Weak approximation of the complex Brownian sheet from a L\'evy sheet and applications to SPDEs

TL;DR

This paper constructs a family of complex-valued random fields from a planar Lévy process that converge to a complex Brownian sheet, enabling weak approximation of solutions to certain stochastic partial differential equations.

Contribution

It introduces a novel approximation method for the complex Brownian sheet using Lévy processes and applies it to approximate solutions of stochastic heat equations.

Findings

Convergence of Lévy-based fields to the complex Brownian sheet

Weak approximation of stochastic heat equation solutions

Application to SPDEs driven by space-time white noise

Abstract

We consider a L\'evy 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.

Equations

$$x_{\epsilon }(s,t):=\epsilon {\int }_0^{\frac{t}{\epsilon }}{\int }_0^{\frac{s}{\epsilon }}\sqrt{xy}(-1{)}^{N(x,y)}dxdy,(s,t)\in [0,S]\times [0,T].$$

$$y_{\epsilon }(t):=\epsilon {\int }_0^{\frac{t}{{\epsilon }^2}}(-1{)}^{N(s)}ds,\epsilon >0,$$

$$X_{\epsilon }(s,t):=\epsilon K{\int }_0^{\frac{t}{\epsilon }}{\int }_0^{\frac{s}{\epsilon }}\sqrt{xy}\{\cos (\theta L(x,y))+i\sin (\theta L(x,y))\}dxdy,$$

$$K=\frac{1}{\sqrt{2}}\frac{a(\theta {)}^2+b(\theta {)}^2}{a(\theta )}.$$

$$\frac{\partial U}{\partial t}(t,x)-\frac{{\partial }^{2}U}{\partial x^2}(t,x)=b(U(t,x))+\dot{W}(t,x),(t,x)\in [0,T]\times [0,1],$$

$${\theta }_n(t,x):=n\sqrt{tx}(-1{)}^{N\left(\sqrt{n}t,\sqrt{n}x\right)},$$

$${\Delta }_{s,t}Y(s^{\prime },t^{\prime }):=Y(s^{\prime },t^{\prime })-Y(s,t^{\prime })-Y(s^{\prime },t)+Y(s,t).$$

$$\mathbb{E}[{\Delta }_{s,t}Y(s^{\prime },t^{\prime })|{\mathcal{F}}_{s,t}]=0,\text{for all}(s,t)<(s^{\prime },t^{\prime }).$$

$$\mathbb{E}[{\Delta }_{s,t}Y(s^{\prime },t^{\prime })|{\mathcal{F}}_{S,t}\vee {\mathcal{F}}_{s,T}]=0,\text{for all}(s,t)<(s^{\prime },t^{\prime }).$$

$$\mathbb{E}\left[e^{i\xi {\Delta }_{Q}L}\right]=e^{-\lambda (Q)\Psi (\xi )},\xi \in \mathbb{R},$$

$$\Psi (\xi )=ia\xi +\frac{1}{2}{\sigma }^2{\xi }^2+{\int }_{\mathbb{R}}\left[1-e^{i\xi x}+\frac{ix\xi }{1+|x{|}^2}\right]\eta (dx),\xi \in \mathbb{R},$$

$${\int }_{\mathbb{R}}\frac{|x{|}^2}{1+|x{|}^2}\eta (dx)<\infty .$$

$$a(\xi ):=\frac{1}{2}{\sigma }^2{\xi }^2+{\int }_{\mathbb{R}}[1-\cos (\xi x)]\eta (dx),$$

$$b(\xi ):=a\xi +{\int }_{\mathbb{R}}\left[\frac{x\xi }{1+|x{|}^2}-\sin (\xi x)\right]\eta (dx).$$

$$\underset{\epsilon >0}{\sup }\mathbb{E}\left[{\left|{\Delta }_{s,t}{X}_{\epsilon }(s^{\prime },t^{\prime })\right|}^4\right]\le C(s^{\prime }-s{)}^2(t^{\prime }-t{)}^2.$$

$$\mathbb{E}\left[{\left|{\Delta }_{s,t}{X}_{\epsilon }(s^{\prime },t^{\prime })\right|}^4\right]$$

$$={\epsilon }^4{K}^4\mathbb{E}\left[{\left|{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{xy}\{\cos (\theta L(x,y))+i\sin (\theta L(x,y))\}dxdy\right|}^4\right]$$

$$={\epsilon }^4{K}^4\mathbb{E}\left[{\left|{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{xy}{e}^{i\theta L(x,y)}dxdy\right|}^4\right]$$

$$={\epsilon }^4{K}^4\mathbb{E}\left[{\left({\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_1{y}_1}{e}^{i\theta L(x_1,y_1)}d{x}_{1}d{y}_1{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_2{y}_2}{e}^{-i\theta L(x_2,y_2)}d{x}_{2}d{y}_2\right)}^2\right]$$

$$={\epsilon }^4{K}^4\mathbb{E}\left[{\left({\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_1{x}_2{y}_1{y}_2}{e}^{i\theta (L(x_2,y_2)-L(x_1,y_1))}d{x}_{1}d{x}_{2}d{y}_{1}d{y}_2\right)}^2\right]$$

$$={\epsilon }^4{K}^4\mathbb{E}[{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\cdots {\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_1{x}_2{x}_3{x}_4}\sqrt{y_1{y}_2{y}_3{y}_4}$$

$$\displaystyle\qquad\qquad\quad\times e^{i\theta(L(x_{4},y_{4})-L(x_{3},y_{3})+L(x_{2},y_{2})-L(x_{1},y_{1}))}dx_{1}\dots dx_{4}dy_{1}\dots dy_{4}\Bigg{]}.$$

$$\mathbb{E}\left[{\left|{\Delta }_{s,t}{X}_{\epsilon }(s^{\prime },t^{\prime })\right|}^4\right]$$

$$\times \mathbb{E}\left[e^{i\theta {\sum }_{j=1}^4(-1{)}^j{\Delta }_{0,0}L(x_j,y_j)}\right]d{x}_1\dots d{x}_{4}d{y}_1\dots d{y}_4.$$

$$\mathbb{E}\left[{\left|{\Delta }_{s,t}{X}_{\epsilon }(s^{\prime },t^{\prime })\right|}^4\right]$$

$$=\sum _{\sigma \in {\mathcal{P}}_4}\sum _{\beta \in {\mathcal{P}}_4}{\epsilon }^4{K}^4{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\cdots {\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_1{x}_2{x}_3{x}_4}\sqrt{y_1{y}_2{y}_3{y}_4}\mathbb{E}\left[e^{i\theta {\sum }_{j=1}^4(-1{)}^j{\Delta }_{0,0}L(x_j,y_j)}\right]$$

$$\times I_{\{x_{\sigma (1)}<x_{\sigma (2)}<x_{\sigma (3)}<x_{\sigma (4)}\}}{I}_{\{y_{\beta (1)}<y_{\beta (2)}<y_{\beta (3)}<y_{\beta (4)}\}}d{x}_1\dots d{x}_{4}d{y}_1\dots d{y}_4$$

$$\le \sum _{\sigma \in {\mathcal{P}}_4}\sum _{\beta \in {\mathcal{P}}_4}{\epsilon }^4{K}^4{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\cdots {\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\sqrt{x_1{x}_2{x}_3{x}_4}\sqrt{y_1{y}_2{y}_3{y}_4}\left|\mathbb{E}\left[e^{i\theta {\sum }_{j=1}^4(-1{)}^j{\Delta }_{0,0}L(x_j,y_j)}\right]\right|$$

$$\times I_{\{x_{\sigma (1)}<x_{\sigma (2)}<x_{\sigma (3)}<x_{\sigma (4)}\}}{I}_{\{y_{\beta (1)}<y_{\beta (2)}<y_{\beta (3)}<y_{\beta (4)}\}}d{x}_1\dots d{x}_{4}d{y}_1\dots d{y}_4.$$

$$\left|\mathbb{E}\left[e^{i\theta {\sum }_{j=1}^4(-1{)}^j{\Delta }_{0,0}L(x_j,y_j)}\right]\right|I_{\{x_{\sigma (1)}<x_{\sigma (2)}<x_{\sigma (3)}<x_{\sigma (4)}\}}{I}_{\{y_{\beta (1)}<y_{\beta (2)}<y_{\beta (3)}<y_{\beta (4)}\}}.$$

$$Q=\left(\cup_{k}\,\bar{Q}_{k}\right)\cup\big{(}\cup_{l}\tilde{Q}_{l}\big{)},$$

$$\left|\mathbb{E}\left[e^{i\theta ({\sum }_k{c}_1^k{\Delta }_{{\bar{Q}}_k}L+{\sum }_l{c}_2^l{\Delta }_{{\tilde{Q}}_l}L)}\right]\right|I_{\{x_1<x_2<x_3<x_4\}}{I}_{\{y_1<y_2<y_3<y_4\}}$$

$${\epsilon }^4{\int }_I{\int }_J\sqrt{x_1{x}_2{x}_3{x}_4}\sqrt{y_1{y}_2{y}_3{y}_4}\exp [-a(\theta )[(x_4-x_3)y_3+(y_4-y_3)x_3$$

$$+(x_2-x_1)y_1+(y_2-y_1)x_1]]d{x}_1...d{x}_{4}d{y}_1...d{y}_4,$$

$$\sqrt{s^{\prime}}\sqrt{t^{\prime}}\varepsilon^{3}\int_{I_{1}}\int_{J_{1}}\frac{\sqrt{x_{1}x_{2}x_{3}}\sqrt{y_{1}y_{2}y_{3}}}{x_{3}y_{3}}\exp\big{[}-a(\theta)[(x_{2}-x_{1})y_{1}+(y_{2}-y_{1})x_{1}]\big{]}dx_{1}...dx_{3}dy_{1}...dy_{3},$$

$$s^{\prime }{t}^{\prime }{\epsilon }^2{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\frac{1}{\sqrt{x_1}}d{x}_1{\int }_{\frac{s}{\epsilon }}^{\frac{s^{\prime }}{\epsilon }}\frac{1}{\sqrt{x_3}}d{x}_3{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}\frac{1}{\sqrt{y_1}}d{y}_1{\int }_{\frac{t}{\epsilon }}^{\frac{t^{\prime }}{\epsilon }}\frac{1}{\sqrt{y_3}}d{y}_3$$

Full text

2000 Mathematics Subject Classification: 60F17; 60G15; 60H15.

Key words and phrases: Brownian sheet; Lévy sheet; stochastic heat equation; weak approximation.

1 Introduction

Let $\{N(x,y);\,x,y\geq 0\}$ be a Poisson process in the plane and $S,T>0$. For any $\varepsilon>0$, define the following random field:

[TABLE]

Then, in [2] (see Theorem 1.1 therein) the authors proved that, as $\varepsilon$ tends to zero, $x_{\varepsilon}$ converges in law, in the Banach space $\mathcal{C}([0,S]\times[0,T])$ of continuous functions, to the Brownian sheet on $[0,S]\times[0,T]$. 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

[TABLE]

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 $\{L(x,y);\,x,y\geq 0\}$ (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 $e^{i\pi N(x,y)}$. Hence, when replacing $N$ by $L$, we will use the form $e^{i\pi L(x,y)}=\cos(\pi L(x,y))+i\sin(\pi L(x,y))$ since the expression $(-1)^{L(x,y)}$ may not be well-defined in $\mathbb{R}$. On the other hand, we will replace $\pi$ by an arbitrary angle $\theta\in(0,2\pi)$. The main result of the paper is the following:

Theorem 1.1**.**

Let $\{L(x,y);\,x,y\geq 0\}$ be a Lévy sheet and $\Psi(\xi):=a(\xi)+ib(\xi)$, $\xi\in\mathbb{R}$, its Lévy exponent. Let $\theta\in(0,2\pi)$ and $S,T>0$, and define, for any $\varepsilon>0$ and $(s,t)\in[0,S]\times[0,T]$,

[TABLE]

where the constant $K$ is given by

[TABLE]

Assume that $a(\theta)a(2\theta)\neq 0$. Then, as $\varepsilon$ tends to zero, $X_{\varepsilon}$ converges in law, in the space of complex-valued continuous functions $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$, to a complex Brownian sheet.

We point out that the processes $X_{\varepsilon}$ 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 $X_{\varepsilon}$ are clearly not independent, for any $\varepsilon>0$, 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:

[TABLE]

where $T>0$ and $b$ is a globally Lipschitz function. We impose some initial datum and Dirichlet boundary 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 $\{U_{\varepsilon}\}_{\varepsilon}$, where $U_{\varepsilon}$ is the mild solution to a stochastic heat equation like (4) but driven by either the real or imaginary part of the noise $X_{\varepsilon}$. 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) 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

[TABLE]

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 $(X_{\varepsilon})_{\varepsilon>0}$ 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 $(\Omega,\mathcal{F},\mathbb{P})$ be a complete probability space. We will use some notation introduced by Cairoli and Walsh in [6]. Namely, let $\{\mathcal{F}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ be a family of sub-$\sigma$-algebras of $\mathcal{F}$ satisfying:

• (i)

$\mathcal{F}_{s,t}\subset\mathcal{F}_{s^{\prime},t^{\prime}}$, for all $s\leq s^{\prime}$ and $t\leq t^{\prime}$.

• (ii)

All zero sets of $\mathcal{F}$ are contained in $\mathcal{F}_{0,0}$ .

• (iii)

For any $z\in[0,S]\times[0,T]$, $\mathcal{F}_{z}=\cap_{z<z^{\prime}}\mathcal{F}_{z^{\prime}}$, where $z=(s,t)<z^{\prime}=(s^{\prime},t^{\prime})$ denotes the partial order in $[0,S]\times[0,T]$, which means that $s<s^{\prime}$ and $t<t^{\prime}$.

If $(s,t)<(s^{\prime},t^{\prime})$ and $Y$ denotes any random field defined in $[0,S]\times[0,T]$, the increment of $Y$ on the rectangle $[(s,t),(s^{\prime},t^{\prime})]$ is defined by

[TABLE]

An adapted process $\{Y(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ with respect to the filtration $\{\mathcal{F}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ is called a martingale if $\mathbb{E}[|Y(s,t)|]<\infty$ for all $(s,t)\in[0,S]\times[0,T]$ and

[TABLE]

It will be called a strong martingale if $\mathbb{E}[|Y(s,t)|]<\infty$ for all $(s,t)\in[0,S]\times[0,T]$, $Y(s,0)=Y(0,t)=0$ for all $s,t$ and

[TABLE]

We recall that a Brownian sheet is an adapted process $\{W(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ such that $W(s,0)=W(0,t)=0$ $\mathbb{P}$-a.s., the increment $\Delta_{s,t}W(s^{\prime},t^{\prime})$ is independent of $\mathcal{F}_{S,t}\vee\mathcal{F}_{s,T}$, for all $(s,t)<(s^{\prime},t^{\prime})$, and it is normally distributed with mean zero and variance $(s^{\prime}-s)(t^{\prime}-t)$. If no filtration is specified, we will consider the one generated by the process itself, namely $\mathcal{F}^{W}:=\sigma\{W(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ (conveniently completed).

A Lévy sheet is defined as follows. In general, if $Q$ is any rectangle in $\mathbb{R}_{+}^{2}$ and $Y$ any random field in $\mathbb{R}_{+}^{2}$, we will also denote by $\Delta_{Q}Y$ the increment of $Y$ on $Q$. It is well-known that, for any nonnegative definite function $\Psi$ in $\mathbb{R}$, there exists a real-valued random field $L=\{L(s,t);\,s,t\geq 0\}$ such that

• (i)

For any family of disjoint rectangles $Q_{1},\dots,Q_{n}$ in $\mathbb{R}_{+}^{2}$, the increments $\Delta_{Q_{1}}L,\dots,\Delta_{Q_{n}}L$ are independent random variables.

• (ii)

For any rectangle $Q$ in $\mathbb{R}_{+}^{2}$, the characteristic function of the increment $\Delta_{Q}L$ is given by

[TABLE]

where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}_{+}^{2}$.

Definition 2.1**.**

A random field $L=\{L(s,t);\,s,t\geq 0\}$ taking values in $\mathbb{R}$ that is continuous in probability and satisfies the above conditions (i) and (ii) is called a Lévy sheet with exponent $\Psi$.

By the Lévy-Khintchine formula, we have

[TABLE]

where $a\in\mathbb{R}$, $\sigma\geq 0$ and $\eta$ is the corresponding Lévy measure, that is a Borel measure on $\mathbb{R}\setminus\{0\}$ that satisfies

[TABLE]

We write $\Psi(\xi)=a(\xi)+ib(\xi)$, where

[TABLE]

and

[TABLE]

Observe that $a(\xi)\geq 0$ and, if $\xi\neq 0$, $a(\xi)>0$ whenever $\sigma>0$ and/or $\eta$ is nontrivial.

3 Tightness

This section is devoted to prove that the family of probability laws of $\{X_{\varepsilon}\}_{\varepsilon>0}$ is tight in $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$. 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 $X_{\varepsilon}$ vanish on the axes.

Proposition 3.1**.**

Let $\{X_{\varepsilon}\}_{\varepsilon>0}$ be the family of random fields defined by (2). There exists a positive constant $C$ such that, for all $(0,0)\leq(s,t)<(s^{\prime},t^{\prime})\leq(S,T)$,

[TABLE]

This implies that the the family of probability laws of $(X_{\varepsilon})_{\varepsilon>0}$ is tight in $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$.

Proof.

By definition of $X_{\varepsilon}$ and the properties of the modulus $|\cdot|$, we have

[TABLE]

Taking into account that we can write $e^{i\theta\sum_{j=1}^{4}(-1)^{j}L(x_{j},y_{j})}=e^{i\theta\sum_{j=1}^{4}(-1)^{j}\Delta_{0,0}L(x_{j},y_{j})}$, we obtain

[TABLE]

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 $\mathcal{P}_{4}$ be the group of permutations of degree 4. Then,

[TABLE]

At this point, we observe that the geometric structure of the resulting increments of $L$ in the expression $\sum_{j=1}^{4}(-1)^{j}\Delta_{0,0}L(x_{j},y_{j})$ of any of the 576 possibilities corresponds to one of the 24 cases drawn in Figure 1; we note that the latter corresponds to all 24 possible orders of the $x$-variables with $y_{1}<y_{2}<y_{3}<y_{4}$. In each of these 24 possible structures, the corresponding increments of $L$ turn out to be multiplied by $c_{1}\in\{-1,1\}$ in the black regions, while they are multiplied by $c_{2}\in\{-2,0,2\}$ in the white regions.

Let us now fix two permutations $\sigma,\beta\in\mathcal{P}_{4}$, and we will focus on the term

[TABLE]

We perform a change of variables in such a way that, making a harmless abuse of notation and using again the same one for the new variables, we have $x_{1}<x_{2}<x_{3}<x_{4}$ and $y_{1}<y_{2}<y_{3}<y_{4}$.

On the other hand, if we denote by $Q$ the region of Figure 1 corresponding the above fixed variables order, we know that $Q$ can be decomposed as a union of black rectangles and white rectangles. More precisely, we can write

[TABLE]

where the increments $\Delta_{\bar{Q}_{k}}L$ are multiplied by $c_{1}^{k}\in\{-1,1\}$ and $\Delta_{\tilde{Q}_{l}}L$ are multiplied by $c_{2}^{l}\in\{-2,0,2\}$. Hence, expression (7) is given by

[TABLE]

where $\bar{Q}:=\cup_{k}\bar{Q}_{k}$. In the above computations, we have used that the real part of the Lévy exponent $\Psi$ is a nonnegative function and satisfies $a(-\theta)=a(\theta)$. We remark that, independently of the constants $c_{1}^{k}$ and $c_{2}^{l}$, we have obtained an estimated of (7) which only involves the black regions multiplied by 1. Recall that $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^{2}$.

Taking into account estimate (8), it is readily checked that, among all 24 possibilities drawn in Figure 1, it suffices to deal with 4 of these cases (see Figure 2). This is because, in the rest of the cases, the area of $\bar{Q}$ is greater than or equal to the corresponding one of one of these 4 possibilities. We point out that, in some cases, one needs to apply a symmetry argument. For instance, the analysis of the figure in the fourth row and third column of Figure 1, by symmetry between the $x$ and $y$ coordinates, is equivalent to that of the figure in the third row and fourth column. Thus, since in (8) the area of $\bar{Q}$ appears with a negative sign, we can focus only on the cases of Figure 2.

Let us start tackling the case corresponding to i) in Figure 2. That is, by estimates (6) and (8), we need to find suitable upper bounds of the term

[TABLE]

where $J=\{\frac{s}{\varepsilon}\leq x_{1}\leq x_{2}\leq x_{3}\leq x_{4}\leq\frac{s^{\prime}}{\varepsilon}\}$ and $I=\{\frac{t}{\varepsilon}\leq y_{1}\leq y_{2}\leq y_{3}\leq y_{4}\leq\frac{t^{\prime}}{\varepsilon}\}$.

First, estimate $x_{4}$ and $y_{4}$ in the square roots above by $\frac{s^{\prime}}{\varepsilon}$ and $\frac{t^{\prime}}{\varepsilon}$, respectively, and then integrate with respect to theses two variables. The resulting expression can be easily bounded by, up to some positive constant,

[TABLE]

where $J_{1}=\{\frac{s}{\varepsilon}\leq x_{1}\leq x_{2}\leq x_{3}\leq\frac{s^{\prime}}{\varepsilon}\}$ and $I_{1}=\{\frac{t}{\varepsilon}\leq y_{1}\leq y_{2}\leq y_{3}\leq\frac{t^{\prime}}{\varepsilon}\}$. Now, we estimate $x_{2}$ and $y_{2}$ by $\frac{s^{\prime}}{\varepsilon}$ and $\frac{t^{\prime}}{\varepsilon}$, respectively, inside the square roots, and then integrate with respect to $x_{2}$ and $y_{2}$. Hence, up to some constant, we obtain an estimate for (3) of the form

[TABLE]

This concludes the analysis of i) in Figure 2.

In the remaining three cases, the above-used argument does not directly work. Instead, we will add some small area in the corresponding drawing in such a way that we will be able to argue similarly as in case i). We remark that some of the integrand’s estimates that will be obtained in the sequel will hold everywhere except of a zero Lebesgue measure set of $\mathbb{R}^{8}$.

Let us start with the analysis of the integral corresponding to ii). We need to bound the following term:

[TABLE]

where $J=\{\frac{s}{\varepsilon}\leq x_{1}\leq x_{2}\leq x_{3}\leq x_{4}\leq\frac{s^{\prime}}{\varepsilon}\}$, $I=\{\frac{t}{\varepsilon}\leq y_{1}\leq y_{2}\leq y_{3}\leq y_{4}\leq\frac{t^{\prime}}{\varepsilon}\}$, and $\bar{Q}$ is the union of black rectangles corresponding to the case ii). Note that $A:=\lambda(\bar{Q})$ is given by

[TABLE]

We split the above integral into two terms:

[TABLE]

When $A\geq 2(x_{2}-x_{1})(y_{2}-y_{1})$, we have

[TABLE]

Hence, the first integral in (10) is less or equal than

[TABLE]

and following the same arguments used in the case i), this term can be estimated by $(s^{\prime}-s)^{2}(t^{\prime}-t)^{2}$, up to some positive constant.

On the other hand, as far as the second integral in (10) is concerned, observe that we have

[TABLE]

In particular, in this region we have

[TABLE]

which implies that

[TABLE]

Thus, the second integral in (10) can be bounded by

[TABLE]

and here again the arguments of the case i) may be applied, yielding an estimate of the form $(s^{\prime}-s)^{2}(t^{\prime}-t)^{2}$, up to some positive constant.

The same idea can be used to deal with the integral corresponding to iii). Indeed, in this case the area $A$ is given by

[TABLE]

and here one splits the underlying integral taking into account the regions $\{A\geq 2(x_{3}-x_{2})y_{1}\}$ and $\{A<2(x_{3}-x_{2})y_{1}\}$. In the former, one has

[TABLE]

and the desired estimated is obtained by using the same computations as for the case i). Note that, in fact, variables which have to be bounded and integrated with respect to are $x_{4},\,y_{4},\,y_{2},\,x_{3}$, following this specific order. On the other hand, in the region $\{A<2(x_{3}-x_{2})y_{1}\}$, we get

[TABLE]

So, in particular, in this region we have

[TABLE]

where we deduce

[TABLE]

At this point, we can follow the arguments of the preceding cases.

Finally, it only remains to estimate the integral involving case iv) in Figure 2. In this case,

[TABLE]

and the splitting regions are $\{A\geq 2(x_{3}-x_{2})y_{2}\}\cup\{A\geq 2(y_{3}-y_{2})x_{2}\}$ and the corresponding complement.

In the first region, condition $A\geq 2(x_{3}-x_{2})y_{2}$ turns out to be equivalent to

[TABLE]

so we will be able to mimic the arguments used so far. Moreover, note that this case is symmetric in $x$ and $y$, which implies that the computations in the case $A\geq 2(y_{3}-y_{2})x_{2}$ will be the same just by exchanging $x$ and $y$.

As far as the case $\{A<2(x_{3}-x_{2})y_{2}\}\cap\{A<2(y_{3}-y_{2})x_{2}\}$ is concerned, we have

[TABLE]

In particular, one has

[TABLE]

which implies

[TABLE]

One can conclude the proof by following the same arguments as in the preceding cases. ∎

4 Limit identification

Let $\{\mathbb{P}_{\varepsilon}\}_{\varepsilon>0}$ be the family of probability laws in $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$ corresponding to $\{X_{\varepsilon}\}_{\varepsilon>0}$. By Proposition 3.1, there exists a subsequence $\{\mathbb{P}_{\varepsilon_{n}}\}_{n\geq 1}$ of $\{\mathbb{P}_{\varepsilon}\}_{\varepsilon>0}$ converging, in the weak sense in the space $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$, to some probability measure $\mathbb{P}$. This section is devoted to prove that $\mathbb{P}$ 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 $X=\{X(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ be a complex-valued and continuous process such that $X(s,0)=X(0,t)=0$ for all $s\in[0,S]$ and $t\in[0,T]$. We write $X=X^{1}+iX^{2}$. Let $\{\mathcal{F}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ 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 $(0,0)<(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$ and $i=1,2$,*

[TABLE]

Proof.

It is obvious that (i) implies (ii) because, for all $(0,0)<(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$ and $i=1,2$, it holds

[TABLE]

and

[TABLE]

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 $(0,0)<(s,t)<(s^{\prime},t^{\prime})\leq(S,T)$ and define the process $Y=\{Y_{u},\,u\in[s,S]\}$ by $Y_{u}:=\Delta_{s,t}X^{1}(u,t^{\prime})$ (note that, indeed, it does not depend on $s^{\prime}$). This process is a martingale with respect to the filtration $\{\mathcal{F}_{u,T},\,u\in[s,S]\}$, since it is adapted to it and, for any $s\leq u<u^{\prime}\leq S$,

[TABLE]

because $X^{1}$ is a strong martingale, by hypothesis. On the other hand, note that, by (11),

[TABLE]

which implies that the quadratic variation of the process $Y$ is $\langle Y\rangle_{u}=(u-s)(t^{\prime}-t)$. Hence, by Lévy’s characterization theorem, we obtain that $Y=\{Y_{u},\,u\in[s,S]\}$ defines a Brownian motion with respect to the filtration $\{\mathcal{F}_{u,T},\,u\in[s,S]\}$ and with variance function $u\mapsto(u-s)(t^{\prime}-t)$.

Thus, the increments $\Delta_{s,t}X^{1}(s^{\prime},t^{\prime})=Y_{s^{\prime}}-Y_{s}$ are normally distributed random variables with mean zero and variance $(s^{\prime}-s)(t^{\prime}-t)$. 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 $\Delta_{s,\varepsilon}X^{1}(s^{\prime},t^{\prime})$, with $\varepsilon>0$. We know that they are Gaussian random variables, and they converge, as $\varepsilon$ tends to zero, to $\Delta_{s,0}X^{1}(s^{\prime},t^{\prime})$, which will thus also be a centered Gaussian random variable with variance equal to the corresponding limit of variances, that is $(s^{\prime}-s)t^{\prime}$. 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 (12), 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 $a,b\in\mathbb{R}$ (not both equal to zero) and $(0,0)<(s,t)<(s^{\prime},t^{\prime})\leq(S,T)$, and define

[TABLE]

Then, one verifies that the process $Z=\{Z_{u},\,u\in[s,S]\}$ is a martingale with respect to the filtration $\{\mathcal{F}_{u,T},\,u\in[s,S]\}$ and, owing to (12), it has quadratic variation $\langle Z\rangle_{u}=(u-s)(t^{\prime}-t)(a^{2}+b^{2})$. Lévy’s characterization theorem implies that $Z$ is a Brownian motion with variance function $u\mapsto(u-s)(t^{\prime}-t)(a^{2}+b^{2})$. In particular, $Z_{u}$ defines a Gaussian random variable, for all $u\in[s,S]$. 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 $\mathbb{P}$ denotes the weak limit in $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$ of a converging subsequence of the family $\{\mathbb{P}_{\varepsilon}\}_{\varepsilon>0}$. Let $X=\{X(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ be the corresponding (complex-valued) canonical process and $\{\mathcal{F}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ its associated natural filtration. Then, the real and imaginary parts of $X$ define strong martingales under the probability $\mathbb{P}$.

Proposition 4.3**.**

Let $X$ be the canonical process defined in the previous proposition. Then, for all $(0,0)<(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$, it holds:

[TABLE]

and

[TABLE]

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 $f:[M,\infty)\rightarrow\mathbb{R}$ is a derivable function, such that $f^{\prime}$ is continuous on $[M,\infty)$, $M\geq 0$, and assume that

[TABLE]

Then,

[TABLE]

The proof of Proposition 4.2 is based on the following lemma.

Lemma 4.5**.**

Let $X_{\varepsilon}=\{X_{\varepsilon}(s,t);\,(s,t)\in[0,S]\times[0,T]\}$ be the (complex-valued) random field defined in (2) and $\{\mathcal{F}^{\varepsilon}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ its natural filtration. Then, for all $(0,0)<(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$,

[TABLE]

where the limit is understood in $L^{2}(\Omega)$.

Proof.

We will use the notation $Y_{\varepsilon}:=\mathbb{E}[\Delta_{s,t}X_{\varepsilon}(s^{\prime},t^{\prime})|\mathcal{F}_{S,t}\vee\mathcal{F}_{s,T}]$. First, note that we can write

[TABLE]

Thus, we have

[TABLE]

and, recalling that $\Psi(\xi)=a(\xi)+ib(\xi)$ is the Lévy exponent,

[TABLE]

In order to compute the expectation inside the integral, we take into account the possible orders of $x_{1},x_{2}$ and $y_{1},y_{2}$, 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

[TABLE]

where

[TABLE]

and

[TABLE]

Applying Fubini theorem, one easily verifies that both $I_{1}$ and $I_{2}$ can be bounded by the term

[TABLE]

This implies that \mathbb{E}\big{[}|Y_{\varepsilon}|^{2}\big{]}\leq 4I. Let us check that $I$ converges to zero as $\varepsilon\to 0$. Indeed, bounding $\sqrt{x_{1}}$ by $\sqrt{x_{2}}$ and $\sqrt{y_{2}}$ by $\sqrt{\frac{t^{\prime}}{\varepsilon}}$, then integrating with respect to $x_{1}$ and $y_{2}$, and finally applying a change of variables, we can infer that

[TABLE]

where $C$ is some positive constant whose value may change from line to line. The latter expression converges to zero as $\varepsilon\to 0$. In order to see this, e.g., one splits the integral with respect to $x_{2}$ on the intervals $(0,\varepsilon)$ and $(\varepsilon,\frac{s^{\prime}-s}{\varepsilon})$. In $(0,\varepsilon)$ one bounds the exponential by 1 and then integrate, while in $(\varepsilon,\frac{s^{\prime}-s}{\varepsilon})$ one first integrates with respect to $y_{1}$ and then with respect to $x_{2}$. 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 $(s,t)\in[0,S]\times[0,T]$. By Skorohod’s representation theorem, there exist a probability space $(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{P}})$, a sequence $\{Y_{\varepsilon}\}_{\varepsilon>0}$ and a random variable $Y$, all of them taking values in $\mathcal{C}([0,S]\times[0,T])$, such that

1. (a)

For all $\varepsilon>0$, $Y_{\varepsilon}$ and $X_{\varepsilon}$ have the same law, 2. (b)

$Y$ and $X$ have the same law, 3. (c)

$Y_{\varepsilon}$ converges to $Y$, as $\varepsilon\to 0$, $\tilde{\mathbb{P}}$-almost surely.

Condition (c) means that

[TABLE]

In particular, for any $t\in[0,T]$, $\lim_{\varepsilon\to 0}Y_{\varepsilon}(0,t)=Y(0,t)$, $\tilde{\mathbb{P}}$-a.s. By (a), it holds $\tilde{\mathbb{P}}(Y_{\varepsilon}(0,t)=0)=\mathbb{P}_{\varepsilon}(X_{\varepsilon}(0,t)=0)=1$. Hence, letting $\varepsilon\to 0$, we get $\tilde{\mathbb{P}}(Y(0,t)=0)=1$. By (b), we end up with $\mathbb{P}(X(0,t)=0)=1$. The same argument let us conclude that $\mathbb{P}(X(s,0)=0)=1$, for all $s\in[0,S]$.

The remaining of the proof is similar to that of [2, Prop. 4.2]. Let $(0,0)<(s,t)<(s^{\prime},t^{\prime})\leq(S,T)$. It suffices to prove that, for any $n\geq 1$ and $(s_{1},t_{1}),\dots,(s_{n},t_{n})$ such that either $s_{i}\leq S$ and $t_{i}\leq t$, or $s_{i}\leq s$ and $t_{i}\leq T$, $i=1,\dots,n$, and for any continuous and bounded function $\varphi:\mathbb{C}^{n}\to\mathbb{R}$, it holds that

[TABLE]

We recall that the notation $|z|$ stands for the modulus of $z\in\mathbb{C}$. Without any loss of generality, the converging subsequence of probability measures to $\mathbb{P}$ will be simply denoted by $\{\mathbb{P}_{\varepsilon}\}_{\varepsilon>0}$. Thus, by Proposition 3.1, it suffices to check that

[TABLE]

For this, we recall that, as in the statement of Lemma 4.5, $\{\mathcal{F}^{\varepsilon}_{s,t};\,(s,t)\in[0,S]\times[0,T]\}$ is the natural filtration associated to the (complex-valued) random field $X^{\varepsilon}$ introduced in (2). Then, we can argue as follows:

[TABLE]

The latter term converges to zero as $\varepsilon\to 0$, 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 $(0,0)\leq(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$, it holds:

[TABLE]

Proof.

We split the proof in three steps.

Step 1. Owing to the definition of $X_{\varepsilon}$ (see (2)) and applying Fubini theorem, we have

[TABLE]

As in the proof of Lemma 4.5, we need to take into account the possible orders of $x_{1},x_{2}$ and $y_{1},y_{2}$, respectively. Then, applying also some suitable changes of variables, we have

[TABLE]

Recalling that $\Psi(\theta)=a(\theta)+ib(\theta)$, where $a(\theta)=a(-\theta)$ and $b(\theta)=-b(-\theta)$, we observe that

[TABLE]

As a consequence, we can infer that

[TABLE]

where

[TABLE]

and

[TABLE]

Step 2. Let us consider the case $s=t=0$. In order to deal with $I_{1}^{\varepsilon}$, we make the changes of variables $z_{i}:=x_{i}y_{i}$ and $v_{i}:=\frac{\varepsilon}{s^{\prime}}x_{i}$, $i=1,2$, and we define $u:=\frac{s^{\prime}t^{\prime}}{\varepsilon^{2}}$. Thus, by l’Hôpital’s rule, we have

[TABLE]

Applying now the changes of variables $v_{2}^{\prime}:=uv_{2}$ and $v_{1}^{\prime}:=uv_{1}$, and again l’Hôpital’s rule, we obtain that the latter limit equals to

[TABLE]

In order to compute the above limit, we use the formula $\cos(\theta)=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$. Hence, the expression inside the limit (19) can be written as the sum $\frac{1}{2}(A_{u}+B_{u})$, where these terms are given by

[TABLE]

[TABLE]

We will only deal with $\lim_{u\to\infty}B_{u}$, because $\lim_{u\to\infty}A_{u}$ can be treated in a similar way. Indeed, rewriting $B_{u}$ as

[TABLE]

and applying l’Hôpital’s rule twice, one easily proves that

[TABLE]

Similarly, one gets

[TABLE]

Thus,

[TABLE]

Now, we are going to compute $\lim_{\varepsilon\to 0}I_{2}^{\varepsilon}$. Recall that the latter term is given in (18). The strategy that we have followed to deal with $I_{1}^{\varepsilon}$ cannot be applied here. More precisely, we have not been able to compute the limit of $I_{2}^{\varepsilon}$ directly, but we will introduce an auxiliary term which will converge to some quantity, and we will prove that the remainder converges to zero.

To start with, we apply the same changes of variables that we performed for $I_{1}^{\varepsilon}$, we set $u:=\frac{s^{\prime}t^{\prime}}{\varepsilon}$ and apply l’Hôpital’s rule, so $\lim_{\varepsilon\to 0}I_{2}^{\varepsilon}$ equals to

[TABLE]

Next, we make the changes of variables $\bar{v}_{1}:=uv_{1}$ and $\bar{v}_{2}:=uv_{2}$ and we apply again l’Hôpital’s rule. Hence, the latter limit becomes

[TABLE]

Finally, performing the changes $x:=\frac{z_{1}}{\bar{v}_{1}}$ and $y:=\frac{\bar{v}_{1}}{u}$, we end up with

[TABLE]

with

[TABLE]

At this point, we introduce the auxiliary term mentioned above:

[TABLE]

where we note that, compared to the right hand-side of (20), we have only replaced $\sqrt{xy}$ by $\sqrt{y}$. For the moment, assume that $\lim_{u\to\infty}(C_{u}-\tilde{C}_{u})=0$. Let us compute the limit of $\tilde{C}_{u}$, recalling that this term has been defined in (21). As in the analysis of the term $I_{1}^{\varepsilon}$, we use the formula $\cos(\theta)=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$, so we split $\tilde{C}_{u}$ as the sum of two terms (multiplied by $\frac{1}{2}$), one of which is given by

[TABLE]

and the other one is the same with $a(\theta)+ib(\theta)$ instead of $a(\theta)-ib(\theta)$. 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

[TABLE]

It is straightforward to check that the latter limit is $\frac{K^{2}s^{\prime}t^{\prime}}{a(\theta)^{2}+b(\theta)^{2}}$. The limit of the term involving $a(\theta)+ib(\theta)$ will also be given by $\frac{K^{2}s^{\prime}t^{\prime}}{a(\theta)^{2}+b(\theta)^{2}}$. Therefore, we have that

[TABLE]

In conclusion, owing to (16) 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 $\lim_{u\to\infty}(C_{u}-\tilde{C}_{u})=0$, that is

[TABLE]

Let us introduce the notation

[TABLE]

Then, it clearly holds that

[TABLE]

In order to apply a sandwich type argument, we will prove that both $-D_{u}$ and $D_{u}$ converge to zero as $u$ tends to infinity. We will only tackle the term $D_{u}$, since the analysis of $-D_{u}$ is analogous. Note that $D_{u}=D_{u}^{1}-D_{u}^{2}$, where

[TABLE]

Regarding $D_{u}^{2}$, observe that the integral in $x$ can be computed explicitly and we can argue as follows:

[TABLE]

In the last equality, we have applied l’Hôpital’s rule. By performing a change of variables, the latter expression equals to

[TABLE]

The second term in the above sum clearly converges to zero as $u\to\infty$, while the limit of the first one equals to, thanks to l’Hôpital’s rule,

[TABLE]

Thus, we have proved that $\lim_{u\to\infty}D_{u}^{2}=\frac{1}{a(\theta)^{2}}$. On the other hand, in order to deal with $D_{u}^{1}$ we will use again a sandwich type argument, as follows. First, note that we trivially have $D_{u}^{1}\leq D_{u}^{2}$. Next, applying the changes of variables $v=uy$ and $z={xv}$, we end up with

[TABLE]

Observe that the latter limit equals to $\frac{1}{a(\theta)^{2}}$ because it corresponds to the limit of $B_{u}$ defined above in the particular case of $s^{\prime}=t^{\prime}=K=1$ and $b=0$. Hence, we obtain that

[TABLE]

and therefore $\lim_{u\to\infty}D_{u}=0$.

Step 3. Assume that either $s\neq 0$ or $t\neq 0$. By step 1, recall that we have

[TABLE]

where the terms on the right hand-side have been defined in (17) and (18), respectively. Set

[TABLE]

where $f(x_{1},x_{2},y_{1},y_{2}):=\sqrt{x_{1}x_{2}y_{1}y_{2}}\,e^{-a(\theta)(x_{2}y_{2}-x_{1}y_{1})}\cos(b(\theta)(x_{2}y_{2}-x_{1}y_{1}))$, and

[TABLE]

where $g(x_{1},x_{2},y_{1},y_{2}):=\sqrt{x_{1}x_{2}y_{1}y_{2}}\,e^{-a(\theta)((y_{2}-y_{1})x_{1}+(x_{2}-x_{1})y_{1})}\cos(b(\theta)((y_{2}-y_{1})x_{1}-(x_{2}-x_{1})y_{1}))$. Observe that $I_{1}^{\varepsilon}$ and $I_{2}^{\varepsilon}$ can be written as follows:

[TABLE]

and

[TABLE]

where $I_{2i}^{\varepsilon}$, $i=1,2,3$, are defined analogously by using the function $g$. By step 2, one verifies that

[TABLE]

In order to conclude the proof, it suffices to check that $I_{ji}^{\varepsilon}$ converges to zero as $\varepsilon\to 0$, for all $j=1,2$ and $i=1,2,3$. For this, we estimate any $I_{ji}^{\varepsilon}$ by $\tilde{I}_{ji}^{\varepsilon}$, where the latter are defined by simply bounding the cosinus by $1$. Next, we note that $\tilde{I}_{1,i}^{\varepsilon}\leq\tilde{I}_{2,i}^{\varepsilon}$, for all $i=1,2,3$, and that any of the $\tilde{I}_{2,i}^{\varepsilon}$ can be bounded by

[TABLE]

In this integral, we perform the changes of variables $\bar{x}_{i}:=\varepsilon x_{i}$ and $\bar{y}_{i}:=\varepsilon y_{i}$, $i=1,2$, we set $u:=\frac{1}{\varepsilon^{2}}$, we use that $\bar{x}_{2}\leq s^{\prime}$ and we integrate with respect to $\bar{x}_{2}$. Thus, (22) can be bounded, up to some positive constant, by (using again the notation $x_{i}$ and $y_{i}$ for the variables)

[TABLE]

Estimating now $y_{2}$ by $t^{\prime}$ inside the square root and integrating in $y_{2}$, the above expression can be bounded by (up to some constant)

[TABLE]

This expression converges to zero as $u\to\infty$, by the Monotone convergence theorem. ∎

Here is the second auxiliary result needed to prove Proposition 4.3.

Lemma 4.7**.**

Let $(0,0)\leq(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$. Then, there exists a sequence $\{C_{\varepsilon}\}_{\varepsilon>0}$ such that $\lim_{\varepsilon\to 0}C_{\varepsilon}=4(s^{\prime}-s)^{2}(t^{\prime}-t)^{2}$ and

[TABLE]

Proof.

We split the proof in four steps.

Step 1. By definition of the random field $X^{\varepsilon}$, we first observe that

[TABLE]

In order to compute the above conditional expectation, we have to consider all possible orders of $x_{1},x_{2}$ and $y_{1},y_{2}$, respectively, which corresponds to a total of 4 possibilities. Hence,

[TABLE]

We have also applied changes of variables in order to have $x_{1}\leq x_{2}$ and $y_{1}\leq y_{2}$ in all terms. We denote by $A_{i}^{\varepsilon}$, $i=1,2,3,4$, the above four terms, respectively. Thus, we have

[TABLE]

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 $\mathbb{E}\left[(A_{1}^{\varepsilon})^{2}\right]$. In fact, by Fubini theorem, we have that

[TABLE]

Note that in the above integral we have $y_{1}\leq y_{2}$ and $y_{3}\leq y_{4}$. However, in order to compute the expectation in (23), we need to consider all possible orders of the variables $y_{1},y_{2},y_{3},y_{4}$, with the restrictions $y_{1}\leq y_{2}$ and $y_{3}\leq y_{4}$. This amounts to take into account 6 different possibilities, which we split in two groups:

• (i)

$y_{1}\leq y_{2}\leq y_{3}\leq y_{4}$ and $y_{3}\leq y_{4}\leq y_{1}\leq y_{2}$,

• (ii)

$y_{1}\leq y_{3}\leq y_{2}\leq y_{4}$, $y_{1}\leq y_{3}\leq y_{4}\leq y_{2}$, $y_{3}\leq y_{1}\leq y_{4}\leq y_{2}$ and $y_{3}\leq y_{1}\leq y_{2}\leq y_{4}$.

Then, we have that

[TABLE]

where $B_{1}^{\varepsilon}(1,1),B_{2}^{\varepsilon}(1,1)$ correspond to (23) with the orders of (i), respectively, while $B_{k}^{\varepsilon}(1,1)$, $k=3,4,5,6$, correspond to (23) with the orders of (ii), respectively. It turns out that we have a similar decomposition of any of the terms \mathbb{E}\big{[}A_{i}^{\varepsilon}A_{j}^{\varepsilon}\big{]}, which we denote by

[TABLE]

Hence

[TABLE]

In the next two steps, we will focus on the analysis of (some of) the terms in the decomposition (24) of $\mathbb{E}\left[(A_{1}^{\varepsilon})^{2}\right]$. As already mentioned, the terms arising from \mathbb{E}\big{[}A_{i}^{\varepsilon}A_{j}^{\varepsilon}\big{]} can be treated analogously. We will come back to expansion (25) later in step 4.

Step 2. We claim that, for any $k=3,4,5,6$, it holds

[TABLE]

where $D:=[\frac{s}{\varepsilon},\frac{s^{\prime}}{\varepsilon}]^{4}\times[\frac{t}{\varepsilon},\frac{t^{\prime}}{\varepsilon}]^{4}$, and we recall that $a(\theta)$ is the real part of $\Psi(\theta)$. We prove this estimate for $B_{3}^{\varepsilon}(1,1)$. For the remaining terms the argument is completely analogous. So, let us assume that in (23) we have the order $y_{1}\leq y_{3}\leq y_{2}\leq y_{4}$. In this case, the expectation in (23) equals to

[TABLE]

Plugging this term in (23) and shifting the modulus inside the integral, we can infer that

[TABLE]

Performing a change of variable, we obtain that the latter term equals to

[TABLE]

In order to obtain (26), it suffices to observe that, in the domain of integration, it holds that $(y_{3}-y_{1})(x_{1}-\frac{s}{\varepsilon})\geq(y_{3}-y_{2})(x_{1}-\frac{s}{\varepsilon})$.

Step 3. Here, we prove that the right hand-side of (26) converges to zero as $\varepsilon\to 0$. Let us introduce the following notation:

[TABLE]

so we want to check that $\lim_{\varepsilon\to 0}\beta^{\varepsilon}=0$.

To start with, in the expression of $\beta^{\varepsilon}$ we bound the two square roots by using the upper limit of any $x_{i}$ and $y_{i}$. Next, we integrate with respect to $x_{4}$, $x_{3}$ and $x_{2}$. We also use the fact that, according to the statement of Proposition 4.3, we may assume that $t>0$. Thus,

[TABLE]

At this point, we integrate with respect to $y_{1}$ and $y_{4}$, thus

[TABLE]

Note that the second term in the latter sum may be bounded, up to some positive constant, by $\varepsilon^{2}$, which converges to zero. Regarding the first term, it can be bounded by

[TABLE]

which also converges to zero as $\varepsilon\to 0$.

Step 4. By (25) in step 1 and steps 2 and 3, we have that

[TABLE]

where we recall that $B_{1}^{\varepsilon}(i,j)$ and $B_{2}^{\varepsilon}(i,j)$ are the terms in the decomposition of \mathbb{E}\big{[}A_{i}^{\varepsilon}A_{j}^{\varepsilon}\big{]} with the orders of (i), respectively, and $\lim_{\varepsilon\to 0}\rho_{\varepsilon}=0$.

Focusing again (only) on the case $i=j=1$, one easily verifies that

[TABLE]

where we recall that $D:=[\frac{s}{\varepsilon},\frac{s^{\prime}}{\varepsilon}]^{4}\times[\frac{t}{\varepsilon},\frac{t^{\prime}}{\varepsilon}]^{4}$. Observing that

[TABLE]

we end up with

[TABLE]

One can get similar estimates for $B_{1}^{\varepsilon}(i,j)+B_{2}^{\varepsilon}(i,j)$ with $i,j\neq 1$. Gathering all the resulting bounds together, it can be verified that

[TABLE]

where

[TABLE]

Note that $\Theta_{\varepsilon}$ coincides with the right hand-side of equality (15) in the proof of Lemma 4.6, where in the latter it was precisely proved that

[TABLE]

Therefore, by (28) and recalling that $\lim_{\varepsilon\to 0}\rho_{\varepsilon}=0$, we conclude the proof by taking $C_{\varepsilon}:=\Theta_{\varepsilon}^{2}+\rho_{\varepsilon}$. ∎

We can now provide the proof of Proposition 4.3.

Proof of Proposition 4.3. We prove that, for all $0\leq s_{1}<\cdots<s_{n}\leq s$ and $0\leq t_{1}<\cdots<t_{n}\leq T$, and any continuous and bounded function $\varphi:\mathbb{C}\to\mathbb{R}$, we have

[TABLE]

and

[TABLE]

Since $\mathbb{P}_{\varepsilon}$ converges to $\mathbb{P}$ weakly in $\mathcal{C}([0,S]\times[0,T];\mathbb{C})$, it suffices to check that

[TABLE]

where

[TABLE]

and

[TABLE]

Indeed, in order to check the validity of the limits in (29), we will prove that

[TABLE]

We will first deal with the limit of $A_{\varepsilon}+B_{\varepsilon}$. More precisely, we have that

[TABLE]

Hence, to prove that $\lim_{\varepsilon\to 0}(A_{\varepsilon}+B_{\varepsilon})=0$, it is enough to check that \mathbb{E}\big{[}|\Delta_{s,t}X^{\varepsilon}(s^{\prime},t^{\prime})|^{2}\,|\mathcal{F}^{\varepsilon}_{s,T}\big{]} converges in $L^{2}(\Omega)$ to $2(s-s^{\prime})(t-t^{\prime})$, as $\varepsilon\to 0$. Indeed, by Lemma 4.7, we have:

[TABLE]

where $\lim_{\varepsilon\to 0}C_{\varepsilon}=4(s^{\prime}-s)^{2}(t^{\prime}-t)^{2}$. So, by Lemma 4.6, the right hand-side of (30) converges to zero as $\varepsilon\to 0$.

Let us now deal with the limit of $A_{\varepsilon}-B_{\varepsilon}$. To start with, note that

[TABLE]

We are going to prove that $\lim_{\varepsilon\to 0}\Lambda_{\varepsilon}=0$, where

[TABLE]

The limit involving the complex conjugate $e^{-i\theta L(x,y)}$ can be tackled using analogous arguments. Expanding the squared integral of $\Lambda_{\varepsilon}$, we end up with

[TABLE]

As we have already done several times throughout the paper, we consider the four possible orders of $x_{1},x_{2}$ and $y_{1},y_{2}$ and, in each of these terms, we apply a change of variables so that we have $x_{1}\leq x_{2}$ and $y_{1}\leq y_{2}$. Thus,

[TABLE]

At this point, the idea is to write $L(x_{1},y_{1})+L(x_{2},y_{2})$ and $L(x_{2},y_{1})+L(x_{1},y_{2})$ 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

[TABLE]

Recalling that $\varphi$ is a bounded function and computing the expectations of complex exponentials in terms of the Lévy exponent $\Psi(\xi)=a(\xi)+ib(\xi)$, one can easily obtain that $|\Lambda_{\varepsilon}|\leq C\,\tilde{\Lambda}_{\varepsilon}$, where $C$ is a positive constant and

[TABLE]

We finally prove that $\lim_{\varepsilon\to 0}\tilde{\Lambda}_{\varepsilon}=0$. Indeed, taking into account the integration limits of all variables and applying Fubini theorem, we have

[TABLE]

We integrate with respect to $x_{2}$ and $y_{2}$, thus

[TABLE]

Hence, we have $\lim_{\varepsilon\to 0}\tilde{\Lambda}_{\varepsilon}=0$, which implies that $\lim_{\varepsilon\to 0}\Lambda_{\varepsilon}=0$, and so $\lim_{\varepsilon\to 0}A_{\varepsilon}-B_{\varepsilon}=0$. 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 (12) in our case.

Note that $\mathsf{Re}(X)$ and $\mathsf{Im}(X)$ satisfy (12) if, for any $(0,0)\leq(s,t)\leq(s^{\prime},t^{\prime})\leq(S,T)$, $0\leq s_{1}<\cdots<s_{n}\leq s$ and $0\leq t_{1}<\cdots<t_{n}\leq t$, and any continuous bounded function $\varphi:\mathbb{C}^{n}\to\mathbb{R}$, we have

[TABLE]

Using the equality $\alpha\beta=\frac{i}{4}\left\{(\alpha-i\beta)^{2}-(\alpha+i\beta)^{2}\right\}$, we obtain that

[TABLE]

We observe that, in the analysis of the term $A_{\varepsilon}-B_{\varepsilon}$ in the proof of Proposition 4.3, we indeed proved that the two terms arising from the difference in (31) converge to zero as $\varepsilon\to 0$. So the proof is complete. ∎

5 Weak convergence for the stochastic heat equation

We consider the following one-dimensional quasi-linear stochastic heat equation:

[TABLE]

where $T>0$ stands for a fixed time horizon, $b:\mathbb{R}\rightarrow\mathbb{R}$ is a globally Lipschitz function and $\dot{W}(t,x)$ denotes the space-time white noise. We impose the initial condition $U(0,x)=u_{0}(x)$, $x\in[0,1]$, where $u_{0}:[0,1]\rightarrow\mathbb{R}$ is a continuous function, and boundary conditions of Dirichlet type:

[TABLE]

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 (32) is interpreted in the mild sense, as follows. Let $\{W(t,x);\,(t,x)\in[0,1]^{2}\}$ be a Brownian sheet defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\{\mathcal{F}_{t};\,t\in[0,1]\}$ its natural filtration. A jointly measurable and adapted random field $U=\{U(t,x);\;(t,x)\in[0,1]^{2}\}$ is a solution of (32) if it holds that

[TABLE]

for all $(t,x)\in(0,1]\times(0,1)$, 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 (33) are a consequence of [15, Thm 3.5]. For the reader’s convenience, we recall that the Green function $G$ is given by

[TABLE]

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 $\mathcal{C}([0,1]^{2})$ of continuous functions, by the family of mild solutions $\{U_{n}\}_{n\geq 0}$, where $U_{n}$ solves a stochastic heat equation perturbed by (the formal derivative of) either the real or imaginary part of the family introduced in (2):

[TABLE]

where we recall that $\{L(s,y);\,s,y\geq 0\}$ denotes a Lévy sheet and its Lévy exponent is given by $\Psi(\xi)=a(\xi)+ib(\xi)$. The constant $K$ is given in (3) and $\theta\in(0,2\pi)$, where we assume that $a(\theta)a(2\theta)\neq 0$. Note that, compared to (2), in the above expression of $X_{\varepsilon}$ 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 $X_{\varepsilon}$ in the following way:

[TABLE]

with $n=\varepsilon^{-2}$. Set

[TABLE]

Let $i\in\{1,2\}$ and consider the stochastic heat equation

[TABLE]

with initial condition $u_{0}$ and Dirichlet boundary conditions. The mild form of this equation is given by

[TABLE]

Owing to [3, Sec. 3], equation (34) admits a unique solution $U^{i}_{n}$ whose paths are continuous almost surely. Here is the main result of the section:

Theorem 5.1**.**

For any $i\in\{1,2\}$, the sequence $\{U^{i}_{n}\}_{n\geq 1}$ converges in law, as $n\to\infty$ and in the space $\mathcal{C}([0,1]^{2})$, to the solution $U$ of (33).

The proof of this theorem is based on [3, Thm. 1.4], where sufficient conditions on a family of random fields $\{\theta_{n}\}_{n\geq 1}$ have been established such that the sequence of solutions to the stochastic heat equation driven by $\theta_{n}$ converges in law to $U$, in the space of continuous functions. The first requirement is that $\theta_{n}\in L^{2}([0,1]^{2})$ 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

[TABLE]

converge in law to those of the Brownian sheet.

• (ii)

For some $q\in[2,3)$, there exists a positive constant $C_{q}$ such that, for any $f\in L^{q}([0,1]^{2})$, it holds:

[TABLE]

• (iii)

There exist $m>8$ and a positive constant $C$ such that the following is satisfied: for all $s_{0},\,s_{0}^{\prime}\in[0,1]$ and $x_{0},\,x_{0}^{\prime}\in[0,1]$ satisfying $0<s_{0}<s_{0}^{\prime}<2s_{0}$ and $0<x_{0}<x_{0}^{\prime}<2x_{0}$, and for any $f\in L^{2}([0,1]^{2})$, it holds:

[TABLE]

Hence, in the proof of Theorem 5.1 we will prove the validity of all above conditions in the case where $\theta_{n}$ is given by $\theta_{n}^{i}$, for any $i\in\{1,2\}$. 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 $\theta_{n}$ are given by the Kac-Stroock processes on the plane:

[TABLE]

where $N_{n}(t,x):=N\left(\sqrt{n}t,\sqrt{n}x\right)$, and $\{N(t,x);\,(t,x)\in[0,1]^{2}\}$ 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 $f\in L^{2}([0,1]^{2})$ and $\alpha\geq 1$. Then, for any $u,u^{\prime}\in(0,1)$ satisfying that $0<u<u^{\prime}\leq 2^{\alpha}u$, it holds

[TABLE]

for any $i\in\{1,2\}$.

Proof.

We will only deal with the case involving $\theta_{n}^{1}$, since the result for $\theta^{2}_{n}$ follows exactly in the same way. Note that we clearly have

[TABLE]

and the latter term equals to

[TABLE]

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

[TABLE]

where

[TABLE]

and

[TABLE]

At this point, we apply the inequality $zw\leq\frac{1}{2}(z^{2}+w^{2})$ in such a way that

[TABLE]

This makes that both $I_{1}^{n}$ and $I_{2}^{n}$ can be bounded by the sum of two terms of the form $I_{j,1}^{n}+I_{j,2}^{n}$, $j=1,2$, respectively, where $I_{j,1}^{n}$ involves $f(t_{1},x_{1})$ and $I_{j,2}^{n}$ involves $f(t_{2},x_{2})$. 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

[TABLE]

Plugging everything together and using (35), 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 $f\in L^{2p}([0,1]^{2})$. Then, there exists a positive constant $C_{p}$ which does not depend on $f$ such that

[TABLE]

for any $i\in\{1,2\}$.

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 $m\in\mathbb{N}$ be an even number and $f\in L^{2}([0,1]^{2})$. Then, there exists a positive constant $C_{m}$ which does not depend on $f$ such that, for all $s_{0},\,s_{0}^{\prime},x_{0},\,x_{0}^{\prime}\in[0,1]$ satisfying $0<s_{0}<s_{0}^{\prime}<2s_{0}$ and $0<x_{0}<x_{0}^{\prime}<2x_{0}$, we have that

[TABLE]

for any $i\in\{1,2\}$.

Proof.

Let $i\in\{1,2\}$. For any $(s_{0},x_{0})\in[0,1]^{2}$, we define

[TABLE]

Observe that, for all $(0,0)\leq(s_{0},x_{0})<(s_{0}^{\prime},x_{0}^{\prime})\leq(1,1)$, we have

[TABLE]

where the random field $\bar{Z}_{n}$, which does not depend on $i$, is complex-valued and given by

[TABLE]

(here $i=\sqrt{-1}$). In order to bound the right hand-side of (36), we can proceed as in the first part of the proof of the tightness result Proposition 3.1, obtaining

[TABLE]

At this point, we apply that $y_{j}<x_{0}^{\prime}<2x_{0}$ and $s_{j}<s_{0}^{\prime}<2s_{0}$, 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

[TABLE]

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. As explained above, we need that $\theta^{i}_{n}\in L^{2}([0,1]^{2})$, a.s., which is clear by definition of the random fields $\theta^{i}_{n}$, $i=1,2$, and that conditions (i), (ii) and (iii) are fulfilled.

First, note that (i) is a consequence of Theorem 1.1. Secondly, Proposition 5.3 implies that (ii) is satisfied and, finally, Proposition 5.4 assures the validity of condition (iii). ∎

Acknowledgement

The authors thank the anonymous referee for a careful reading of the manuscript.