Proof. Note that in our proof we follow ideas of the proof of Lemma 1 in Birkner [Reference Birkner2].

(a) We have the collection of independent Poisson point processes $\{N_{x,y}^{\textrm{RW}_{\xi}}\}_{x,y\in S},$ $\{N_{x,y}^{\textrm{RW}_{\eta}}\}_{x,y\in S},\quad \{N_{x,k}^{{\mathrm{br}}_{\xi}}\}_{x\in S,k\in\mathbb{Z}_{+}},\quad \{N_{x,k}^{{\mathrm{br}}_{\eta}}\}_{x\in S,k\in\mathbb{Z}_{+}}$ . Therefore, with probability 1 there is no more than one jump simultaneously. Then we can define a stopping time $T_{1}$ : the first time a jump happens. The stopping time $T_1$ is strictly positive a.s., since $(\xi_{0},\eta_{0})\in E_{\mathrm{fin}}\times E_{\mathrm{fin}}$ , and therefore the indicators $1_{\{\xi_{s-}(y)\geq u\}}, 1_{\{\eta_{s-}(y)\geq u\}},1_{\{\gamma\xi_{s-}(x)\eta_{s-}(x)\geq u\}}$ make the rate of the first jump finite. Then $(\xi_{t},\eta_{t})=(\xi_{0},\eta_{0})$ for $t\in[0,T_{1})$ . In the same way, we define a sequence of stopping times: $0=T_{0} < T_{1} < T_{2} < \cdots < \infty$ , and again the rate of jumps is finite since for any $i\geq 1$ , $(\xi_{T_{i}},\eta_{T_{i}})\in E_{\mathrm{fin}}\times E_{\mathrm{fin}}$ a.s., and this makes the rate of the $(i+1)$ th jump a.s. finite by the same reasoning as for the first jump. Clearly, by construction, the process $(\xi_{t},\eta_{t})$ is constant on the intervals $[T_{i},T_{i+1})$ . In order to show that this construction defines the process properly (that is, it does not explode in finite time), it is enough to show that $\lim_{n\rightarrow\infty}T_{n}=\infty$ , a.s. To this end define $M_{n}=\sum_{x\in S}\left(\xi_{T_{n}}(x)+\eta_{T_{n}}(x)\right)$ . Here $M_{n}$ denotes the total number of particles in both populations at time $T_{n}$ . Since the branching mechanism is critical (see (2.2)), it is easy to see that $\left\{ M_{n}\right\} _{n\geq0}$ is a non-negative martingale.

Indeed, suppose that $T_{n}$ is the stopping time originated from a ‘random walk’ jump, that is, from a jump of one of the processes

\begin{equation*}R_{t}^{\xi,x}= \sum_{y\neq x}\left\{ \mathop\int\limits_{0}^{t}\int_{\mathbb{R}_{+}}1_{\{\xi_{s-}(y)\geq u\}}N_{y,x}^{\textrm{RW}_{\xi}}({\mathrm{d}} s\,{\mathrm{d}} u)-\mathop\int\limits_{0}^{t}\int_{\mathbb{R}_{+}}1_{\{\xi_{s-}(x)\geq u\}}N_{x,y}^{\textrm{RW}_{\xi}}({\mathrm{d}} s\,{\mathrm{d}} u)\right\},\!\!\quad t\geq 0, \ x\in S,\end{equation*}

or

\begin{equation*}R_{t}^{\eta,x}=\sum_{y\neq x}\left\{ \mathop\int\limits_{0}^{t}\int_{\mathbb{R}_{+}}1_{\{\eta_{s-}(y)\geq u\}}N_{y,x}^{\textrm{RW}_{\eta}}({\mathrm{d}} s\,{\mathrm{d}} u)-\mathop\int\limits_{0}^{t}\int_{\mathbb{R}_{+}}1_{\{\eta_{s-}(x)\geq u\}}N_{x,y}^{\textrm{RW}_{\eta}}({\mathrm{d}} s\,{\mathrm{d}} u)\right\},\!\!\quad t\geq 0, \ x\in S.\end{equation*}

In that case the total number of particles does not change (this can also be readily seen from our (2.4) since $\sum_{x\in S} R_{t}^{\xi,x}=\sum_{x\in S} R_{t}^{\eta,x}=0$ ) and thus we have $M_{n}=M_{n-1}$ . Alternatively $T_{n}$ can be originated from the ‘branching’, that is from the jump of one of the processes $B_{t}^{\xi}= \sum_{x\in S}\sum_{k\geq0}\int_{0}^{t}\int_{\mathbb{R}_{+}}(k-1)1_{\{\gamma\eta_{s-}(x)\xi_{s-}(x)\geq u\}}N_{x,k}^{{\mathrm{br}}_{\xi}}({\mathrm{d}} s\,{\mathrm{d}} u)$ or $B_{t}^{\eta}= \sum_{x\in S}\sum_{k\geq0}\int_{0}^{t}\int_{\mathbb{R}_{+}}(k-1)1_{\{\gamma\eta_{s-}(x)\xi_{s-}(x)\geq u\}}N_{x,k}^{{\mathrm{br}}_{\eta}}({\mathrm{d}} s\,{\mathrm{d}} u)$ . In this case one can easily get that

\begin{equation*}{\mathrm{E}}\left(M_{n}|\, M_{0},\ldots ,M_{n-1}\right)=M_{n-1}+{\mathrm{E}}\left(Z-1\right)=M_{n-1},\end{equation*}

where Z is distributed according to the branching law $\nu$ .

Therefore, by the well-known martingale convergence theorem $\sup_{n\geq1}M_{n}<\infty$ a.s. [Reference Ikeda and Watanabe24, Theorem 1.6.4]. This implies that $\sup_{n}T_{n}=\infty$ a.s.

Now let us turn to the proof of uniqueness. Let $(\tilde{\xi},\tilde{\eta})_{t}$ be another solution to (2.4) starting from the same initial conditions $\tilde{\xi}_{0}=\xi^{0},\ \tilde{\eta}_{0}=\eta^{0}$ . We see from (2.4) that $\tilde{\xi}_{t}(x)=\xi_{t}(x),\ \tilde{\eta}_{t}(x)=\eta_{t}(x)$ for all $x\in S$ and $t\in[0,T_{1})$ , and also that $\tilde{\xi}_{T_{1}}=\xi_{T_{1}},\ \tilde{\eta}_{T_{1}}=\eta_{T_{1}}$ . Then, by induction, $(\xi,\eta)$ and $(\tilde{\xi},\tilde{\eta})$ agree on $[T_{n},T_{n+1})$ for all $n\in\mathbb{N}$ .

(b) Poisson processes have independent and stationary increments. Therefore, by construction described in (a), we can immediately see that the distribution of $(\xi_{t+h},\eta_{t+h})$ , given $\mathcal{F}_{t}$ , depends only on $(\xi_{t},\eta_{t})$ , and hence the process $(\xi_{t},\eta_{t})_{t\geq0}$ is Markov.

(c) Now we show that $(\xi_{t},\eta_{t})_{t\geq0}$ satisfies (3.1). Define a sequence of stopping times

(3.5) \begin{equation}T_{n}\,:\!=\,\inf\left\{ t\geq0\,:\,\begin{array}{c}\sum_{x\in S}\left(\xi_{t}(x)+\eta_{t}(x)\right)>n\end{array}\right\},\ n\geq 1.\end{equation}

Choose arbitrarily $m\in\mathbb{N}$ such that

\begin{equation*}\sum_{k}k^{m}\nu_{k}<\infty.\end{equation*}

For $\varphi, \psi\in E_{\mathrm{fin}}$ , define

\begin{equation*}h_{m}(\varphi,\psi)\,:\!=\,\left\Vert (\varphi,\psi)\right\Vert _{m}^{m}.\end{equation*}

Then, by an appropriate version of the Itô formula, we have that $\{ M_{t\wedge T_{n}}^{h_{m}}\}_{t\geq 0} $ is a martingale. Let $\varphi,\psi\in E_{\mathrm{fin}}$ . Apply the $L^{(2)}$ operator on $h_{m}(\varphi,\psi)$ to get

\begin{align*}L^{(2)}h_{m}(\varphi,\psi) & = \kappa\sum_{x,y}\varphi(x)p_{xy}\left\{ \left((\varphi(y)+1)^{m}-\varphi(y)^{m}\right)\right.\\ & \quad \left.+\left((\varphi(x)-1)^{m}-\varphi(x)^{m}\right)\right\} \\ & \quad +\kappa\sum_{x,y}\psi(x)p_{xy}\left\{\left((\psi(y)+1)^{m}-\psi(y)^{m}\right)\right.\\ & \quad \left.+\left((\psi(x)-1)^{m}-\psi(x)^{m}\right)\right\} \\ & \quad +\sum_{x}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\left\{ (\varphi(x)+k-1)^m-\varphi(x)^{m}\right\}\\ & \quad +\sum_{x}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\left\{ (\psi(x)+k-1)^m-\psi(x)^{m}\right\} \\ & = \kappa\sum_{x}\varphi(x)\sum_{j=1}^{m}\binom{m}{j}({-}1)^{j}\varphi(x)^{m-j}\\ & \quad +\kappa\sum_{x,y}\varphi(x)p_{xy}\sum_{j=1}^{m}\binom{m}{j}\varphi(y)^{m-j}\\ & \quad +\kappa\sum_{x}\psi(x)\sum_{j=1}^{m}\binom{m}{j}({-}1)^{j}\psi(x)^{m-j} \\ & \quad +\kappa\sum_{x,y}\psi(x)p_{xy}\sum_{j=1}^{m}\binom{m}{j}\psi(y)^{m-j}\\ & \quad +\sum_{x}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\sum_{j=2}^{m}\binom{m}{j}\varphi(x)^{m-j}(k-1)^{\,j}\\ & \quad +\sum_{x}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\sum_{j=2}^{m}\binom{m}{j}\psi(x)^{m-j}(k-1)^{\,j}.\end{align*}

where in the last equality we used the binomial expansion, the fact that $\sum_{y}p_{xy}=1$ and our assumptions $\sum_{k\geq0}(k-1)\nu_{k}=0.$ Now we can estimate

(3.6) \begin{align}\big|L^{(2)}h_{m}(\varphi,\psi)\big| & \leq \kappa\Bigg(\sum_{j=1}^{m}\binom{m}{j}\Bigg)\big(h_{m}(\varphi,\psi)+\sum_{x,y}p_{xy}\big[\varphi(x)^m + \varphi(y)^{m}+\psi(x)^m+ \psi(y)^{m}\big]\big)\nonumber\\ & \quad +\Bigg(3\gamma\sum_{j=2}^{m}\binom{m}{j}\sum_{k}\nu_{k}(k-1)^{j}\Bigg)h_{m}(\varphi,\psi)\end{align}

where we used the following simple inequalities: for $m\geq j\geq1$ ,

\begin{equation*}\varphi(x)^{m-j}\leq\varphi(x)^{m-1}\end{equation*}

(recall that $\varphi(x)$ is a non-negative integer number), and

\begin{equation*} ab^{m-1}\leq a^m+ b^m,\quad \forall\ a,b\geq 0. \end{equation*}

Since, by our assumptions, $p_{xy}=p_{yx}$ for all $x,y\in S$ , we have

(3.7) \begin{equation}\sum_{x,y\in S}p_{xy}\left[\varphi(x)^m + \varphi(y)^{m}+\psi(x)^m+ \psi(y)^{m}\right]\leq 4h_{m}(\varphi,\psi).\end{equation}

Denote

\begin{equation*}c_{m}\,:\!=\,\sum_{j=1}^{m}\binom{m}{j}=2^{m}-1,\quad c'_{m}\,:\!=\,3\gamma\sum_{j=2}^{m}\binom{m}{j}\sum_{k\geq 0}\nu_{k}(k-1)^{j}<\infty.\end{equation*}

Then by (3.6) and (3.7) we get

(3.8) \begin{align}\big|L^{(2)}h_{m}(\varphi,\psi)\big| \leq (5 \kappa c_{m}+c'_{m})h_{m}(\varphi,\psi)\,=\!:\,C_m h_{m}(\varphi,\psi).\end{align}

Now recall that for any n, $\left\{M^{h_{m}}(t\wedge T_{n})\right\}_{t\geq0}$ is a martingale. Therefore,

\begin{align*}{\mathrm{E}}\big[h_{m}\big(\xi_{t\wedge T_{n}},\eta_{t\wedge T_{n}}\big)\big] & = h_{m}(\xi^{0},\eta^{0})+{\mathrm{E}}\left[\mathop\int\limits_{0}^{t\wedge T_{n}}L^{(2)}h_{m}(\xi_{s},\eta_{s})\,{\mathrm{d}} s\right]\\ & \leq h_{m}(\xi^{0},\eta^{0})+C_{m}\mathop\int\limits_{0}^{t}{\mathrm{E}}\left[\mathbf{1}_{\left\{ s\leq T_{n}\right\} }h_{m}(\xi_{s},\eta_{s})\right]\,{\mathrm{d}} s\\ & \leq h_{m}(\xi^{0},\eta^{0})+C_{m}\mathop\int\limits_{0}^{t}{\mathrm{E}}\left[h_{m}(\xi_{s\wedge T_{n}},\eta_{s\wedge T_{n}})\right]\,{\mathrm{d}} s.\end{align*}

Thus, from Gronwall’s lemma we get that

\begin{equation*}{\mathrm{E}}\left[h_{m}(\xi_{t\wedge T_{n}},\eta_{t\wedge T_{n}})\right]\leq\exp(C_{m}t)h_{m}(\xi^{0},\eta^{0}),\end{equation*}

uniformly in n. It is easy to see from (a) that $T_n\to \infty$ as $n\to\infty$ , a.s. Thus, inequality (3.1) follows from Fatou’s lemma by letting $n\rightarrow\infty$ .

(d) Let $f\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ . We wish to show that $M^{\,f}$ is indeed a martingale. In order to do that, first we show that for any such f there is a constant $C=C(\kappa,p,\sigma,\nu,f)$ such that

(3.9) \begin{equation}\big|L^{(2)}f(\varphi,\psi)\big|\leq C\left\Vert (\varphi,\psi)\right\Vert _{2}^{2}\quad \mathrm{for\ all}\ \varphi,\psi\in E_{\mathrm{fin}}.\end{equation}

We decompose $L^{(2)}f(\varphi,\psi)$ into two parts corresponding to motion and branching mechanisms:

(3.10) \begin{equation}L^{(2)}f(\varphi,\psi)=L_{\,\mathrm{RW}}\,f(\varphi,\psi)+L_{\,\mathrm{br}}\,f(\varphi,\psi),\end{equation}

where

(3.11) \begin{align}L_{\,\mathrm{RW}}\,f(\varphi,\psi) & = \kappa\sum_{x,y\in S}\varphi(x)p_{xy}\left(\,f\big(\varphi^{x\rightarrow y},\psi\big)-f(\varphi,\psi)\right)\nonumber\\[4pt] & \quad +\kappa\sum_{x,y\in S}\psi(x)p_{xy}\left(\,f\big(\varphi,\psi^{x\rightarrow y}\big)-f(\varphi,\psi)\right), \end{align}

(3.12) \begin{align} L_{\,\mathrm{br}}\,f(\varphi,\psi) & = \sum_{x\in S}\gamma\varphi(x)\psi(x)\sum_{k\geq0}\nu_{k}\left(\,f(\varphi+(k-1)\delta_{x},\psi)-f(\varphi,\psi)\right)\nonumber\\[4pt] & \quad +\sum_{x\in S}\gamma\varphi(x)\psi(x)\sum_{k\geq0}\nu_{k}\left(\,f(\varphi,\psi+(k-1)\delta_{x})-f(\varphi,\psi)\right).\end{align}

Using the Lipshitz property of f we obtain

\begin{align*}\left|L_{\,\mathrm{RW}}\,f(\varphi,\psi)\right| & = \Bigg|\sum_{x,y\in S}\varphi(x)p_{xy}\left(\,f\big(\varphi^{(x,y)},\psi\big)-f(\varphi,\psi)\right)\nonumber\\ & \quad +\sum_{x,y\in S}\psi(x)p_{xy}\big(\,f\big(\varphi,\psi^{(x,y)}\big)-f(\varphi,\psi)\big)\Bigg|\\ & \quad\leq C_{f}\sum_{x,y\in S}\varphi(x)p_{xy}\big\Vert \big(\varphi^{(x,y)},\psi\big)-(\varphi,\psi)\big\Vert _{1}\\ & \quad +C_{f}\sum_{x,y\in S}\psi(x)p_{xy}\big\Vert \big(\varphi,\psi^{(x,y)}\big)-(\varphi,\psi)\big\Vert _{1}\\ & \quad\leq 2C_{f}\sum_{x,y\in S}\varphi(x)p_{xy}+2C_{f}\sum_{x,y\in S}\psi(x)p_{xy}\\ & = 2C_{f}\left\Vert (\varphi,\psi)\right\Vert _{1}\leq 2C_{f}\left\Vert (\varphi,\psi)\right\Vert _{2}^{2},\end{align*}

where in the last inequality we used $\left\Vert (\varphi,\psi)\right\Vert _{1}\leq\left\Vert (\varphi,\psi)\right\Vert _{2}^{2}$ , which holds since the functions $\left\{ \varphi(x)\right\} _{x\in S}$ and $\left\{ \psi(x)\right\} _{x\in S}$ are integer valued. Turning to $L_{\,\mathrm{br}}\,f(\varphi,\psi)$ we get

\begin{align*}\left|L_{\,\mathrm{br}}\,f(\varphi,\psi)\right| & = \bigg|\sum_{x\in S}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\left(\left[f(\varphi+(k-1)\delta_{x},\psi)-f(\varphi,\psi)\right]\right.\\ & \quad \left.+\left[f(\varphi,\psi+(k-1)\delta_{x})-f(\varphi,\psi)\right]\right)\bigg|\\ & \leq C_{f}\sum_{x\in S}\gamma\varphi(x)\psi(x)\sum_{k\geq 0}\nu_{k}\left(2\left\Vert (k-1)\delta_{x}\right\Vert _{1}\right)\\ & = \bigg(2C_{f}\sum_{k\geq 0}\nu_{k}|k-1|\bigg)\sum_{x\in S}\gamma\varphi(x)\psi(x)\\ & \leq 2\gamma C_{f}\sum_{k\geq0}2\nu_{k}|k-1|\cdot\left\Vert (\varphi,\psi)\right\Vert _{2}^2,\end{align*}

where in the last inequality we used the fact that $\varphi(x)\psi(x)\leq\varphi(x)^{2}+\psi(x)^{2}$ . Thus, (3.9) holds with

\begin{equation*}C\,:\!=\,C_{f}\bigg(2+2\gamma\sum_{k\geq0}\nu_{k}|k-1|\bigg).\end{equation*}

Consider a bounded $f\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ , then $M^{f}$ is a local martingale, so we have for all $t,h\geq0$

(3.13) \begin{equation}{\mathrm{E}}\big[M^{f}((t+h)\wedge T_{n})\big|\mathcal{F}_{t}\big]=M^{f}\left(\left(t+h\right)\wedge T_{n}\right).,\end{equation}

where $T_n$ is defined in (3.5). The right-hand side converges to $M^{f}(t)$ a.s., as $n\rightarrow\infty$ . Then

\begin{equation*}{\mathrm{E}}\left[\left.\mathop\int\limits_{0}^{(t+h)\wedge T_{n}}L^{(2)}f(\xi_{s},\eta_{s})\,{\mathrm{d}} s\right|\mathcal{F}_{t}\right]\rightarrow{\mathrm{E}}\left[\left.\mathop\int\limits_{0}^{t+h}L^{(2)}f(\xi_{s},\eta_{s})\,{\mathrm{d}} s\right|\mathcal{F}_{t}\right],\ \mathrm{a.s},\end{equation*}

as $n\rightarrow\infty$ . Here, we used again the dominated convergence, since by (3.9)

\begin{equation*}\mathop\int\limits_{0}^{(t+h)\wedge T_{n}}L^{(2)}f(\xi_{s},\eta_{s})\,{\mathrm{d}} s\leq C\mathop\int\limits_{0}^{t+h}\left\Vert (\xi_{s},\eta_{s})\right\Vert _{2}^{2}\,{\mathrm{d}} s\end{equation*}

and the expectation on the right-hand side of the above inequality is bounded due to (3.1) and finite initial conditions, for which clearly $\left\Vert (\xi_0,\eta_0)\right\Vert _{2}^{2}<\infty$ . Thus, $M^{f}$ is indeed a martingale in the case of bounded $f\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ .

Next, consider $f\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ which is non-negative, but not necessarily bounded. Define $f_{n}(\varphi,\psi)\,:\!=\,f(\varphi,\psi)\wedge n$ . Note that $f_{n}$ is bounded and $f_{n}\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ with Lipshitz constant $C_{f_{n}}\leq C_{f}$ . As $n\rightarrow\infty$ , we have

\begin{equation*}\begin{array}{c}M^{f_{n}}(t)\rightarrow M^{f}(t),\ \mathrm{a.s.},\\[5pt]{\mathrm{E}}\left[f_{n}(\xi_{t+h},\eta_{t+h})\,|\,\mathcal{F}_{t}\right]\rightarrow{\mathrm{E}}\left[f(\xi_{t+h},\eta_{t+h})\,|\,\mathcal{F}_{t}\right],\ \mathrm{a.s.},\end{array}\end{equation*}

by monotone convergence. Observe that $\left|L^{(2)}f_{n}(\varphi,\eta)\right|\leq C\left\Vert (\varphi,\eta)\right\Vert _{2}^{2}$ uniformly in n. We thus obtain

\begin{equation*}{\mathrm{E}}\bigg[\int_{0}^{t+h}L^{(2)}f_{n}(\xi_{s},\eta_{s})\,{\mathrm{d}} s\bigg|\mathcal{F}_{t}\bigg]\rightarrow{\mathrm{E}}\bigg[\int_{0}^{t+h}L^{(2)}f(\xi_{s},\eta_{s})\,{\mathrm{d}} s\bigg|\mathcal{F}_{t}\bigg]\end{equation*}

a.s. as $n\rightarrow\infty$ by the dominated convergence theorem. Therefore, $M^{f}$ is a martingale for non-negative Lipschitz f. For the general case, we use the decomposition of $f\in\mathrm{Lip}(E_{\mathrm{fin}}\times E_{\mathrm{fin}})$ as $f=f^{+}-f^{-}$ , where $f^{+}\,:\!=\,\max(f,0)$ and $f^{-}\,:\!=\,\max({-}f,0)$ .

The same proof holds for $N^{f}$ too, since $\frac{\partial}{\partial s}f$ is bounded by (3.3).

Proof. The proof goes through application of Lemma 3.1 and Itô’s formula to functions in the form of $f(s,\xi_s,\eta_s)=\left\langle \xi_{s},P_{t-s}\varphi\right\rangle$ and $f(s,\xi_s,\eta_s)=\left\langle \eta_{s},P_{t-s}\varphi\right\rangle$ . The proof is pretty standard and we leave details to the enthusiastic reader.

In the end the orthogonality of the martingales $N_{\cdot}^{\xi}(t,\varphi)$ and $N_{\cdot}^{\eta}(t,\psi)$ follows from independence of driving Poisson point processes.

Proof. Since $\Lambda_n$ is finite, (5.4) follows immediately from Lemma 5.1.

As for (5.5), recalling again that $\Lambda_n$ is finite, from Lemma 5.1 we get

\begin{eqnarray*}{\mathrm{E}}(\langle \xi_{t},\varphi\rangle \langle \eta_{t},\psi\rangle ) & \,=\, & {\mathrm{E}}\big(\big[\langle \xi_{0},P_{t}\varphi\rangle +N_{t}^{\xi}(t,\varphi)\big]\big[\langle \eta_{0},P_{t}\psi\rangle +N_{t}^{\eta}(t,\psi)\big]\big)\\ & =\, & \left\langle \xi_{0},P_{t}\varphi\right\rangle \left\langle \eta_{0},P_{t}\psi\right\rangle +\left\langle \eta_{0},P_{t}\psi\right\rangle {\mathrm{E}}\big(N_{t}^{\xi}(t,\varphi)\big)\\ & & +\left\langle \xi_{0},P_{t}\varphi\right\rangle {\mathrm{E}}(N_{t}^{\eta}(t,\psi))+{\mathrm{E}}\big(N_{t}^{\xi}(t,\varphi)N_{t}^{\eta}(t,\psi)\big)\\ & =\, & \left\langle \xi_{0},P_{t}\varphi\right\rangle \left\langle \eta_{0},P_{t}\psi\right\rangle ,\end{eqnarray*}

where the second and the third terms on the right-hand side are equal to zero since $N_{t}^{\xi}(t,\varphi)$ and $N_{t}^{\eta}(t,\psi)$ are martingales, and the last term vanishes because of the orthogonality of $N_{t}^{\xi}(t,\varphi)$ and $N_{t}^{\eta}(t,\psi)$ .

Proof. The result follows easily from Corollary 5.1.

Proof. The proof follows a standard procedure, using the evolution equation for the transition densities of a continuous-time nearest-neighbor random walk. As we could not find an exact reference, we have included the derivation here for completeness.

The evolution of the transition probabilities of a continuous-time Markov chain is governed by the first-order differential equation

\begin{equation*}\frac{\partial}{\partial t}p_{t}(x)=Qp_{t}(x),\quad \forall\ t\geq0,\ x\in S.\end{equation*}

In our case $Q=(q_{xy})_{x,y\in S}$ where

\begin{equation*}q_{xx}=-\kappa,\quad q_{xy}=\frac{1}{2d}\kappa\quad \mathrm{for}\ x\in S,\ y=x\pm e_{i},\ i=1,\ldots ,d\quad \mathrm{and}\quad q_{xy}=0\quad \mathrm{otherwise}.\end{equation*}

Therefore, we have

\begin{eqnarray*}\frac{\partial}{\partial t}p_{t}(x) & \,=\, & \sum_{y\in S}q_{xy}p_{t}(y)=q_{xx}(p_{t}(x)-\frac{1}{2d}\sum_{i=1}^{d}(p_{t}(x+e_{1})+p_{t}(x-e_{1})))\\ & =\, & -\kappa\left(p_{t}(x)-\frac{1}{2d}\sum_{i=1}^{d}\left[p_{t}(x+e_{i})+p_{t}(x-e_{i})\right]\right).\end{eqnarray*}

Now, integrate both sides over the interval [0, t], which gives

\begin{eqnarray*}p_{t}(x)-p_{0}(x) & \,=\, & -\kappa\mathop\int\limits_{0}^{t}\left(p_{s}(x)-\frac{1}{2d}\sum_{i=1}^{d}\big[p_{s}(x+e_{i})+p_{s}(x-e_{i})\big]\right)\,{\mathrm{d}} s\\ & =\, & -\kappa\left(g_{t}(x)-\frac{1}{2d}\sum_{i=1}^{d}\big[g_{t}(x+e_{i})+g_{t}(x-e_{i})\big]\right).\end{eqnarray*}

The result follows immediately once we recall that $p_0(x)=\delta_{x,0}$ .

Proof. We prove only (5.6), since the proof of (5.7) is the same. Again we use the representation of the process from Lemma 5.1, with $\varphi\left(\cdot\right)=\delta_{x}\left(\cdot\right)$ and use notation $\phi_{r}({\cdot})=p_{t-r}\delta_{x}({\cdot})=p_{t-r}(\cdot-x).$

For $0\leq s\leq t$ denote

\begin{eqnarray*}N_{s}^{\xi}(y) & = & N_{s}^{\xi}(t,y)=N_{s}^{\xi}(t,\delta_{y}).\\{\mathrm{E}}\big(\xi_{t}(x)^{2}\big) & = & \left(P_{t}\xi_{0}(x)\right)^{2}+{\mathrm{E}}\big(\big(N_{t}^{\xi}(x)\big)^{2}\big)\\ & = & v^{2}+{\mathrm{E}}\left(\left\langle N_{\cdot}^{\xi}(x)\right\rangle _{t}\right)\\ & = & v^{2}+v\kappa\sum_{y\in S}\mathop\int\limits_{0}^{t}\sum_{z\in S}(\phi_{r}(z)-\phi_{r}(y))^{2}p{}_{y,z}\,{\mathrm{d}} r\\ & & +\,\sigma^{2}\gamma uv\sum_{y\in S}\mathop\int\limits_{0}^{t}\phi_{r}(y)^{2}\,{\mathrm{d}} r\\ & \,=\!:\, & v^{2}+v\kappa J_{1}(t)+\sigma^{2}\gamma uv J_{2}(t),\quad t\geq0\end{eqnarray*}

where in the third equality we used again Lemma 5.1, the Fubini theorem, and Corollary 5.1 which implies ${\mathrm{E}}(\xi_{r-}(y)\eta_{r-}(y))=P_{t}\xi_{0}(y)P_{t}\eta_{0}(y)=vu$ , ${\mathrm{E}}(\xi_{r-}(y))=P_{t}\xi_{0}(y)=v.$ Now we will compute each term separately.

First, let us evaluate $J_{2}(t)$ . Recall that $\phi_{r}(y)=p_{t-r}(y-x)$ . Then we have

(5.8) \begin{align}J_{2}(t) = & \mathop\int\limits_{0}^{t}\sum_{y\in S}p_{t-r}(y-x)^{2}\,{\mathrm{d}} r =\mathop\int\limits_{0}^{t}p_{2(t-r)}(0)\,{\mathrm{d}} r =\frac{1}{2}\mathop\int\limits_{0}^{2t}p_{\tau}(0)\,{\mathrm{d}} \tau=0.5g_{2t}(0),\end{align}

where $p_{s}(x,x)=p_{s}(0,0)=p_{s}(0),$ for all $x\in S$ .

Now we will handle $J_{1}(t)$ :

(5.9) \begin{align}J_{1}(t) & = \sum_{y\in S}\mathop\int\limits_{0}^{t}\sum_{z\in S}(\phi_{r}(z)-\phi_{r}(y))^{2}p{}_{y,z}\,{\mathrm{d}} r\\ & = \sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(z-x)^{2}p_{y,z}\,{\mathrm{d}} r+\sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(y-x)^{2}p_{y,z}\,{\mathrm{d}} dr\nonumber \\ & -2\sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(z-x)p_{t-r}(y-x)p_{y,z}\,{\mathrm{d}} r.\nonumber\end{align}

We will treat each of the three terms above separately. For the first term we have

(5.10) \begin{align}\sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(z,x)^{2}p_{y,z}\,{\mathrm{d}} r & = \sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(z,x)^{2}\,{\mathrm{d}} r = 0.5g_{2t}(0)\end{align}

where the last equality follows as in (5.8).

Similarly we get

(5.11) \begin{align}\sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(y,x)^{2}p_{y,z}\,{\mathrm{d}} dr &=0.5g_{2t}(0).\end{align}

Finally. it is easy to obtain

(5.12) \begin{align}& \sum_{y\in S}\sum_{z\in S}\mathop\int\limits_{0}^{t}p_{t-r}(z,x)p_{t-r}(y,x)p_{y,z}\,{\mathrm{d}} r\nonumber\\& \qquad\qquad\qquad\qquad\qquad =0.5\frac{1}{2d}\sum_{i=1}^{d}\left[g_{2t}(e_{i})+g_{2t}({-}e_{i})\right]=\frac{1}{2d}\sum_{i=1}^{d}g_{2t}(e_{i}).\end{align}

By putting (5.10), (5.11), (5.12), and (5.8) together we have

(5.13) \begin{equation}{\mathrm{E}}\big(\xi_{t}(x)^{2}\big)=v^{2}+\sigma^{2}\gamma uv\frac{1}{2}g_{2s}(0)+v\kappa\left(g_{2t}(0)-\frac{1}{2d}\sum_{i=1}^{d}g_{2t}(e_{i})\right),\quad t\geq0,\ x\in S.\end{equation}

Now use (5.13) and Lemma 5.2 with $x=y$ to get

\begin{equation*}{\mathrm{E}}\big(\xi_{t}(x)^{2}\big)=v^{2}+\frac{1}{2}\sigma^{2}\gamma uvg_{2t}(0)+v(1-p_{2t}(0)),\quad \forall\ t\geq0,\ x\in S.\end{equation*}

Proof. The proof goes along similar lines as the proof of Lemma 5.3 and thus is omitted.

Proof of Theorem 2.3. By easy adaptation of Lemma 2.5 of [Reference Mytnik33] one gets that the mixed Laplace–Fourier transform

\begin{equation*}{\mathrm{E}}\big({\mathrm{e}}^{-(X+Y)(a+b)-{\mathrm{i}}(X-Y)(a-b)}\big),\quad a,b\geq0,\end{equation*}

determines the distribution of non-negative two-dimensional random variables (X, Y). Therefore, Theorem 2.3 follows easily from Proposition 6.1 and properties of weak convergence.

The rest of the section is organized as follows. Section 6.1 is devoted to the proof of Proposition 6.1, and the proof of one of the technical propositions is deferred to Section 6.2.

Proof. The proof is postponed until the end of this section. It is proved via a series of other results.

Proof of Proposition 6.1. Fix arbitrary $\theta_{1},\theta_{2}\geq 0$ and $a,b\geq0$ . For any $n\geq1$ , let $(u_{t}^{n},v_{t}^{n})_{t\geq0}$ be the solution to (1.3) with $Q^{n}$ being a Q-matrix of the nearest-neighbor random walk on $\Lambda_{n}$ , and initial conditions $(u_{0}^{n},v_{0}^{n})=(\xi_{0}^{n},\eta_{0}^{n})=\boldsymbol{\bar{\theta}}$ . Recall that

\begin{equation*}\boldsymbol{U}_{t}^{n}=\sum_{x\in\Lambda_{n}}u_{t}^{n}(x),\quad \boldsymbol{V}_{t}^{n}=\sum_{x\in\Lambda_{n}}v_{t}^{n}(x).\end{equation*}

Note that

(6.4) \begin{align}\lim_{n\rightarrow\infty}{\mathrm{E}}\big[F_{0,\beta_{n}(T)}^{n}\big] & = \lim_{n\rightarrow\infty}{\mathrm{E}}\Big({\mathrm{e}}^{-\big\langle \xi_{0}^{n}+\eta_{0}^{n},\tilde{u}_{\beta_{n}(T)}^{n}+\tilde{v}_{\beta_{n}(T)}^{n}\big\rangle -{\mathrm{i}}\langle \xi_{0}^{n}-\eta_{0}^{n},\tilde{u}_{\beta_{n}(T)}^{n}-\tilde{v}_{\beta_{n}(T)}^{n}\rangle }\Big)\nonumber \\& = \lim_{n\rightarrow\infty}{\mathrm{E}}\Big({\mathrm{e}}^{-\big(\mathbf{U}_{\beta_{n}(T)}^{n}+\mathbf{V}_{\beta_{n}(T)}^{n}\big) \frac{1}{\left|\Lambda_{n}\right|}(a+b)-{\mathrm{i}}\big(\mathbf{U}_{\beta_{n}(T)}^{n}-\mathbf{V}_{\beta_{n}(T)}^{n}\big) \frac{1}{\left|\Lambda_{n}\right|}(a-b)}\Big)\nonumber \\& = {\mathrm{E}}\big(e^{-(X_{T}+Y_{T})(a+b)-i(X_{T}-Y_{T})(a-b)}\big),\end{align}

where the second equality follows by a self-duality relation in Lemma 6.1, and the third equality follows by Theorem 1.1. This means that

(6.5) \begin{align}\lim_{n\rightarrow\infty}{\mathrm{E}}\left({\mathrm{e}}^{-(\boldsymbol{\xi}_{\beta(T)}^{n}+\boldsymbol{\eta}_{\beta_{n}(T)}^{n})\frac{1}{\left|\Lambda_{n}\right|}(a+b)-{\mathrm{i}}(\boldsymbol{\xi}_{\beta_{n}(T)}^{n}-\boldsymbol{\eta}_{\beta_{n}(T)}^{n})\frac{1}{\left|\Lambda_{n}\right|}(a-b)}\right)\end{align}

(6.6) \begin{align}& = \lim_{n\rightarrow\infty}{\mathrm{E}}\big[F_{\beta_{n}(T),0}^{n}\big]=\lim_{n\rightarrow\infty}{\mathrm{E}}\big[F_{0,\beta_{n}(T)}^{n}\big]\nonumber\\[3pt]& = {\mathrm{E}}\big({\mathrm{e}}^{-(X_{T}+Y_{T})(a+b)-{\mathrm{i}}(X_{T}-Y_{T})(a-b)}\big),\end{align}

where the second equality follows by Proposition 6.2, and the last equality follows by (6.4). This finishes the proof of Proposition 6.1.

Proof. The result is immediate by Lemma 3.1(d).

Proof. The result is immediate by [Reference Dawson and Perkins16, Theorem 2.2(c)(iv)], Itô’s lemma [Reference Ikeda and Watanabe24, Theorem II.5.1]), and simple algebra.

Proof. Equation (6.11) is verified in the proof of Theorem 2.4(b) in [Reference Dawson and Perkins16].

Now let us check (6.12). First, by simple algebra it is trivial to see that for any $z\in\mathbb{R}_{+}$ and $y\in\mathbb{R}$ ,

(6.13) \begin{align}\big|{\mathrm{e}}^{-z+{\mathrm{i}} y}-1\big| & \leq \left(\left|z\right|+\left|y\right|\right).\end{align}

Hence,

\begin{align*}& \sup_{\begin{array}{c}0\leq s\leq t\\0\leq r\leq t\end{array}}{\mathrm{E}}\left|g\big(\xi_{r}^{n},\eta_{r}^{n},\tilde{u}_{s}^{n},\tilde{v}_{s}^{n}\big)\right|\\& \qquad\qquad \leq \sup_{0\leq s,r\leq t}C{\mathrm{E}}\Bigg\{ \kappa\sum_{x,y\in\Lambda_{n}}\xi_{r}^{n}(x)p_{xy}^{n}\left[\tilde{u}_{s}^{n}(y)+\tilde{v}_{s}^{n}(y)+\tilde{u}_{s}^{n}(x)+\tilde{v}_{s}^{n}(x)\right]\\ & \qquad\qquad +\kappa\sum_{x,y\in\Lambda_{n}}\eta_{r}^{n}(x)p_{xy}^{n}\left[\tilde{u}_{s}^{n}(y)+\tilde{v}_{s}^{n}(y)+\tilde{u}_{s}^{n}(x)+\tilde{v}_{s}^{n}(x)\right]\\ & \qquad\qquad +\gamma\sum_{x\in\Lambda_{n}}\xi_{r}^{n}(x)\eta_{r}^{n}(x)\sum_{k\geq0}\nu_{k}\left|k-1\right|\left[\tilde{u}_{s}^{n}(x)+\tilde{v}_{s}^{n}(x)\right]\\ & \qquad\qquad +\gamma\sum_{x\in\Lambda_{n}}\xi_{r}^{n}(x)\eta_{r}^{n}(x)\sum_{k\geq0}\nu_{k}\left|k-1\right|\left[\tilde{u}_{s}^{n}(x)+\tilde{v}_{s}^{n}(x)\right]\Bigg\} ,\end{align*}

where $C>0$ is a constant and the last inequality follows from (6.13). Recall that, by Corollary 5.2, ${\mathrm{E}}[\xi_{s}^{n}(x)]=\theta_{1},\ {\mathrm{E}}[\eta_{s}^{n}(x)]=\theta_{2}$ and, ${\mathrm{E}}[\xi_{s}^{n}(x)\eta_{s}^{n}(x)]=\theta_{1}\theta_{2}$ . By [Reference Dawson and Perkins16, Theorem 2.2b(iii)],

\begin{align*}{\mathrm{E}}\left[\left\langle \tilde{u}_{s}^{n},1\right\rangle \right] = a<\infty,\quad {\mathrm{E}}\left[\left\langle \tilde{v}_{s}^{n},1\right\rangle \right]=b<\infty,\quad \forall\ s\geq0,\end{align*}

since initial conditions have a finite mass. Also note that ${{\sum_{k\geq0}\left|k-1\right|\nu_{k}<\infty}}$ . Then (6.12) holds.

Proof. By Lemmas 6.3, 6.4, and 6.5 we can apply Lemma 6.2 to the function

\begin{equation*}F_{r,s}^{n}={\mathrm{E}}\left[H\left(\xi_{r}^{n},\eta_{r}^{n},\tilde{u}_{s}^{n},\tilde{v}_{s}^{n}\right)\right],\end{equation*}

and immediately see that (6.14) holds for almost every $t>0$ . However, again by Lemmas 6.3, 6.4, and 6.5 one can see that both left-hand and right-hand sides of (6.14) are continuous in t. Hence, (6.14) holds for all $t>0$ .

Proof of Proposition 6.2. The proof is immediate by Lemma 6.6 and Proposition 6.3.

Proof. The proof is technical, however it follows easily from the proof of Lemma 2.2 in [Reference Cox, Greven and Shiga7]. Since $u^n, v^n$ have constant initial conditions it is easy to see that ${\mathrm{E}}(u_{t}^{n}(x)^p)$ , $ {\mathrm{E}}(v_{t}^{n}(x)^p)$ , ${\mathrm{E}}(u_{t}^{n}(x)^p v_t^n(x)^p)$ are constant functions in x for $p>0$ . Thus, denote $f^n_t={\mathrm{E}}((u_{t}^{n}(0))^{4}), r^n_t={\mathrm{E}}((v_{t}^{n}(0))^{4}),\ t\geq 0$ . Then following the argument in the proof of Lemma 2.2 in [Reference Cox, Greven and Shiga7, pp.175--176], one gets that there exists a constant $C>0$ such that

(6.20) \begin{align}\nonumber f^n(t) &\leq C \vartheta_1^4 + 3^5 \int_0^t p^n_{2s}(0)\,{\mathrm{d}} s \int_0^t p^n_{2s}(0)\tilde\gamma^2{\mathrm{E}}\big(\big(u_{t-s}^{n}(0)v_{t-s}^{n}(0)\big)^2\big)\,{\mathrm{d}} s\\[5pt] &\leq C \vartheta_1^4 + 3^5\int_0^t p^n_{2s}(0)\,{\mathrm{d}} s\int_0^t p^n_{2s}(0)\tilde\gamma^2\frac{1}{2}(f^n(t-s) + r^n(t-s))\,{\mathrm{d}} s,\end{align}

and, similarly,

\begin{align}\nonumber r^n(t) &\leq C \vartheta_2^4 + 3^5\int_0^t p^n_{2s}(0)\,{\mathrm{d}} s\int_0^t p^n_{2s}(0)\tilde\gamma^2\frac{1}{2}(f^n(t-s) + r^n(t-s))\,{\mathrm{d}} s,\end{align}

Letting $J^n(t)=\int_0^t p^n_{2s}(0)\,{\mathrm{d}} s,$ and

\begin{equation*} \bar h^n_t = \sup_{s\leq t} f^n_s + \sup_{s\leq t} r^n_s,\quad t\geq 0,\end{equation*}

we have

\begin{align}\nonumber \bar h^n(t) &\leq C\vartheta_1^4 \vartheta_2^4+ 3^5J^n(t)^2 \tilde\gamma^2\bar h^n(t),\quad t\geq 0.\end{align}

From Lemma 6.7 (b) we get that

\begin{equation*}\lim_{n\to\infty} J^n(\beta_n(t))=\frac{1}{2}g_\infty(0)+t\leq \frac12 g_\infty(0)+1,\quad\textrm{for}\ t\leq T \leq 1.\end{equation*}

Recalling that $\tilde\gamma<\frac{1}{\sqrt{3^{5}}(\frac12 g_{\infty}(0)+1)}$ we get that

\begin{equation*} \limsup_{n\to\infty} \bar h^n(\beta_n(T))\leq \frac{C\vartheta_1^4 \vartheta_2^4}{1-3^5\tilde\gamma^2 ( \frac12 g_\infty(0)+1)^2}<\infty, \quad T\leq 1.\end{equation*}

Since $\bar h^n(\beta_n(T))<\infty$ for each finite n, we are done.

Proof. By Lemmas 5.3 and 5.4 it is enough to show that.

\begin{equation*}\sup_{n}\sup_{t\leq\beta_{n}(T)}\sup_{x,y\in\Lambda_{n}}g_{t}^{n}(x,y)<\infty,\end{equation*}

where $\left\{ g_{t}^{n}(\cdot,\cdot)\right\} _{t\geq0}$ is the Green function of the symmetric nearest-neighbor random walk on $\Lambda_{n}$ . For any $t\geq0$ , $x,y\in\Lambda_{n}$ , we have

\begin{equation*}g_{t}^{n}(x,y)\leq g_{\beta_{n}(T)}^{n}(x,y)\leq g_{\beta_{n}(T)}^{n}(0,0).\end{equation*}

By Lemma 6.7 (b) $\sup_{n}g_{\beta_{n}(T)}^{n}(0,0)$ is finite, and we are done.

Proof. We will show only (6.21), since the proof of (6.22) follows along the similar lines:

(6.23) \begin{align}\nonumber|e_{\mathrm{br}}(T,n)| & = \Bigg|{\mathrm{E}}\mathop\int\limits_{0}^{\beta_{n}(T)}\tilde f_{T,s}^{n}\sum_{x\in\Lambda_{n}}\gamma \xi_{s}^{n}(x)\eta_{s}^{n}(x)\Bigg(\sum_{j=1}^2 G^{2,j}(\tilde{u}_{\beta_{n}(T)-s}^{n},\tilde{v}_{\beta_{n}(T)-s}^{n},x) \Bigg)\,{\mathrm{d}} s\Bigg|\\\nonumber&\leq C_{(6.17)} {\mathrm{E}}\mathop\int\limits_{0}^{\beta_{n}(T)}|\tilde f_{T,s}^{n}|\sum_{x\in\Lambda_{n}}\gamma \xi_{s}^{n}(x)\eta_{s}^{n}(x){\mathrm{e}}^{\tilde{u}_{\beta_{n}(T)-s}^{n}(x)+\tilde{v}_{\beta_{n}(T)-s}^{n}(x)}\\\nonumber&\times\left(\tilde{u}_{\beta_{n}(T)-s}^{n}(x)^3+\tilde{v}_{\beta_{n}(T)-s}^{n}(x)^3\right)\,{\mathrm{d}} s\\&\leq C_{(6.17)}\gamma {\mathrm{E}}\mathop\int\limits_{0}^{\beta_{n}(T)}\sum_{x\in\Lambda_{n}} \xi_{s}^{n}(x)\eta_{s}^{n}(x)\left(\tilde{u}_{\beta_{n}(T)-s}^{n}(x)^3+\tilde{v}_{\beta_{n}(T)-s}^{n}(x)^3\right)\,{\mathrm{d}} s,\end{align}

where the first inequality follows by (6.17) and the second inequality follows by the trivial inequality

\begin{equation*}|\tilde f_{T,s}^{n} | \xi_{s}^{n}(x)\eta_{s}^{n}(x) {\mathrm{e}}^{\tilde{u}_{\beta_{n}(T)-s}^{n}(x)+\tilde{v}_{\beta_{n}(T)-s}^{n}(x)}\leq \xi_{s}^{n}(x)\eta_{s}^{n}(x),\end{equation*}

for all $x\in \Lambda_n$ (recall the definition of $\tilde f_{T,s}^{n}$ in (6.18)).

Consider the process $(\hat u^{n},\hat v^{n})$ that solves (1.3) equations with initial conditions

\begin{equation*}\hat u_{0}^{n}(x)=|\Lambda_{n}|\tilde{u}_{0}^{n}(x),\quad \hat v_{0}^{n}(x)=|\Lambda_{n}|\tilde{v}_{0}^{n}(x),\quad \forall\ x\in\Lambda_{n}.\end{equation*}

Then for any $s>0$ ,

\begin{equation*}\hat u_{s}^{n}(x)=|\Lambda_{n}|\tilde{u}_{s}^{n}(x),\quad \hat v_{s}^{n}(x)=|\Lambda_{n}|\tilde{v}_{s}^{n}(x),\quad \forall\ x\in\Lambda_{n}.\end{equation*}

Therefore, by the above, (6.23), and Fubini’s theorem, we get

\begin{eqnarray*}|e_{\mathrm{br}}(T,n)|&\,\leq\,&|\Lambda_{n}|^{-3}C_{(6.17)}\gamma \mathop\int\limits_{0}^{\beta_{n}(T)} {\mathrm{E}}\Bigg[\sum_{x\in\Lambda_{n}} \xi_{s}^{n}(x)\eta_{s}^{n}(x)\left(\hat{u}_{\beta_{n}(T)-s}^{n}(x)^3+\hat{v}_{\beta_{n}(T)-s}^{n}(x)^3\right)\Bigg]\,{\mathrm{d}} s\\ & \leq\, & C_{(6.17)}\gamma T |\Lambda_{n}|^{-1}\theta_{1}\theta_{2}\sup_{s\leq\beta_{n}(T)}\frac{1}{\left|\Lambda_{n}\right|}{\mathrm{E}}\Bigg[\sum_{x\in\Lambda_{n}}\Big(\Big(\hat u_{\beta_{n}(T)-s}^{n}(x)\Big)^{3}+\Big(\hat v_{\beta_{n}(T)-s}^{n}(x)\Big)^{3}\Big)\Bigg]\\ & \leq\, & C_{(6.17)}\gamma T |\Lambda_{n}|^{-1}\theta_{1}\theta_{2}\sup_{x\in\Lambda_{n}}\sup_{s\leq\beta_{n}(T)}{\mathrm{E}}\Big[\Big(\left(\hat u_{\beta_{n}(T)-s}^{n}(x)\right)^{3}+\Big(\hat v_{\beta_{n}(T)-s}^{n}(x)\Big)^{3}\Big)\Big],\end{eqnarray*}

where the second inequality follows by Corollary 5.2, and the third inequality is trivial. With this, to obtain (6.21), it is enough to show that

\begin{equation*}\sup_{n\geq 1}\sup_{x\in\Lambda_{n}}\sup_{s\leq\beta_{n}(T)}{\mathrm{E}}\Big[\Big(\Big(\hat u_{\beta_{n}(T)-s}^{n}(x)\Big)^{3}+\Big(\hat v_{\beta_{n}(T)-s}^{n}(x)\Big)^{3}\Big)\Big]<\infty.\end{equation*}

However, this follows from Lemma 6.8 and Jensen’s inequality.

Proof. We will take care of $e_{\xi,\textrm{RW},1}(T,n)$ ; the proof for $e_{\eta,\textrm{RW},1}(T,n)$ is the same:

(6.24) \begin{multline}\left|e_{\xi,\textrm{RW},1}(T,n)\right|\\\begin{aligned}\quad \quad \quad \leq\, & C\kappa{\mathrm{E}}\Bigg(\mathop\int\limits_{0}^{\beta_{n}(T)}\left| \tilde f_{T,s}^{n}\right|\Bigg\{ \Bigg|\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)\Bigg|\\ & +\Bigg|\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)^{2}-\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)^{2}\Bigg|\,{\mathrm{d}} s\Bigg\} \Bigg)\\ \quad \quad \quad \leq & C{\mathrm{E}}\Bigg(\mathop\int\limits_{0}^{\beta_{n}(T)}J_{1}^{n}(s)\,{\mathrm{d}} s+\mathop\int\limits_{0}^{\beta_{n}(T)}J_{2}^{n}(s)\,{\mathrm{d}} s\Bigg),\end{aligned}\end{multline}

where

\begin{eqnarray*}J_{1}^{n}(s) & \,=\, & \Bigg|\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)\Bigg|,\\J_{2}^{n}(s) & \,=\, & \Bigg|\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)^{2}-\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)^{2}\right)\Bigg|.\end{eqnarray*}

Let us bound the expected value of $J_{1}^{n}$ :

\begin{eqnarray*}{\mathrm{E}}(J_{1}^{n}(s)) & \,=\, & {\mathrm{E}}\Bigg|\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)\Bigg|\\ & \leq\, & \sqrt{{\mathrm{E}}\Bigg[\Bigg(\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)\Bigg)^{2}\Bigg]}.\end{eqnarray*}

Now we will recall the following representation from [Reference Dawson and Perkins16, Theorem 2.2] with $\varphi=\delta_{x}$ :

\begin{equation*}\begin{cases}\tilde{u}_{t}^{n}(x)=P_{t}^{n}\tilde{u}_{0}^{n}(x)+\sum\limits_{z\in\Lambda_{n}}\int_{0}^{t}p_{t-s}^{n}(x-z)\sqrt{\tilde{\gamma}\tilde{u}_{s}^{n}(z)\tilde{v}_{s}^{n}(z)}\,{\mathrm{d}} B_{s}(z), & x\in\Lambda_{n},\\[12pt]\tilde{v}_{t}^{n}(x)=P_{t}^{n}\tilde{v}_{0}^{n}(x)+\sum\limits_{z\in\Lambda_{n}}\int_{0}^{t}p_{t-s}^{n}(x-z)\sqrt{\tilde{\gamma}\tilde{u}_{s}^{n}(z)\tilde{v}_{s}^{n}(z)}\,{\mathrm{d}} W_{s}(z), & x\in\Lambda_{n},\end{cases}\end{equation*}

to get

(6.25) \begin{align}N_{r}^{t}(x,y) & \,:\!=\, P_{t-r}^{n}\tilde{u}_{r}^{n}(x)-P_{t-r}^{n}\tilde{u}_{r}^{n}(y)\\[5pt] & = P_{t}^{n}\tilde{u}_{0}^{n}(x)-P_{t}^{n}\tilde{u}_{0}^{n}(y)\nonumber \\[5pt] & \quad +\sum_{z\in\Lambda_{n}}\mathop\int\limits_{0}^{r}\left(p_{t-s}^{n}(x-z)-p_{t-s}^{n}(y-z)\right)\sqrt{\tilde{\gamma}\tilde{u}_{s}^{n}(z)\tilde{v}_{s}^{n}(z)}\,{\mathrm{d}} B_{s}(z),\quad r\leq t,\nonumber\end{align}

where the last equality follows from the Chapman–Kolmogorov formula. Similarly, for $r\leq t$ we get

(6.26) \begin{align}M_{r}^{t}(x,y) \,:\!=\, & P_{t-r}^{n}\tilde{v}_{t}^{n}(x)-P_{t-r}^{n}\tilde{v}_{t}^{n}(y)\\[5pt] = & P_{t}^{n}\tilde{v}_{0}^{n}(x)-P_{t}^{n}\tilde{v}_{0}^{n}(y)\nonumber \\[5pt] & +\sum_{z\in\Lambda_{n}}\mathop\int\limits_{0}^{r}\left(p_{t-s}^{n}(x-z)-p_{t-s}^{n}(y-z)\right)\sqrt{\tilde{\gamma}\tilde{u}_{s}^{n}(z)\tilde{v}_{s}^{n}(z)}\,{\mathrm{d}} W_{s}(z)\nonumber\end{align}

where $\left\{ B_{\cdot}(z)\right\} _{z\in\Lambda_{n}}$ , $\left\{ W_{\cdot}(z)\right\} _{z\in\Lambda_{n}}$ are orthogonal Brownian motions.

Note that $\left\{ N_{r}^{t}(x,y)\right\} _{0\leq r\leq t}$ and $\left\{ M_{r}^{t}(x,y)\right\} _{0\leq r\leq t}$ are martingales; in addition

(6.27) \begin{equation}\begin{array}{c}\tilde{u}_{t}^{n}(x)-\tilde{u}_{t}^{n}(y)=\left.P_{t-r}^{n}\tilde{u}_{r}^{n}(x)-P_{t-r}^{n}\tilde{u}_{r}^{n}(y)\right|_{r=t}=\left.N_{r}^{t}(x,y)\right|_{r=t},\\[5pt]\tilde{v}_{t}^{n}(x)-\tilde{v}_{t}^{n}(y)=\left.P_{t-r}^{n}\tilde{v}_{r}^{n}(x)-P_{t-r}^{n}\tilde{v}_{r}^{n}(y)\right|_{r=t}=\left.M_{r}^{t}(x,y)\right|_{r=t}.\end{array}\end{equation}

Then by orthogonality of the Brownian motions $B_{\cdot}(z)$ and $W_{\cdot}(z)$ for all ${{z\in\Lambda_{n}}}$ , and the Itô formula we get

\begin{multline*}\sum_{x,y\in\Lambda_{n}}p_{x,y}^{n}\xi_{\beta_{n}(T)-s}^{n}M_{s}^{s}(x,y)N_{s}^{s}(x,y)\\\begin{aligned}= & \sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\left(\tilde{u}_{s}^{n}(x)-\tilde{u}_{s}^{n}(y)\right)\left(\tilde{v}_{s}^{n}(x)-\tilde{v}_{s}^{n}(y)\right)\\= & \sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\sum_{z\in\Lambda_{n}}\mathop\int\limits_{0}^{s}\left(P_{s-r}^{n}\tilde{u}_{r}^{n}(x)-P_{s-r}^{n}\tilde{u}_{r}^{n}(y)\right)\\ & \times\left(p_{s-r}^{n}(x-z)-p_{s-r}^{n}(y-z)\right)\sqrt{\tilde\gamma\tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)}\,{\mathrm{d}} W_{r}(z)\\ & +\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\sum_{z\in\Lambda_{n}}\mathop\int\limits_{0}^{s}\left(P_{s-r}^{n}\tilde{v}_{r}^{n}(x)-P_{s-r}^{n}\tilde{v}_{r}^{n}(y)\right)\\ & \times\left(p_{s-r}^{n}(x-z)-p_{s-r}^{n}(y-z)\right)\sqrt{\tilde\gamma\tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)}\,{\mathrm{d}} B_{r}(z)\\\,=\!:\, & \sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\sum_{z\in\Lambda_{n}}\tilde{I}_{1,1}^{n}(s,x,y,z)\\ & +\sum_{x,y\in\Lambda_{n}}p_{xy}^{n}\xi_{\beta_{n}(T)-s}^{n}(x)\sum_{z\in\Lambda_{n}}\tilde{I}_{1,2}^{n}(s,x,y,z)\\\,=\!:\,&I_{1,1}^{n}(s)+I_{1,2}^{n}(s).\end{aligned}\end{multline*}

Note that

(6.28) \begin{equation}{\mathrm{E}}\left(J_{1}^{n}(s)\right)\leq C\sqrt{\big({\mathrm{E}}\big[I_{1,1}^{n}(s)\big)^{2}\big]+{\mathrm{E}}\big[\big(I_{1,2}^{n}(s)\big)^{2}\big]}.\end{equation}

Thus, let us bound ${\mathrm{E}}[(I_{1,1}^{n}(s))^{2}]$ : for all $ s\leq\beta_{n}(T)$ , we have

(6.29) \begin{align}{\mathrm{E}}\big[\big(I_{1,1}^{n}(s)\big)^{2}\big] = & \sum_{x_{1},y_{1}\in \Lambda_n}\sum_{x_{2},y_{2}\in \Lambda_n}{\mathrm{E}}\big(\xi_{\beta_{n}(T)-s}^{n}(x_{1})\xi_{\beta_{n}(T)-s}^{n}(x_{2})\big)\\\nonumber & \times p_{x_{1},y_{1}}^{n}p_{x_{2},y_{2}}^{n}\sum_{z_{1}\in\Lambda_{n}}\sum_{z_{2}\in\Lambda_{n}}{\mathrm{E}}\left[\tilde{I}_{1,1}^{n}(s,x_{1},y_{1},z_{1})\tilde{I}_{1,1}^{n}(s,x_{2},y_{2},z_{2})\right].\nonumber\end{align}

Note that for $z_{1}\neq z_{2}$ , $\tilde{I}_{1,1}^{n}(r,x_{1},y_{1},z_{1})$ and $\tilde{I}_{1,1}^{n}(r,x_{2},y_{2},z_{2})$ are orthogonal square integrable martingales for $r\leq s$ and, hence,

\begin{align*} & \sum_{z_{1}\in\Lambda_{n}}\sum_{z_{2}\in\Lambda_{n}}{\mathrm{E}}\left[\tilde{I}_{1,1}^{n}(s,x_{1},y_{1},z_{1})\tilde{I}_{1,1}^{n}(s,x_{2},y_{2},z_{2})\right]\\ & \quad = \sum_{z\in\Lambda_{n}}{\mathrm{E}}\left[\tilde{I}_{1,1}^{n}(s,x_{1},y_{1},z)\tilde{I}_{1,1}^{n}(s,x_{2},y_{2},z)\right]\\ & \quad = \tilde\gamma\sum_{z\in\Lambda_{n}}{\mathrm{E}}\Bigg[\mathop\int\limits_{0}^{s}\left(P_{s-r}^{n}\tilde{u}_{r}^{n}(x_{1})-P_{s-r}^{n}\tilde{u}_{r}^{n}(y_{1})\right)\left(p_{s-r}^{n}(x_{1}-z)-p_{s-r}^{n}(y_{1}-z)\right)\\ & \qquad \times\left(P_{s-r}^{n}\tilde{u}_{r}^{n}(x_{2})-P_{s-r}^{n}\tilde{u}_{r}^{n}(y_{2})\right)\left(p_{s-r}^{n}(x_{2}-z)-p_{s-r}^{n}(y_{2}-z)\right)\tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)\,{\mathrm{d}} r\Bigg]\\ & \quad = \tilde\gamma\sum_{z\in\Lambda_{n}}{\mathrm{E}}\Bigg[\mathop\int\limits_{0}^{s}\sum_{z_{1}\in\Lambda_{n}}\left(p_{s-r}^{n}(x_{1}-z_{1})-p_{s-r}^{n}(y_{1}-z_{1})\right)\tilde{u}_{r}^{n}(z_{1})\\[3pt] & \qquad \times\sum_{z_{2}\in\Lambda_{n}}\left(p_{s-r}^{n}(x_{2}-z_{2})-p_{s-r}^{n}(y_{2}-z_{2})\right)\tilde{u}_{r}^{n}(z_{2})\\[3pt] & \qquad \times\left(p_{s-r}^{n}(x_{1}-z)-p_{s-r}^{n}(y_{1}-z)\right)\left(p_{s-r}^{n}(x_{2}-z)-p_{s-r}^{n}(y_{2}-z)\right)\\[3pt] & \qquad \times \tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)\,{\mathrm{d}} r\Bigg]\\[3pt] & \quad\leq \tilde\gamma\sum_{z\in \Lambda_n}\mathop\int\limits_{0}^{s}\sum_{z_{1}\in \Lambda_n}\sum_{z_{2}\in \Lambda_n} \hat J_{1,1}(\vec{x},\vec{y},\vec{z},s-r){\mathrm{E}}\left[\tilde{u}_{r}^{n}(z_{1})\tilde{u}_{r}^{n}(z_{2})\tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)\right]\,{\mathrm{d}} r,\end{align*}

where $\vec{x}=(x_{1},x_{2}),\vec{y}=(y_{1},y_{2}),\vec{z}=(z_{1},z_{2},z)$ and

\begin{equation*}\hat J_{1,1}(\vec{x},\vec{y},\vec{z},s-r)=\mathop\Pi\limits_{i=1,2}\left|p_{s-r}^{n}(x_{i}-z_{i})-p_{s-r}^{n}(y_{i}-z_{i})\right|\left|p_{s-r}^{n}(x_{i}-z)-p_{s-r}^{n}(y_{i}-z)\right|.\end{equation*}

By Lemma 6.8 and assumption on the initial conditions of $(\tilde{u},\tilde{v})$ ,

\begin{equation*}{\mathrm{E}}\left[\tilde{u}_{r}^{n}(z_{1})\tilde{u}_{r}^{n}(z_{2})\tilde{u}_{r}^{n}(z)\tilde{v}_{r}^{n}(z)\right]\end{equation*}

is bounded by $C\left|\Lambda_{n}\right|^{-4}$ uniformly on $z,z_{1},z_{2}\in\Lambda_{n}$ , $r\leq\beta_{n}(T)$ and $n\geq 1$ . Therefore,

(6.30) \begin{align}\sum_{z_{1}\in\Lambda_{n}}&\sum_{z_{2}\in\Lambda_{n}}{\mathrm{E}}\left[\tilde{I}_{1,1}^{n}(s,x_{1},y_{1},z_{1})\tilde{I}_{1,1}^{n}(s,x_{2},y_{2},z_{2})\right]\nonumber\\[3pt]&\leq C\left|\Lambda_{n}\right|^{-4}\sum_{z\in \Lambda_n}\mathop\int\limits_{0}^{s}\sum_{z_{1}\in \Lambda_n}\sum_{z_{2}\in \Lambda_n} \hat J_{1,1}(\vec{x},\vec{y},\vec{z},s-r)\,{\mathrm{d}} r.\end{align}

Denote

\begin{eqnarray*}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r) & \,:\!=\, & \sum_{z\in\Lambda_{n}}\sum_{z_{1}\in\Lambda_{n}}\sum_{z_{2}\in\Lambda_{n}} \hat J_{1,1}(\vec{x},\vec{y},\vec{z},s-r). \end{eqnarray*}

Now we decompose the term on the right-hand side of (6.30) into two terms

(6.31) \begin{equation}C\left|\Lambda_{n}\right|^{-4}\mathop\int\limits_{0}^{\left(s-n^{\delta}\right)_{+}}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r)dr+C\left|\Lambda_{n}\right|^{-4}\mathop\int\limits_{\left(s-n^{\delta}\right)_{+}}^{s}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r)dr\end{equation}

for some $\delta\in(2,d)$ .

By Lemma 6.7 (a) we get

\begin{equation*}\lim_{n\rightarrow\infty}\sup_{t>n^{\delta}}\sup_{z_{1},z_{2}\in\Lambda_{n}}\left|\Lambda_{n}\right|\left|p_{t}^{n}(z_{1},z_{2})-(2n)^{-d}\right|=0,\end{equation*}

for any $\delta>2$ . This implies that, for any $\delta>2$ , there exists a sequence $a_{n}=a_{n}(\delta)$ , such that

(6.32) \begin{equation}\sup_{s\geq n^{\delta}}\sup_{w_{1},w_{2},w_{3}\in\Lambda_{n}}\left|p_{s}^{n}(w_{1},w_{2})-p_{s}^{n}(w_{3},w_{2})\right|\leq\frac{a_{n}}{\left|\Lambda_{n}\right|},\end{equation}

where $a_{n}\rightarrow0$ , as $n\rightarrow\infty$ .

By (6.32) we immediately get

(6.33) \begin{equation}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r)\leq\left|\Lambda_{n}\right|^{3}a_{n}^{4}\left|\Lambda_{n}\right|^{-4}=\left|\Lambda_{n}\right|^{-1}a_{n}^{4},\end{equation}

for $s>n^{\delta}$ and $r\leq s-n^{\delta}$ . Hence for $s\leq \beta_n(T)$ , we get

(6.34) \begin{align}\left|\Lambda_{n}\right|^{-4}\mathop\int\limits_{0}^{(s-n^{\delta})_{+}}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r)dr & \leq\left|\Lambda_{n}\right|^{-4} a_{n}^{4}\left|\Lambda_{n}\right|^{-1}\mathop\int\limits_{0}^{(s-n^{\delta})_{+}}1dr \leq C\left|\Lambda_{n}\right|^{-4}a_{n}^{4},\end{align}

where the last inequality follows since $s\leq \beta_n(T)=\left|\Lambda_{n}\right|T$ .

Let us treat the second term in (6.31). Note that

\begin{equation*}\sum_{z_{i}\in\Lambda_{n}}\left|p_{s}^{n}(x_{i}-z_{i})-p_{s}^{n}(y_{i}-z_{i})\right|\leq\sum_{z_{i}\in\Lambda_{n}}p_{s}^{n}(x_{i}-z_{i})+p_{s}^{n}(y_{i}-z_{i})\leq2, \forall\ i=1,2,\ s\geq 0.\end{equation*}

In addition,

\begin{multline*}\sum_{z\in\Lambda_{n}}\left|p_{s-r}^{n}(x_{1}-z)-p_{s-r}^{n}(y_{1}-z)\right|\left|p_{s-r}^{n}(x_{2}-z)-p_{s-r}^{n}(y_{2}-z)\right|\\\begin{aligned}\quad & \leq\sum_{z\in\Lambda_{n}} \left(p_{s-r}^{n}(x_{1}-z)+p_{s-r}^{n}(y_{1}-z)\right)\left(p_{s-r}^{n}(x_{2}-z)+p_{s-r}^{n}(y_{2}-z)\right)\\ & =p_{2(s-r)}^{n}(x_{1}-x_{2})+p_{2(s-r)}^{n}(y_{1}-y_{2})+p_{2(s-r)}^{n}(x_{1}-y_{2})+p_{2(s-r)}^{n}(y_{1}-x_{2}),\end{aligned}\end{multline*}

where the last equality follows from the Chapman–Kolmogorov formula. Then

(6.35) \begin{multline}C\left|\Lambda_{n}\right|^{-4}\mathop\int\limits_{(s-n^{\delta})_+}^{s}\tilde{J}_{1,1}(\vec{x},\vec{y},s-r)\,{\mathrm{d}} r\\\quad\leq C\left|\Lambda_{n}\right|^{-4}\mathop\int\limits_{0}^{n^{\delta}}\left(p_{2r}^{n}(x_{1}-x_{2})+p_{2r}^{n}(y_{1}-y_{2}) +p_{2r}^{n}(x_{1}-y_{2})+p_{2r}^{n}(y_{1}-x_{2})\right)\,{\mathrm{d}} r.\end{multline}

By (6.30), (6.31), (6.34), and (6.35) we get

\begin{multline*}\sum_{z_{1}\in\Lambda_{n}}\sum_{z_{2}\in\Lambda_{n}}{\mathrm{E}}\left[\tilde{I}_{1,1}^{n}(s,x_{1},y_{1},z_{1})\tilde{I}_{1,1}^{n}(s,x_{2},y_{2},z_{2})\right]\\\begin{aligned}\quad\,\leq & C\left|\Lambda_{n}\right|^{-4}\left(a_{n}^{4}+\int_{0}^{n^{\delta}}\left(p_{2r}^{n}(x_{1}-x_{2})+p_{2r}^{n}(y_{1}-y_{2}) +p_{2r}^{n}(x_{1}-y_{2})+p_{2r}^{n}(y_{1}-x_{2})\right)dr\right).\end{aligned}\end{multline*}

Use the above inequality, (6.29), and also Corollary 6.1 and Lemma 6.10 to get

(6.36) \begin{align}{\mathrm{E}}\big[\big(I_{1,1}^{n}(s)\big)^{2}\big] & \leq C\bigg(\sum_{x_{1},y_{1}\in \Lambda_n}\sum_{x_{2},y_{2}\in \Lambda_n}p_{x_{1},y_{1}}^{n}p_{x_{2},y_{2}}^{n}\left|\Lambda_{n}\right|^{-4}a_{n}^{4}+\left|\Lambda_{n}\right|^{-4}n^{\delta}\left|\Lambda_{n}\right|\bigg)\nonumber \\ & \leq C\big(\left|\Lambda_{n}\right|^{-2}a_{n}^{4}+\left|\Lambda_{n}\right|^{-3}n^{\delta}\big).\end{align}

In the same way, we handle $I_{1,2}^{n}(s)$ and get

(6.37) \begin{align}{\mathrm{E}}\big[\big(I_{1,2}^{n}(s)\big)^{2}\big] & \leq C\bigg(\sum_{x_{1},y_{1}\in \Lambda_n}\sum_{x_{2},y_{2}\in \Lambda_n}p_{x_{1},y_{1}}^{n}p_{x_{2},y_{2}}^{n}\left|\Lambda_{n}\right|^{-4}a_{n}^{4}+\left|\Lambda_{n}\right|^{-4}n^{\delta}\left|\Lambda_{n}\right|\bigg)\nonumber \\ & \leq C\big(\left|\Lambda_{n}\right|^{-2}a_{n}^{4}+\left|\Lambda_{n}\right|^{-3}n^{\delta}\big).\end{align}

By (6.36), (6.37), and (6.28), we have

\begin{eqnarray*}{\mathrm{E}}\left[J_{1}^{n}(s)\right] & \leq & \sqrt{{\mathrm{E}}\Big[\Big(I_{1,1}^{n}(s)\Big)^{2}+\Big(I_{1,2}^{n}(s)\Big)^{2}\Big]}\\ & \leq & C\sqrt{\left|\Lambda_{n}\right|^{-2}a_{n}^{4}+\left|\Lambda_{n}\right|^{-3}n^{\delta}}\\ & \leq & C\left|\Lambda_{n}\right|^{-1}a_{n}^{2}+C\left|\Lambda_{n}\right|^{-3/2}n^{\delta/2}.\end{eqnarray*}

Thus,

(6.38) \begin{align}\int_{0}^{\beta_{n}(T)}{\mathrm{E}}\left(J_{1}^{n}(s)\right)\,{\mathrm{d}} s & \leq C\int_{0}^{\beta_{n}(T)}\left(\left|\Lambda_{n}\right|^{-1}a_{n}^{2}+\left|\Lambda_{n}\right|^{-3/2}n^{\delta/2}\right)\,{\mathrm{d}} s\\ & \leq C\left|\Lambda_{n}\right|\left(\left|\Lambda_{n}\right|^{-1}a_{n}^{2}+\left|\Lambda_{n}\right|^{-3/2}n^{\delta/2}\right)\nonumber \\ & \leq Ca_{n}^{2}+C\left(\frac{n^{\delta}}{\left|\Lambda_{n}\right|}\right)^{1/2}\rightarrow0,\quad\mathrm{as}\ n\rightarrow\infty,\nonumber\end{align}

where the last convergence holds since $\delta < d$ and $\left|\Lambda_{n}\right|=(2n+1)^{d}$ .

Now we are ready to treat $J_{2}^{n}(s)$ in a similar way.

By the Itô formula,

\begin{equation*}\left(M_{r}^{s}(x,y)\right)^{2}=\mathop\int\limits_{0}^{s}M_{r}^{s}(x,y)\,{\mathrm{d}} M_{r}^{s}(x,y)+\left\langle M_{\cdot}^{s}(x,y)\right\rangle _{s}\end{equation*}

and

\begin{equation*}\left(N_{r}^{s}(x,y)\right)^{2}=\mathop\int\limits_{0}^{s}N_{r}^{s}(x,y)\,{\mathrm{d}} N_{r}^{s}(x,y)+\left\langle N_{\cdot}^{s}(x,y)\right\rangle _{s}.\end{equation*}

Note that $\left\langle M_{\cdot}^{t}(x,y)\right\rangle _{t}=\left\langle N_{\cdot}^{t}(x,y)\right\rangle _{t}$ , and recall (6.27); therefore

\begin{equation*}J_{2}^{n}(s)=\sum_{x,y\in\Lambda_{n}}\frac{1}{2}p_{x,y}^{n}\xi_{\beta_{n}(T)-s}^{n}\left[\mathop\int\limits_{0}^{s}M_{r}^{s}(x,y)\,{\mathrm{d}} M_{r}^{s}(x,y)-\mathop\int\limits_{0}^{s}N_{r}^{s}(x,y)\,{\mathrm{d}} N_{r}^{s}(x,y)\right].\end{equation*}

If we follow the steps of computations for $J_{1}^{n}(s)$ , we get that

\begin{equation*}\lim_{n\rightarrow\infty} {\mathrm{E}}\left( \mathop\int\limits_{0}^{\beta_{n}(T)} J_{2}^{n}(s)\,{\mathrm{d}} s\right)=0.\end{equation*}

Use this, (6.38), and (6.24) to finish the proof.