1. 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