2. Main results

In this section we give our main result. For later use, we first introduce some notation and definitions. We denote by $\{S_n\}_{n\geq 0}$ the discrete-time simple random walk on $\mathbb{Z}^d$ , i.e. $\mathbb{P}(S_{n+1}=y\mid S_n=x)={1}/{2d}$ for any $n\geq 0$ , $x\in \mathbb{Z}^d$ , and $y\sim x$ . We denote by $\{Y_t\}_{t\geq 0}$ the continuous-time simple random walk on $\mathbb{Z}^d$ with generator L given by $Lh(x)=({1}/{2d})\sum_{y\sim x}(h(y)-h(x))$ for any bounded h from $\mathbb{Z}^d$ to $\mathbb{R}$ and $x\in \mathbb{Z}^d$ . As defined in Section 1, we use $\gamma_d$ to denote the probability that $\{S_n\}_{n\geq 1}$ never returns to O again conditioned on $S_0=O$ , i.e.

\begin{align*} \gamma_d = \mathbb{P}(S_n\neq O\,\text{for all}\,n\geq 1 \mid S_0=O) = \mathbb{P}(Y_t\neq O\,\text{for any}\,t\geq 0\mid Y_0=x)\end{align*}

for any $x\sim O$ . For any $t\geq 0$ , we use $p_t(\,\cdot\,, \cdot\,)$ to denote the transition probabilities of $Y_t$ , i.e. $p_t(x,y)=\mathbb{P}(Y_t=y\mid Y_0=x)$ for any $x,y\in \mathbb{Z}^d$ . For any $x\in \mathbb{Z}^d$ , we define

(2.1) \begin{equation} \Phi(x)=\mathbb{P}(S_n=x\,\text{for some}\,n\geq 0\mid S_0=O).\end{equation}

We use $\vec{1}$ to denote the configuration in $\mathbb{Y}$ where all vertices take the value 1. For any $\eta\in \mathbb{Y}$ , as defined in Section 1, we denote by $\mathbb{E}_\eta$ the expectation operator of $\{\eta_t\}_{t\geq 0}$ conditioned on $\eta_0=\eta$ . Furthermore, for any probability measure $\mu$ on $\mathbb{Y}$ , we use $\mathbb{E}_\mu$ to denote the expectation operator of $\{\eta_t\}_{t\geq 0}$ conditioned on $\eta_0$ being distributed with $\mu$ .

Now we recall more properties of the NBCPP $\{\eta_t\}_{t\geq 0}$ proved in [Reference Liggett11, Chapter 9] and [Reference Xue and Zhao18, Section 2].

Proposition 2.1. (Liggett, [Reference Liggett11].) For any $\eta\in \mathbb{Y}$ , $t\geq 0$ , and $x\in \mathbb{Z}^d$ ,

$$\mathbb{E}_\eta\eta_t(x)=\sum_{y\in \mathbb{Z}^d}p_t(x, y)\eta(y).$$

Furthermore, $\mathbb{E}_{\vec{1}}\,\eta_t(x)=1$ for any $x\in \mathbb{Z}^d$ and $t\geq 0$ .

Proposition 2.1 follows from (1.1) directly. In detail, since the generator L of the simple random walk on $\mathbb{Z}^d$ satisfies

\begin{align*} L(x,y)= \begin{cases} -1 & \text{if}\ x=y, \\ \dfrac{1}{2d} & \text{if}\ x\sim y, \\ 0 & \,\text{otherwise}, \end{cases}\end{align*}

we have $\mathbb{E}_\eta\eta_t(x)=\sum_{y\in \mathbb{Z}^d}{\mathrm{e}}^{tL}(x,y)\mathbb{E}_\eta\eta_t(y)$ by (1.1), where ${\mathrm{e}}^{tL}=\sum_{n=0}^{+\infty}{t^nL^n}/{n!}=p_t$ .

Proposition 2.2. (Liggett, [Reference Liggett11].) There exist a series of functions $\{q_t\}_{t\geq 0}$ from $(\mathbb{Z}^d)^2$ to $\mathbb{R}$ and a series of functions $\{\hat{q}_t\}_{t\geq 0}$ from $(\mathbb{Z}^d)^4$ to $\mathbb{R}$ such that

(2.2) \begin{align} \mathbb{E}_{\vec{1}}\,(\eta_t(O)\eta_t(x)) & = \mathbb{E}_{\vec{1}}\,(\eta_t(y)\eta_t(x+y)) = \sum_{z\in \mathbb{Z}^d}q_t(x,z), \end{align}

(2.3) \begin{align} \mathbb{E}_\eta(\eta_t(x)\eta_t(y)) & = \sum_{z,w\in \mathbb{Z}^d}\hat{q}_t((x,y), (z,w))\eta(z)\eta(w) \end{align}

for any $t\geq 0$ , $x,y\in \mathbb{Z}^d$ . and $\eta\in \mathbb{Y}$ . Furthermore,

(2.4) \begin{equation} q_t(O,w)=\sum_{(y,z):\,y-z=w}\hat{q}_t((x,x), (y,z)) \end{equation}

for any $x,w\in \mathbb{Z}^d$ .

Proposition 2.2 follows from (1.2). In detail, (2.3) holds by taking $\hat{q}_t={\mathrm{e}}^{tH}$ . When $\eta_0=\vec{1}$ , applying the spatial homogeneity of $\eta_t$ , we have

(2.5) \begin{equation} F_{\vec{1}, t}(O,x)=F_{\vec{1}, t}(O, -x)=F_{\vec{1}, t}(y, y+x)=F_{\vec{1}, t}(y+x, y).\end{equation}

Denote by $A_t(x)$ the expression $F_{\vec{1}, t}(O,x)$ , applying (1.2) and (2.5), we have

\begin{align*}\frac{{\mathrm{d}}}{{\mathrm{d}} t}A_t=QA_t\end{align*}

for any $t\geq 0$ and a $\mathbb{Z}^d\times \mathbb{Z}^d$ matrix $Q=\{q(x,y)\}_{x,y\in \mathbb{Z}^d}$ independent of t. Consequently, (2.2) holds by taking $q_t={\mathrm{e}}^{tQ}$ . Since

\begin{align*} \frac{{\mathrm{d}}}{{\mathrm{d}} t}A_t(x) & = \frac{{\mathrm{d}}}{{\mathrm{d}} t}F_{\vec{1}, t}(z+x,z) \\ & = \sum_u\sum_vH((z+x,z), (u,v))F_{\vec{1}, t}(u,v) = \sum_{y\in \mathbb{Z}^d}\sum_{(u,v):u-v=y}H((z+x,z), (u,v))A_t(y),\end{align*}

the $\mathbb{Z}^d\times \mathbb{Z}^d$ matrix Q can be chosen such that $q(x,y)=\sum_{(u, v): u-v=y}H((z+x, z), (u, v))$ for any $x,y,z\in \mathbb{Z}^d$ . Applying this equation repeatedly, we have

\begin{align*} q^n(x,y)=\sum_{(u, v): u-v=y}H^n((z+x, z), (u, v))\end{align*}

for all $n\geq 1$ by induction, and then

(2.6) \begin{equation} q_t(x,y)=\sum_{(u, v): u-v=y}\hat{q}_t((z+x, z), (u, v)).\end{equation}

As a result, (2.4) holds.

Proposition 2.3 follows from [Reference Liggett11, Theorem 2.8.13, Corollary 2.8.20, Theorem 9.3.17]. In detail, let $\{\overline{p}(x.y)\}_{x,y\in \mathbb{Z}^d}$ be the one-step transition probabilities of the discrete-time simple random walk on $\mathbb{Z}^d$ , and $\{\overline{q}(x,y)\}_{x,y\in \mathbb{Z}^d}$ be defined as

\begin{align*} \overline{q}(x,y)= \begin{cases} \frac12{q(x,y)} & \text{if}\ x\neq y, \\[5pt] \frac12{q(x,y)}+1 & \text{if}\ x=y. \end{cases}\end{align*}

Furthermore, let $\overline{h}\colon \mathbb{Z}^d\rightarrow (0,+\infty)$ be defined as

\begin{align*} \overline{h}(x)=\frac{\Phi(x)+h_{\lambda,d}}{h_{\lambda,d}}.\end{align*}

Then, direct calculations show that $\overline{p}$ , $\overline{q}$ , and $\overline{h}$ satisfy the assumption of [Reference Liggett11, Theorem 2.8.13]. Consequently, applying [Reference Liggett11, Theorem 2.8.13], $\lim_{t\rightarrow+\infty}\overline{q}_t(x,y)=0$ for all $x,y\in \mathbb{Z}^d$ , where $\overline{q}_t={\mathrm{e}}^{-t}{\mathrm{e}}^{t\overline{q}}$ . It is easy to check that $q_t=\overline{q}_{2t}$ and hence the first limit in (i) is zero. By (2.6), $\hat{q}_t(x,y,z,w)\leq q_t((x-y), (z-w))$ and hence the second limit in (i) is zero. Therefore, (i) holds. It is not difficult to show that $\overline{p}$ , $\overline{q}$ , and $\overline{h}$ also satisfy the assumption of [Reference Liggett11, Corollary 2.8.20]. Applying this corollary, we have $\lim_{t\rightarrow+\infty}\sum_{y}q_t(x,y)=\overline{h}(x)$ . Equation (2.2) shows that $\sum_{y}q_t(x,y)=\mathbb{E}(\eta_t(z)\eta_t(z+x))$ , and hence

\begin{align*} \lim_{t\rightarrow+\infty}\mathbb{E}_{\vec{1}}\,((\eta_t(x))^2) & = \overline{h}(0)=1+\frac{1}{h_{\lambda, d}}, \\ \lim_{t\rightarrow+\infty}{\rm Cov}_{\vec{1}}\,(\eta_t(x), \eta_t(y)) & = \overline{h}(x,y) - 1 = \frac{\Phi(x-y)}{h_{\lambda,d}}.\end{align*}

Note that the second limit also applies Proposition 2.1. These last two limits along with (i) ensure that $\{q_t\}_{t\geq 0}$ and $\vec{1}$ satisfy the assumption of [Reference Liggett11, Theorem 9.3.17]. By this theorem, (ii) and (iii) hold.

Here we briefly recall how to obtain Proposition 2.4 from the main results given in [Reference Xue and Zhao18]. Note that, in [Reference Xue and Zhao18], limit behaviors are discussed for $\eta_{tN^2}$ as $N\rightarrow+\infty$ and t is fixed. All these behaviors can be equivalently given for $\eta_t$ as $t\rightarrow+\infty$ . In [Reference Xue and Zhao18], a discrete-time symmetric random walk $\{S_n^{\lambda,d}\}_{n\geq 0}$ on $(\mathbb{Z}^d)^4$ and function $\overline{H}^{\lambda,d}\colon (\mathbb{Z}^d)^4\times(\mathbb{Z}^d)^4\rightarrow [1, +\infty)$ are introduced such that

\begin{align*} \mathbb{E}^{\lambda,d}_{\vec{1}}\Bigg(\prod_{i=1}^4\eta_t(x_i)\Bigg) \leq \mathbb{E}_{\vec{x}}\Bigg(\prod_{n=0}^{+\infty}\overline{H}(S_n, S_{n+1})\Bigg)\end{align*}

for any $\vec{x}=(x_1, x_2, x_3, x_4)\in (\mathbb{Z}^d)^4$ and $t\geq 0$ . We refer the reader to [Reference Xue and Zhao18] for precise expressions of $\overline{H}$ and the transition probabilities of $\{S_n\}_{n\geq 0}$ . It is shown in [Reference Xue and Zhao18] that there exist $\,\hat{\!d}_0$ and $\hat{\lambda}_0$ such that

(2.7) \begin{equation} \sup_{\vec{x}\in(\mathbb{Z}^d)^4} \mathbb{E}_{\vec{x}}\Bigg(\Bigg(\prod_{n=0}^{+\infty}\overline{H}(S_n,S_{n+1})\Bigg)^{1+\varepsilon}\Bigg) < +\infty\end{equation}

for some $\varepsilon=\varepsilon(\lambda, d)>0$ when $d\geq \,\hat{\!d}_0$ and $\lambda\geq \hat{\lambda}_0$ . Taking

$$d_0=\max\{5, \,\hat{\!d}_0\}, \qquad \lambda_0=\max\bigg\{\hat{\lambda}_0, 1+\frac{1}{2d_0(2\gamma_{d_0}-1)}\bigg\},$$

(i) holds. By Proposition 2.3(ii) and Fatou’s lemma, we have

\begin{align*} \mathbb{E}_{\nu_{\lambda,d}}((\eta_0(O)^4)) \leq \liminf_{t\rightarrow+\infty}\mathbb{E}^{\lambda,d}_{\vec{1}}((\eta_t(x))^4).\end{align*}

Then, Proposition 2.4(ii) follows from (2.7). It is further shown in [Reference Xue and Zhao18] that

\begin{align*} |{\rm Cov}_{\vec{1}}\,(\eta_t(x)\eta_t(v),\eta_t(y)\eta_t(z))| \leq \mathbb{E}_{(x,v,y,z)}\Bigg(\prod_{n=0}^{+\infty}\overline{H}(S_n, S_{n+1}), \sigma<+\infty\Bigg)\end{align*}

for any $x,v,y,z\in \mathbb{Z}^d$ and $t\geq 0$ , where $\sigma=\inf\{n\colon\{S_n(1), S_n(2)\}\bigcap\{S_n(3), S_n(4)\}\neq\emptyset\}$ and $S_n(j)$ is the jth component of $S_n$ . Applying Proposition 2.4(ii) and (iii), the left-hand side in the last inequality can be replaced by $|{\rm Cov}_{\nu_{\lambda,d}}(\eta_0(x)\eta_0(v),\eta_0(y)\eta_0(z))|$ . By (2.7) and the Hölder inequality, (iii) holds by checking that

(2.8) \begin{equation} \lim_{M\rightarrow+\infty}\sup_{\stackrel{(x,y,z)\in(\mathbb{Z}^d)^3:}{\|x-y\|_1\wedge\|x-z\|_1\geq M}} \mathbb{P}_{(x,x,y,z)}(\sigma<+\infty)=0.\end{equation}

By the transition probabilities of $\{S_n\}_{n\geq 0}$ , $\{S_n(i)-S_n(j)\}_{n\geq 0}$ is a lazy version of the simple random walk on $\mathbb{Z}^d$ for any $1\leq i\neq j\leq 4$ . Hence, (2.8) follows from the transience of the simple random walk on $\mathbb{Z}^d$ for $d\geq 3$ , and then (iii) holds.

Now we give our main results. The first one is about the law of large numbers of the occupation time process $\big\{\int_0^t \eta_u(O)\,{\mathrm{d}} u\big\}_{t\geq 0}$ .

It is natural to ask what happens for the occupation time for small $\lambda$ and $d=1,2$ . Currently, we have the following result.

Theorem 2.2. Assuming that $d\geq 1$ and $\lambda<{1}/{2d}$ , if the NBCPP on $\mathbb{Z}^d$ starts from $\vec{1}$ , then

\begin{align*} \lim_{N\rightarrow+\infty}\mathbb{E}_{\vec{1}}\bigg(\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg)^2\bigg)=+\infty \end{align*}

for any $t>0$ .

Our third main result is about the CLT of the occupation time process. To state our result, we first introduce some notation and definitions. For any $t\geq 0$ and $N\geq 1$ , we define

\begin{align*} X_t^N=\frac{1}{\sqrt{N}\,}\int_0^{tN}(\eta_u(O)-1)\,{\mathrm{d}} u.\end{align*}

We write $X_t^N$ as $X_{t,\lambda,d}^N$ when we need to distinguish d and $\lambda$ .

Here we recall the definition of ‘weak convergence’. Let S be a topological space. Assuming that $\{Y_n\}_{n\geq 1}$ is a sequence of S-valued random variables and Y is an S-valued random variable, then we say that $Y_n$ converges weakly to Y when and only when $\lim_{n\rightarrow+\infty}\mathbb{E}f(Y_n)=\mathbb{E}f(Y)$ for any bounded and continuous f from S to $\mathbb{R}$ .

Now we give our central limit theorem.

Theorem 2.3. Let $d_0$ and $\lambda_0$ be defined as in Proposition 2.4. Assume that $d\geq d_0$ , $\lambda>\lambda_0$ , and the NBCPP on $\mathbb{Z}^d$ starts from $\nu_{\lambda, d}$ . Then, for any integer $m\geq 1$ and $t_1,t_2,\ldots,t_m\geq 0$ , $\big(X_{t_1, \lambda, d}^N, X_{t_2, \lambda, d}^N,\ldots, X_{t_m, \lambda, d}^N\big)$ converges weakly, with respect to the Euclidean topology of $\mathbb{R}^m$ , to $\sqrt{C_1(\lambda, d)}(B_{t_1}, B_{t_2},\ldots,B_{t_m})$ as $N\rightarrow+\infty$ , where $\{B_t\}_{t\geq 0}$ is a standard Brownian motion and

\begin{align*} C_1(\lambda, d) = \frac{2\int_0^{+\infty}\int_0^{+\infty}p_{r+\theta}(O, O)\,{\mathrm{d}} r\,{\mathrm{d}}\theta} {h_{\lambda, d}\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta}. \end{align*}

Remark 2.3. Theorem 2.3 is consistent with a calculation of variance. Applying Propositions 2.1–2.3 and the Markov property of $\{\eta_t\}_{t\geq 0}$ ,

\begin{align*} \lim_{N\rightarrow+\infty}{\rm Var}_{\nu_{\lambda,d}}(X_t^N) = \frac{2t}{h_{\lambda, d}}\int_0^{+\infty}\sum_x\Phi(x)p_r(O,x)\,{\mathrm{d}} r. \end{align*}

Applying the strong Markov property of the simple random walk,

\begin{align*} \int_0^{+\infty}p_\theta(x,O)\,{\mathrm{d}}\theta = \Phi(x)\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta \end{align*}

and hence

\begin{align*} \int_0^{+\infty}\sum_x\Phi(x)p_\theta(O,x)\,{\mathrm{d}}\theta & = \frac{\int_0^{+\infty}\int_0^{+\infty}\sum_xp_r(O,x)p_\theta(x,O)\,{\mathrm{d}} r\,{\mathrm{d}}\theta} {\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta} \\ & = \frac{\int_0^{+\infty}\int_0^{+\infty}p_{r+\theta}(O, O)\,{\mathrm{d}} r\,{\mathrm{d}}\theta} {\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta}. \end{align*}

Note that $p_t(O,O)=O(t^{-{d}/{2}})$ . Hence,

$$ \frac{\int_0^{+\infty}\int_0^{+\infty}p_{r+\theta}(O, O)\,{\mathrm{d}} r\,{\mathrm{d}}\theta} {\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta} \in (0, +\infty) $$

when and only when $d\geq 5$ . That is why we guess that Theorem 2.3 holds for all $d\geq 5$ and sufficiently large $\lambda$ . Note that

$$\gamma_d=\frac{1}{\int_0^{+\infty}p_\theta(O, O)\,{\mathrm{d}}\theta}$$

by the strong Markov property, and hence

\begin{align*} C_1(\lambda, d)=\frac{1}{h_{\lambda, d}}C(x)\big|_{x=O}, \end{align*}

where $\{C(x)\}_{x\in \mathbb{Z}^d}$ are constants given in [Reference Cox and Griffeath3, Theorem 1] for cases where $d\geq 5$ . Explicit calculations of the constants occurring in the main theorems and their dependency on d are given in [Reference Birkner and Zähle1, Section 1] and [Reference Cox and Griffeath3, Sections 0 and 2].

Theorem 2.3 shows that the central limit theorem of the NBCPP is an analogue of that of voter models and branching random walks given in [Reference Birkner and Zähle1, Reference Cox and Griffeath3] when the dimension d is sufficiently large. Using Proposition 2.3 and the fact that $p_t(O,O)=O(t^{-{d}/{2}})$ , calculations of variances similar to that in Remark 2.3 show that

\begin{align*} \lim_{N\rightarrow+\infty}\frac{1}{N\log N}{\rm Cov}_{\nu_{\lambda,d}} \bigg(\int_0^{tN}(\eta_u(O)-1)\,{\mathrm{d}} u,\int_0^{sN}(\eta_u(O)-1)\,{\mathrm{d}} u\bigg) = K_4\min\{t,s\}\end{align*}

when $d=4$ and

\begin{align*} \lim_{N\rightarrow+\infty}\frac{1}{N^{{3}/{2}}}{\rm Cov}_{\nu_{\lambda, d}} \bigg(\int_0^{tN}\!\!(\eta_u(O)-1)\,{\mathrm{d}} u,\int_0^{sN}\!(\eta_u(O)-1)\,{\mathrm{d}} u\bigg) = K_3\big(t^{{3}/{2}}+s^{{3}/{2}}-|t-s|^{{3}/{2}}\big)\end{align*}

when $d=3$ for any $t,s>0$ and some constants $K_3, K_4\in (0, +\infty)$ . These two limits provide evidence for us to guess that, as N grows to infinity,

$$\bigg\{\frac{1}{N^{{3}/{4}}}\int_0^{tN}(\eta_u(O)-1)\,{\mathrm{d}} u\bigg\}_{t\geq 0}$$

converges weakly to $\sqrt{K_3}$ times the fractional Brownian motion with Hurst parameter $\frac34$ when $d=3$ , and

$$\bigg\{\frac{1}{\sqrt{N\log N}\,}\int_0^{tN}(\eta_u(O)-1)\,{\mathrm{d}} u\bigg\}_{t\geq 0}$$

converges weakly to $\{\sqrt{K_4}B_t\}_{t\geq 0}$ when $d=4$ , where $\{B_t\}_{t\geq 0}$ is the standard Brownian motion. Furthermore, Remark 2.3 provides evidence for us to guess that Theorem 2.3 holds for all $d\geq 5$ . That is to say, central limit theorems of occupation times of NBCPPs on $\mathbb{Z}^3$ , $\mathbb{Z}^4$ , and $\mathbb{Z}^d$ for $d\geq 5$ are guessed to be analogues of those of voter models and branching random walks in each case. However, our current proof of Theorem 2.3 relies heavily on the fact that $\sup_{t\geq 0}\mathbb{E}_{\vec{1}}^{\lambda, d}((\eta_t(O))^4)<+\infty$ when d and $\lambda$ are sufficiently large, which we have not managed to prove yet for small d. That is why we currently only discuss the NBCPP on $\mathbb{Z}^d$ with d sufficiently large.

Another possible way to extend Theorem 2.3 to cases with small d is to check the ergodicity of the NBCPP, as mentioned in Remark 2.1. If the NBCPP is ergodic, then we can apply Birkhoff’s theorem for f with the form $f(\eta)=\eta^2(O)$ and then an upper bound of fourth moments of $\{\eta_t(O)\}_{t\geq 0}$ is not required. We will work on this way as a further investigation.

Proofs of Theorems 2.1, 2.2, and 2.3 are given in Section 3. The proof of Theorem 2.1 is relatively easy, where it is shown that

\begin{align*} \lim_{N\rightarrow+\infty}\mathbb{E}_{\nu_{\lambda, d}}\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg)=t, \qquad \lim_{N\rightarrow+\infty}{\rm Var}_{\nu_{\lambda, d}}\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg)=0\end{align*}

according to Propositions 2.1 and 2.3. The proof of Theorem 2.2 utilizes the coupling relationship between the NBCPP and the contact process to show that

\begin{align*} \mathbb{E}_{\vec{1}}\,((\eta_t(O))^2) \geq \exp\bigg\{\frac{(1-2\lambda d)t}{(1+2\lambda d)}\bigg\}.\end{align*}

The proof of Theorem 2.3 is inspired by [Reference Birkner and Zähle1], which gives CLTs of occupation times of critical branching random walks. The core idea of the proof is to decompose $\int_0^t(\eta_u(O)-1)\,{\mathrm{d}} u$ as a martingale $M_t$ plus a remainder $R_t$ such that the quadratic variation process $({1}/{N})\langle M\rangle_{tN}$ of $({1}/{\sqrt{N}})M_{tN}$ converges to $C_1 t$ in $L^2$ and $({1}/{\sqrt{N}})R_{tN}$ converges to 0 in probability as $N\rightarrow+\infty$ . To give the above decomposition, a resolvent function of the simple random walk $\{Y_t\}_{t\geq 0}$ on $\mathbb{Z}^d$ is utilized. There are two main technical difficulties in the execution of the above strategy for the NBCPP. The first one is to prove that ${\rm Var}(({1}/{N})\langle M\rangle_{tN})$ converges to 0 as $N\rightarrow+\infty$ , which ensures that the limit of $({1}/{N})\langle M\rangle_{tN}$ is deterministic. The second one is to show that $\mathbb{E}(({1}/{N})\sup_{0\leq s\leq tN}(M_s-M_{s-})^2)$ converges to 0 as $N\rightarrow+\infty$ , which ensures that the limit process of $\{({1}/{\sqrt{N}})M_{tN}\}_{t\geq 0}$ is continuous. To overcome the first difficulty, we relate the calculation of ${\rm Var}(\langle M\rangle_{tN})$ to a random walk on $\mathbb{Z}^{2d}$ according to the Kolmogorov–Chapman equation of linear systems introduced in [Reference Liggett11]. To overcome the second difficulty, we bound $\mathbb{P}(({1}/{N})\sup_{0\leq s\leq tN}(M_s-M_{s-})^2>\varepsilon)$ from above by $O\big(({1}/{N^2})\int_0^{tN}\mathbb{E}(\eta^4_s(O))\,{\mathrm{d}} s\big)$ according to the strong Markov property of our process. For the mathematical details, see Section 3.

3. Theorem proofs

In this section we prove Theorems 2.1, 2.2, and 2.3. We first give the proof of Theorem 2.1.

Proof of Theorem 2.1. Throughout this proof we assume that $d\geq 3$ and $\lambda>{1}/({2d(2\gamma_d-1)})$ . We only prove (i), since (ii) follows from an analysis similar to that leading to the $L^2$ convergence in (i). So in this proof we further assume that $\eta_0$ is distributed with $\nu_{\lambda,d}$ . By Proposition 2.3, the NBCPP starting from $\nu_{\lambda, d}$ is stationary. Hence, applying Birkhoff’s ergodic theorem,

\begin{align*} \lim_{N\rightarrow+\infty}\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u = t\mathbb{E}_{\nu_{\lambda,d}}(\eta(O)\mid\mathcal{I}) \quad \text{a.s.}, \end{align*}

where $\mathcal{I}$ is the set of all invariant events. As a result, we only need to show that the convergence in Theorem 2.1 is in $L^2$ .

Applying Proposition 2.3, $\{\eta_u(O)\}_{u\geq 0}$ are uniformly integrable and hence

\begin{align*} \mathbb{E}_{\nu_{\lambda, d}}\eta(O)=\lim_{t\rightarrow+\infty}\mathbb{E}_{\vec{1}}\eta_t(O)=1. \end{align*}

Therefore,

\begin{align*} \mathbb{E}_{\nu_{\lambda, d}}\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg) = t \end{align*}

for any $N\geq 1$ and $t>0$ . Consequently,

\begin{align*} \mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u - t\bigg)^2\bigg) = \frac{1}{N^2}{\rm Var}_{\nu_{\lambda, d}}\bigg(\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg). \end{align*}

Hence, to prove the $L^2$ convergence in Theorem 2.1 we only need to show that

(3.1) \begin{equation} \lim_{N\rightarrow+\infty}\frac{1}{N^2}{\rm Var}_{\nu_{\lambda, d}}\bigg(\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg) = 0 \end{equation}

for any $t\geq 0$ . Since $\nu_{\lambda,d}$ is an invariant measure of our process,

(3.2) \begin{align} {\rm Var}_{\nu_{\lambda, d}}\bigg(\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg) & = 2\int_0^{tN}\bigg(\int_0^\theta{\rm Cov}_{\nu_{\lambda, d}}(\eta_r(O),\eta_{\theta}(O))\,{\mathrm{d}} r\bigg)\,{\mathrm{d}}\theta \notag \\ & = 2\int_0^{tN}\bigg(\int_0^\theta{\rm Cov}_{\nu_{\lambda, d}}(\eta_0(O),\eta_{\theta-r}(O))\,{\mathrm{d}} r\bigg)\,{\mathrm{d}}\theta. \end{align}

Applying Propositions 2.1 and 2.3,

(3.3) \begin{align} {\rm Cov}_{\nu_{\lambda,d}}(\eta_0(O),\eta_t(O)) & = \mathbb{E}_{\nu_{\lambda,d}}(\eta_0(O)\eta_t(O)) - 1 \notag \\ & = \mathbb{E}_{\nu_{\lambda,d}}\Bigg(\eta_0(O)\Bigg(\sum_{y\in \mathbb{Z}^d}p_t(O, y)\eta_0(y)\Bigg)\Bigg)-1 \notag \\ & = \sum_{y\in\mathbb{Z}^d}p_t(O,y)(\mathbb{E}_{\nu_{\lambda,d}}(\eta_0(O)\eta_0(y)) - 1) \notag \\ & = \sum_{y\in\mathbb{Z}^d}p_t(O,y){\rm Cov}_{\nu_{\lambda,d}}(\eta_0(O),\eta_0(y)) \notag\\ & = \frac{1}{h_{\lambda, d}}\sum_{y\in \mathbb{Z}^d}p_t(O, y)\Phi(y). \end{align}

Therefore, by (3.3), for any $t\geq 0$ ,

(3.4) \begin{equation} -1 \leq {\rm Cov}_{\nu_{\lambda,d}}(\eta_0(O),\eta_t(O)) \leq \frac{1}{h_{\lambda,d}}\sum_{y\in\mathbb{Z}^d}p_t(O,y) = \frac{1}{h_{\lambda, d}}, \end{equation}

and hence $\sup_{t\geq 0}|{\rm Cov}_{\nu_{\lambda, d}}(\eta_0(O), \eta_t(O))|<+\infty$ . By (3.2) and (3.4), to prove (3.1) we only need to show that

(3.5) \begin{equation} \lim_{t\rightarrow+\infty}{\rm Cov}_{\nu_{\lambda, d}}(\eta_0(O), \eta_t(O))=0. \end{equation}

Since $d\geq 3$ , for any $\varepsilon>0$ there exists $M>0$ such that $\Phi(y)<\varepsilon$ when $\|y\|_1\geq M$ . As a result, by (3.3),

\begin{align*} \limsup_{t\rightarrow+\infty}{\rm Cov}_{\nu_{\lambda, d}}(\eta_0(O), \eta_t(O)) & \leq \limsup_{t\rightarrow+\infty}\frac{\varepsilon}{h_{\lambda,d}}\sum_{y:\|y\|_1\geq M}\! p_t(O,y) + \frac{1}{h_{\lambda, d}}\sum_{y:\|y\|_1\leq M}\!\!\Big(\lim_{t\rightarrow+\infty}p_t(O, y)\Big) \\ & = \limsup_{t\rightarrow+\infty}\frac{\varepsilon}{h_{\lambda,d}}\sum_{y:\|y\|_1\geq M}p_t(O,y) \leq \frac{\varepsilon}{h_{\lambda, d}}. \end{align*}

Since $\varepsilon$ is arbitrary, let $\varepsilon\rightarrow 0$ and then (3.5) holds. As we have explained, Theorem 2.1 follows from (3.5), and the proof is complete.

Now we prove Theorem 2.2.

Proof of Theorem 2.2. Throughout this proof we assume that $\lambda<1/2d$ . According to a calculation similar to that leading to (3.3),

(3.6) \begin{align} \mathbb{E}_{\vec{1}}\bigg(\bigg(\frac{1}{N}\int_0^{tN}\eta_u(O)\,{\mathrm{d}} u\bigg)^2\bigg) & = \frac{2}{N^2}\int_0^{tN}\int_0^s\sum_{x}p_{s-r}(O,x) \mathbb{E}_{\vec{1}}\,(\eta_r(O)\eta_r(x))\,{\mathrm{d}} r\,{\mathrm{d}} s \notag \\ & \geq \frac{2}{N^2}\int_0^{tN}\int_0^sp_{s-r}(O,O)\mathbb{E}_{\vec{1}}\,(\eta_r^2(O))\,{\mathrm{d}} r\,{\mathrm{d}} s. \end{align}

Let $\xi_t$ be the contact process defined as in Section 1, then a well-known property of $\xi_t$ is that

\begin{align*} \mathbb{P}_{\vec{1}}\,(\xi_t(O)=1) \leq \exp\bigg\{\frac{(2\lambda d-1)t}{1+2d\lambda}\bigg\}. \end{align*}

This property can be proved by coupling the contact process with a branching process, the detail of which we omit here. According to the coupling relationship of the NBCPP and the contact process given in Section 1, utilizing the Hölder inequality and Proposition 2.1, we have

\begin{equation*} \mathbb{P}_{\vec{1}}\,(\xi_t(O)=1) = \mathbb{P}_{\vec{1}}\,(\eta_t(O)>0) \geq \frac{(\mathbb{E}_{\vec{1}}\,\eta_t(O))^2}{\mathbb{E}_{\vec{1}}\,(\eta_t^2(O))} = \frac{1}{\mathbb{E}_{\vec{1}}\,(\eta_t^2(O))}. \end{equation*}

Hence,

(3.7) \begin{equation} \mathbb{E}_{\vec{1}}\,(\eta_t^2(O)) \geq \exp\bigg\{\frac{(1-2\lambda d)t}{1+2d\lambda}\bigg\}. \end{equation}

Since $p_t(O,O)=\Theta(t^{-{d}/{2}})$ as $t\rightarrow+\infty$ , Theorem 2.2 follows directly from (3.6) and (3.7).

From now on we assume that $d\geq d_0$ and $\lambda\geq \lambda_0$ , where $d_0$ and $\lambda_0$ are defined as in Proposition 2.4. As mentioned at the end of Section 2, our proof follows the strategy introduced in [Reference Birkner and Zähle1] where $X_t^N$ is decomposed as a martingale plus a remainder term. As $N\rightarrow+\infty$ , the martingale converges weakly to a Brownian motion and the remainder converges to 0 in probability. In detail, for any $\theta>0$ , we define

\begin{align*} G_\theta(\eta)=\sum_{x\in \mathbb{Z}^d}g_\theta(x)(\eta(x)-1),\end{align*}

where $g_\theta$ is the resolvent function of the simple random walk given by

\begin{align*} g_\theta(x)=\int_0^{+\infty}{\mathrm{e}}^{-\theta u}p_u(O,x)\,{\mathrm{d}} u.\end{align*}

According to the Markov property of the simple random walk,

\begin{align*} \sum_{x} g_0^2(x)=\int_0^{+\infty}\int_0^{+\infty}p_{u+\theta}(O,O)\,{\mathrm{d}} u\,{\mathrm{d}}\theta.\end{align*}

Then, applying $p_t(O, O)=\Theta(t^{-d/2})$ as $t\rightarrow+\infty$ , $\sum_{x} g_0^2(x)$ is finite when and only when $d\geq 5$ . Hence, $\sum_{x} g_0^k(x)<+\infty$ when $k\geq 2$ and $d\geq 5$ . Let

\begin{align*} M_t^\theta=G_\theta(\eta_t)-G_\theta(\eta_0)-\int_0^t \mathcal{L}G_\theta(\eta_s)\,{\mathrm{d}} s,\end{align*}

then by Dynkin’s martingale formula, $\{M_t^\theta\}_{t\geq 0}$ is a martingale with quadratic variation process $\{\langle M^\theta\rangle_t\}_{t\geq 0}$ given by

\begin{align*} \langle M^\theta\rangle_t = \int_0^t(\mathcal{L}((G_\theta(\eta_s))^2) - 2G_\theta(\eta_s)\mathcal{L}G_\theta(\eta_s))\,{\mathrm{d}} u.\end{align*}

Applying the definition of $\mathcal{L}$ ,

(3.8) \begin{equation} \langle M^\theta\rangle_t = \frac{1}{2\lambda d} \int_0^t\sum_{x\in\mathbb{Z}^d}\eta_s^2(x)\Bigg(g_\theta^2(x)+\lambda\sum_{y\sim x}g_\theta^2(y)\Bigg)\,{\mathrm{d}} s.\end{equation}

Applying $\theta g_\theta({\cdot})-Lg_\theta({\cdot})=\textbf{1}_{O}({\cdot})$ ,

\begin{align*} \mathcal{L}G_\theta(\eta) = \theta G_\theta(\eta) - \Bigg(\eta(O)-\theta\sum_xg_\theta(x)\Bigg) = \theta G_\theta(\eta)-(\eta(O)-1).\end{align*}

As a result,

(3.9) \begin{equation} X_t^N=\frac{1}{\sqrt{N}\,}M_{tN}^{1/N}+\frac{1}{\sqrt{N}\,}R_{tN}^{1/N},\end{equation}

where $R_t^\theta=-G_\theta(\eta_t)+G_\theta(\eta_0)+\int_0^t\theta G_\theta(\eta_s)\,{\mathrm{d}} s$ . To prove Theorem 2.3, we need the following lemmas.

Lemma 3.3. For any $t\geq 0$ ,

\begin{align*} \lim_{N\rightarrow+\infty}\mathbb{E}\bigg(\sup_{0\leq s\leq tN}\frac{1}{N}\big(M^{1/N}_{s}-M^{1/N}_{s-}\big)^2\bigg) = 0, \end{align*}

where $M^{1/N}_{s-}=\lim_{u<s, u\rightarrow s}M^{1/N}_u$ , i.e. the state of $M^{1/N}_{\cdot}$ at the moment just before s.

We first utilize Lemmas 3.1, 3.2, and 3.3 to prove Theorem 2.3.

Proof of Theorem 2.3. We denote by $\mathcal{D}[0,+\infty)$ the set of càdlàg functions from $[0,+\infty)$ to $\mathbb{R}$ endowed with the Skorokhod topology. According to Lemmas 3.1 and 3.3,

\begin{align*} \bigg(\bigg\{\frac{1}{\sqrt{N}\,}M_{tN}^{1/N}\bigg\}_{t\geq 0}, \bigg\{\frac{1}{N}\langle M^{1/N}\rangle_{tN}\bigg\}_{t\geq 0}\bigg) \end{align*}

satisfies condition (b) of [Reference Ethier and Kurtz5, Theorem 1.4, Chapter 7]. Hence, using that theorem, $\big\{({1}/{\sqrt{N}})M_{tN}^{1/N}\colon t\geq 0\big\}_{N\geq 1}$ converges weakly, with respect to the Skorokhod topology of $\mathcal{D}[0,+\infty)$ , to a continuous martingale $\{W_t\}_{t\geq 0}$ as $N\rightarrow+\infty$ , where $\{W_t\}_{t\geq 0}$ satisfies that $\{W_t^2-C_1(\lambda, d)t\}_{t\geq 0}$ is also a martingale. Hence, $\{W_t\}_{t\geq 0}$ is a standard Brownian motion times $\sqrt{C_1(\lambda, d)}$ . Then, for given $0<t_1<t_2<\cdots<t_m$ , since $\{\eta_t\}_{t\geq 0}$ is continuous at each $t_i$ with probability 1, we have that $({1}/{\sqrt{N}})\big(M_{t_1N}^{1/N},\ldots,M_{t_mN}^{1/N}\big)$ converges weakly to $\sqrt{C_1(\lambda, d)}(B_{t_1},\ldots,B_{t_m})$ as $N\rightarrow+\infty$ . Consequently, Theorem 2.3 follows from (3.9) and Lemma 3.2.

Now we only need to prove Lemmas 3.1, 3.2, and 3.3. We first give the proof of Lemma 3.1.

Proof of Lemma 3.1. Conditioned on $Y_0=O$ , the number of times $\{Y_t\}_{t\geq 0}$ visits O follows a geometric distribution with parameter $\gamma_d$ , and at each time $Y_t$ stays at O for an exponential time with mean 1. Therefore,

\begin{align*} \int_0^{+\infty}p_s(O,O)\,{\mathrm{d}} s = \mathbb{E}_O\int_0^{+\infty}\textbf{1}_{\{X_s=O\}}\,{\mathrm{d}} s = 1 \times \frac{1}{\gamma_d} = \frac{1}{\gamma_d}. \end{align*}

Then, by (3.8) and Proposition 2.3,

\begin{align*} \lim_{N\rightarrow+\infty}\mathbb{E}_{\nu_{\lambda,d}}\bigg(\frac{1}{N}\langle M^{1/N}\rangle_{tN}\bigg) & = \frac{t(1+2\lambda d)}{2\lambda d}\bigg(1+\frac{1}{h_{\lambda,d}}\bigg)\sum_{x}g_0^2(x) \\ & = \frac{2\gamma_d t}{h_{\lambda,d}}\int_0^{+\infty}\int_0^{+\infty}p_{\theta+r}(O,O)\,{\mathrm{d}}\theta\,{\mathrm{d}} r = C_1(\lambda,d)t. \end{align*}

Hence, to complete the proof we only need to show that

(3.10) \begin{equation} \lim_{N\rightarrow+\infty}\frac{1}{N^2}{\rm Var}_{\nu_{\lambda,d}}(\langle M^{1/N}\rangle_{tN}) = 0. \end{equation}

By Proposition 2.2 and the Markov property of $\{\eta_t\}_{t\geq 0}$ , for $s<u$ and $x, w\in \mathbb{Z}^d$ ,

\begin{align*} {\rm Cov}_{\nu_{\lambda,d}}(\eta_s^2(x),\eta_u^2(w)) = \sum_{y,z}\hat{q}_{u-s}((w,w),(y,z)) {\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z)). \end{align*}

Therefore, to prove (3.10) we only need to show that

(3.11) \begin{equation} \lim_{r\rightarrow+\infty}\sum_{x,w,y,z}V_0(x)V_0(w)\hat{q}_r((w,w),(y,z)) |{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))| = 0, \end{equation}

where $V_\theta(x)=g_\theta^2(x)+\sum_{y\sim x}g^2_\theta(y)$ , which is decreasing in $\theta$ .

According to Propositions 2.2–2.4 and the Cauchy–Schwarz inequality,

\begin{align*} \sum_{y,z}\hat{q}_r((w,w),(y,z)) = \sum_{y}q_r(O,y) = \mathbb{E}_{\vec{1}}\,((\eta_r(O))^2) & \leq 1 + \frac{1}{h_{\lambda,d}}, \\ \sup_{w,y,z}|{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(w),\eta_0(y)\eta_0(z))| < 2\mathbb{E}_{\nu_{\lambda,d}}((\eta_0(O))^4) & < +\infty. \end{align*}

Then, using the fact that $\sum_xV_0(x)<+\infty$ and the dominated convergence theorem, to prove (3.11) we only need to show that

(3.12) \begin{equation} \lim_{r\rightarrow+\infty}\sum_{y,z}\hat{q}_r((w,w),(y,z)) |{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))| = 0 \end{equation}

for any $x,w\in\mathbb{Z}^d$ . By Proposition 2.4, for any $\varepsilon>0$ , there exists $M>0$ such that $|{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))|<\varepsilon$ when $\|x-y\|_1\wedge\|x-z\|_1\geq M$ , and hence

\begin{multline*} \sum_{(y,z):\|x-y\|_1\wedge\|x-z\|_1\geq M}\hat{q}_r((w,w),(y,z)) |{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))| \\ \leq \varepsilon\sum_{(y,z):\|x-y\|_1\wedge\|x-z\|_1\geq M}\hat{q}_r((w,w),(y,z)) \leq \varepsilon\bigg(1+\frac{1}{h_{\lambda, d}}\bigg). \end{multline*}

We claim that

(3.13) \begin{equation} \lim_{t\rightarrow+\infty}\sum_{v}\hat{q}_t((w,w),(y,v)) = \lim_{t\rightarrow+\infty}\sum_{v}\hat{q}_t((w,w),(v,z)) = 0 \end{equation}

for any $w,y,z\in \mathbb{Z}^d$ . The proof of (3.13) is given later. By (3.13), Proposition 2.4, and the Cauchy–Schwarz inequality,

\begin{align*} \lim_{r\rightarrow+\infty}\sum_{\substack{(y,z):\|x-y\|_1\leq M\,\text{or}\\\|x-z\|_1\leq M}} \hat{q}_r((w,w),(y,z))|{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))| = 0 \end{align*}

and hence

\begin{align*} \limsup_{r\rightarrow+\infty}\sum_{y,z}\hat{q}_r((w,w),(y,z)) |{\rm Cov}_{\nu_{\lambda,d}}(\eta_0^2(x),\eta_0(y)\eta_0(z))| \leq \varepsilon\bigg(1+\frac{1}{h_{\lambda, d}}\bigg). \end{align*}

Since $\varepsilon$ is arbitrary, let $\varepsilon\rightarrow 0$ and then (3.12) holds.

Now we prove (3.13) to complete the proof of Lemma 3.1.

Proof of (3.13). By [Reference Liggett11, Theorem 9.3.1] and the definition of $\mathcal{L}$ ,

\begin{align*} \hat{q}_t((w,w),(y,v)) = {\mathrm{e}}^{-2t}\sum_{n=0}^{+\infty}\frac{t^n}{n!}H^n((w,w),(y,v)), \end{align*}

where H is a $(\mathbb{Z}^d)^2\times (\mathbb{Z}^d)^2$ matrix given by

\begin{align*} H((x,y), (u,v))= \begin{cases} {1}/{2d} & \text{if}\ x\neq y,\ u\sim x, \,\text{and}\,v=y, \\ {1}/{2d} & \text{if}\ x\neq y,\ u=x, \,\text{and}\,v\sim y, \\ {1}/{2d\lambda} & \text{if}\ x=y \,\text{and}\,(u,v)=(x,x), \\ {1}/{2d} & \text{if}\ x=y,\ u\sim x, \,\text{and}\,v=u, \\ {1}/{2d} & \text{if}\ x=y,\ u=x, \,\text{and}\,v\sim x, \\ {1}/{2d} & \,\text{if}\ x=y,\ u\sim x. \,\text{and}\,v=x, \\ 0 & \text{otherwise} \end{cases} \end{align*}

and

\begin{align*} H^n((w,w),(y,v)) = \sum_{\stackrel{\{(u_i,v_i)\}_{0\leq i\leq n}:}{(u_0,v_0)=(w,w),(u_n,v_n)=(y,v)}} \prod_{i=0}^{n-1}H((u_i,v_i),(u_{i+1},v_{i+1})). \end{align*}

As a result,

\begin{align*} \sum_{v\in\mathbb{Z}^d}\hat{q}_t((w,w),(y,v)) = {\mathrm{e}}^{-2t}\sum_{n=0}^{+\infty}\frac{(2t)^n}{n!} \mathbb{E}_{(w,w)}\Bigg(\prod_{i=0}^{n-1}\Theta(\beta_{i},\beta_{i+1})\textbf{1}_{\{\beta_n(1)=y\}}\Bigg), \end{align*}

where $\{\beta_n=(\beta_n(1), \beta_n(2))\colon n\geq 0\}$ is a random walk on $(\mathbb{Z}^d)^2$ such that

\begin{align*} \mathbb{P}(\beta_{n+1}=(u,v)\mid\beta_n=(x,y)) = \begin{cases} {1}/{4d} & \text{if}\ x\neq y,\ u\sim x. \,\text{and}\,v=y, \\ {1}/{4d} & \text{if}\ x\neq y,\ u=x, \,\text{and}\,v\sim y, \\ {1}({6d+1}) & \text{if}\ x=y\,\text{and}\,(u,v)=(x,x), \\ {1}/({6d+1}) & \text{if}\ x=y,\ u\sim x. \,\text{and}\,v=u, \\ {1}/({6d+1}) & \text{if}\ x=y,\ u\sim x, \,\text{and}\,v=x, \\ {1}/{6d+1} & \text{if}\ x=y,\ u=x, {and}\,v\sim x, \\ 0 & \text{otherwise} \end{cases} \end{align*}

and $\Theta$ is a function from $(\mathbb{Z}^d)^2$ to $\mathbb{R}$ such that $\Theta((x,y),(u,v)) =\frac12{H((x,y),(u,v)){\rm deg}(x,y)}$ , where

\begin{align*} {\rm deg}(x,y) = \begin{cases} 4d & \text{if}\ x\neq y, \\ 6d+1 & \text{if}\ x=y. \end{cases} \end{align*}

Note that ${\rm deg}({\cdot})$ is the degree function of the graph generated by adding edges on $\mathbb{Z}^{2d}$ to connect (x,x) with itself and (y, y) for all $x\in \mathbb{Z}^d$ and all y such that $y\sim x$ on $\mathbb{Z}^d$ .

According to the definition of $\Theta$ , for each $i\geq 0$ , $\Theta(\beta_i,\beta_{i+1})\geq 1$ and $\Theta(\beta_i,\beta_{i+1})=1$ when and only when $\beta_i(1)=\beta_i(2)$ . Hence, by Hölder’s inequality,

(3.14) \begin{multline} \sum_{v\in\mathbb{Z}^d}\hat{q}_t((w,w),(y,v)) \\ \leq \Bigg(\mathbb{E}_{(w,w)}\Bigg(\prod_{i=0}^{+\infty} \Theta^{1+\varepsilon}(\beta_{i},\beta_{i+1})\Bigg)\Bigg)^{{1}/({1+\varepsilon})}{\mathrm{e}}^{-2t} \sum_{n=0}^{+\infty}\frac{(2t)^n}{n!}(\mathbb{P}(\beta_n(1)=y))^{\varepsilon/(1+\varepsilon)} \end{multline}

for any $\varepsilon>0$ . Since $\{\beta_n(1)-\beta_n(2)\}_{n\geq 0}$ is a lazy version of the simple random walk on $\mathbb{Z}^d$ and $H(\beta_i,\beta_{i+1})>1$ only when $\beta_i(1)=\beta_i(2)$ , by the strong Markov property of $\{\beta_n\}_{n\geq 0}$ ,

\begin{align*} \mathbb{E}_{(w,w)}\Bigg(\prod_{i=0}^{+\infty}\Theta^{1+\varepsilon}(\beta_{i},\beta_{i+1})\Bigg) = \frac{4d}{6d+1}\bigg(\frac{6d+1}{4d}\bigg)^{1+\varepsilon}\gamma_d + \sum_{k=1}^{+\infty}(\alpha_\varepsilon(d,\lambda))^k\gamma_d, \end{align*}

where

\begin{align*} \alpha_\varepsilon(d,\lambda) = \frac{1}{6d+1}\bigg(\frac{6d+1}{2\cdot2d\lambda}\bigg)^{1+\varepsilon} + \frac{2d}{6d+1}\bigg(\frac{6d+1}{2\cdot2d}\bigg)^{1+\varepsilon} + \frac{4d}{6d+1}\bigg(\frac{6d+1}{2\cdot2d}\bigg)^{1+\varepsilon}(1-\gamma_d). \end{align*}

Since $\lambda>{1}/{2d(2\gamma_d-1)}$ , $\lim_{\varepsilon\rightarrow 0}\alpha_\varepsilon(d,\lambda)={1}/{4d\lambda}+\frac12+1-\gamma_d<1$ and hence there exists $\varepsilon_0>0$ such that $\mathbb{E}_{(w,w)}\big(\prod_{i=0}^{+\infty}\Theta^{1+\varepsilon_0}(\beta_{i},\beta_{i+1})\big)<+\infty$ . By (3.14), to complete the proof we only need to show that

(3.15) \begin{equation} \lim_{t\rightarrow+\infty}{\mathrm{e}}^{-2t}\sum_{n=0}^{+\infty}\frac{(2t)^n}{n!} (\mathbb{P}(\beta_n(1)=y))^{\varepsilon_0/(1+\varepsilon_0)} = 0. \end{equation}

Since $d\geq 3$ and $\{\beta_n(1)\}_{n\geq 1}$ is a lazy version of the simple random walk on $\mathbb{Z}^d$ ,

\begin{align*} \lim_{n\rightarrow+\infty}\mathbb{P}(\beta_n(1)=y)=0. \end{align*}

Hence, for any $\varepsilon>0$ , there exists $M\geq 1$ such that $(\mathbb{P}(\beta_n(1)=y))^{\varepsilon_0/(1+\varepsilon_0)}\leq \varepsilon$ when $n\geq M$ , and then

\begin{align*} {\mathrm{e}}^{-2t}\sum_{n=M}^{+\infty}\frac{(2t)^n}{n!} (\mathbb{P}(\beta_n(1)=y))^{\varepsilon_0/(1+\varepsilon_0)} \leq \varepsilon{\mathrm{e}}^{-2t}\sum_{n=0}^{+\infty}\frac{(2t)^n}{n!} = \varepsilon. \end{align*}

Since $\lim_{t\rightarrow+\infty}\sum_{n=0}^{M-1}{\mathrm{e}}^{-2t}{(2t)^n}/{n!}=0$ , we have

\begin{align*} \limsup_{t\rightarrow+\infty}{\mathrm{e}}^{-2t}\sum_{n=0}^{+\infty}\frac{(2t)^n}{n!} (\mathbb{P}(\beta_n(1)=y))^{\varepsilon_0/(1+\varepsilon_0)} \leq \varepsilon. \end{align*}

Since $\varepsilon$ is arbitrary, let $\varepsilon\rightarrow0$ and then (3.15) holds.

Now we prove Lemma 3.2.

Proof of Lemma 3.2. Applying Propositions 2.1 and 2.3, for any moment $c\geq 0$ ,

\begin{align*} \mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{\sqrt{N}\,}G_{1/N}(\eta_{c})\bigg)^2\bigg) & = {\rm Var}_{\nu_{\lambda,d}}\bigg(\frac{1}{\sqrt{N}\,}G_{1/N}(\eta_{c})\bigg) \\ & = {\rm Var}_{\nu_{\lambda,d}}\bigg(\frac{1}{\sqrt{N}\,}G_{1/N}(\eta_0)\bigg) = \frac{1}{N}\sum_{x}\sum_{y}g_{1/N}(x)g_{1/N}(y)\frac{\Phi(y-x)}{h_{\lambda, d}}. \end{align*}

As shown in Remark 2.3,

\begin{align*} \Phi(y-x) = \frac{\int_0^{+\infty}p_s(y-x,O)\,{\mathrm{d}} s}{\int_0^{+\infty}p_s(O,O)\,{\mathrm{d}} s} = \int_0^{+\infty}p_s(x,y)\,{\mathrm{d}} s\gamma_d. \end{align*}

Hence,

(3.16) \begin{align} \mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(&\frac{1}{\sqrt{N}}G_{1/N}(\eta_{c})\bigg)^2\bigg) \notag \\ & = \frac{\gamma_d}{Nh_{\lambda,d}}\int_0^{+\infty}\int_0^{+\infty}\int_0^{+\infty}{\mathrm{e}}^{-(r+s)/N} \sum_x\sum_yp_r(0,x)p_s(x,y)p_u(y,0)\,{\mathrm{d}} r\,{\mathrm{d}} s\,{\mathrm{d}} u \notag \\ & = \frac{\gamma_d}{Nh_{\lambda,d}}\int_0^{+\infty}\int_0^{+\infty}\int_0^{+\infty}{\mathrm{e}}^{-(r+s)/N} p_{r+s+u}(O,O)\,{\mathrm{d}} r\,{\mathrm{d}} s\,{\mathrm{d}} u \notag \\ & = \frac{\gamma_d}{Nh_{\lambda,d}}\int_0^{+\infty}p_{\theta}(O,O) \bigg(\int_0^{\theta}{\mathrm{e}}^{-v/N}\bigg(\int_0^v1\,{\mathrm{d}} u\bigg)\,{\mathrm{d}} v\bigg)\,{\mathrm{d}}\theta \notag \\ & = \frac{N\gamma_d}{h_{\lambda,d}}\int_0^{+\infty}p_{\theta}(O,O) \bigg(1-{\mathrm{e}}^{-\theta/N}\bigg(1+\frac{\theta}{N}\bigg)\bigg)\,{\mathrm{d}}\theta. \end{align}

Since $p_\theta(O,O)=O(\theta^{-d/2})$ as $\theta\rightarrow+\infty$ and $\lim_{x\rightarrow+\infty}1-{\mathrm{e}}^{-x}(1+x)=1$ , there exist $0<M_1,\widehat{C}_4<+\infty$ such that $p_{\theta}(O,O)\leq\widehat{C}_4 \theta^{-d/2}$ and $1-{\mathrm{e}}^{-x}(1+x)\leq \widehat{C}_4$ when $\theta,x\geq M_1$ . Since $\lim_{x\rightarrow0}({1-{\mathrm{e}}^{-x}(1+x)})/{x^2}=\frac12$ , we have

\begin{align*} \widetilde{C}_4 \,:\!=\, \sup_{0<x\leq M_1}\frac{1-{\mathrm{e}}^{-x}(1+x)}{x^2}<+\infty. \end{align*}

We denote by $C_4$ the maximum of $\widehat{C}_4$ and $\widetilde{C}_4$ . Then,

\begin{align*} p_{\theta}(O,O)\bigg(1-{\mathrm{e}}^{-\theta/N}\bigg(1+\frac{\theta}{N}\bigg)\bigg) \leq \begin{cases} {\theta^2 C_4}/{N^2} & \text{when}\,0\leq \theta\leq M_1, \\ {C_4^2\theta^{2-({d}/{2})}}/{N^2} & \text{when}\,M_1\leq \theta\leq NM_1, \\ C_4^2\theta^{-{d}/{2}} & \text{when}\, \theta>NM_1. \end{cases} \end{align*}

Hence, for $d\geq 5$ ,

\begin{align*} \frac{N\gamma_d}{h_{\lambda, d}}\int_0^{M_1}p_{\theta}(O,O) \bigg(1-{\mathrm{e}}^{-\theta/N}\bigg(1+\frac{\theta}{N}\bigg)\bigg)\,{\mathrm{d}}\theta & \leq \frac{C_4\gamma_d}{Nh_{\lambda,d}}\int_0^{M_1}\theta^2\,{\mathrm{d}}\theta = O(N^{-1}), \\ \frac{N\gamma_d}{h_{\lambda,d}}\int_{M_1}^{NM_1}p_{\theta}(O,O) \bigg(1-{\mathrm{e}}^{-\theta/N}\bigg(1+\frac{\theta}{N}\bigg)\bigg)\,{\mathrm{d}}\theta & \leq \frac{C_4^2\gamma_d}{Nh_{\lambda,d}}\int_{M_1}^{NM_1}\theta^{2-({d}/{2})}\,{\mathrm{d}}\theta \\ & \leq \frac{C_4^2\gamma_d}{Nh_{\lambda,d}}\int_{M_1}^{NM_1}\theta^{2-{5}/{2}}\,{\mathrm{d}}\theta = O(N^{-1/2}), \\ \frac{N\gamma_d}{h_{\lambda,d}}\int_{NM_1}^{+\infty}p_{\theta}(O,O) \bigg(1-{\mathrm{e}}^{-\theta/N}\bigg(1+\frac{\theta}{N}\bigg)\bigg)\,{\mathrm{d}}\theta & \leq \frac{C_4^2N\gamma_d}{h_{\lambda,d}}\int_{NM_1}^{+\infty}\theta^{-d/2}\,{\mathrm{d}}\theta = O(N^{2-({d}/{2})}). \end{align*}

Consequently, applying (3.16),

(3.17) \begin{equation} \lim_{N\rightarrow+\infty}\mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{\sqrt{N}\,}G_{1/N}(\eta_{0})\bigg)^2\bigg) = \lim_{N\rightarrow+\infty}\mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{\sqrt{N}\,}G_{1/N}(\eta_{tN^2})\bigg)^2\bigg) = 0. \end{equation}

By the Cauchy–Schwarz inequality,

\begin{align*} \mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{\sqrt{N}\,} \int_0^{tN}\frac{1}{N}G_{1/N}(\eta_s)\,{\mathrm{d}} s\bigg)^2\bigg) & \leq \frac{tN}{N^3}\int_0^{tN}(\mathbb{E}_{\nu_{\lambda,d}}((G_{1/N}(\eta_s))^2))\,{\mathrm{d}} s \\ & = \frac{t^2N^2}{N^3}\mathbb{E}_{\nu_{\lambda,d}}((G_{1/N}(\eta_0))^2). \end{align*}

Hence, applying (3.17),

(3.18) \begin{equation} \lim_{N\rightarrow+\infty}\mathbb{E}_{\nu_{\lambda,d}}\bigg(\bigg(\frac{1}{\sqrt{N}\,} \int_0^{tN}\frac{1}{N}G_{1/N}(\eta_s)\,{\mathrm{d}} s\bigg)^2\bigg) = 0. \end{equation}

Lemma 3.2 follows from (3.17) and (3.18).

Finally, we prove Lemma 3.3.

Proof of Lemma 3.3. By (3.9) and the fact that $\{X_t^N\}_{t\geq 0}$ is continuous in t,

(3.19) \begin{equation} \sup_{0\leq s\leq tN}\frac{1}{N}\big(M^{1/N}_{s}-M^{1/N}_{s-}\big)^2 \leq \sup_{0\leq s\leq tN}\frac{1}{N}\sup_{x}\max_{y\sim x}\{g^2_{1/N}(x)\eta^2_s(x), g^2_{1/N}(x)\eta^2_s(y)\}. \end{equation}

For any $M>0$ , let

\begin{align*} \tau_M = \inf\bigg\{s>0\colon \frac{1}{N}\sup_{x}\max_{y\sim x}\{g^2_{1/N}(x)\eta^2_s(x), g^2_{1/N}(x)\eta^2_s(y)\}>M\bigg\}; \end{align*}

then

$$ \bigg\{\sup_{0\leq s\leq tN}\frac{1}{N}\sup_{x}\max_{y\sim x}\{g^2_{1/N}(x)\eta^2_s(x),g^2_{1/N}(x)\eta^2_s(y)\}>M\bigg\} = \{\tau_M\leq tN\}. $$

Conditioned on $\{\tau_M\leq tN\}$ , there exists $x_0\in \mathbb{Z}^d$ such that

\begin{align*} \max_{y\sim x_0}\{g^2_{1/N}(x_0)\eta^2_{\tau_M}(x_0), g^2_{1/N}(x_0)\eta^2_{\tau_ M}(y)\}>NM. \end{align*}

If $\{\eta_s(y)\}_{y\sim x_0}$ and $\eta_s(x_0)$ do not jump to 0 during $s\in [\tau_M,\tau_M+1]$ , then

\begin{align*} \max_{y\sim x_0}\{g^2_{1/N}(x)\eta^2_{s}(x_0),g^2_{1/N}(x_0)\eta^2_{s}(y)\} \geq NM\exp\bigg\{\min\bigg\{\frac{1}{2\lambda d}-1,0\bigg\}\bigg\} \end{align*}

for $s\in [\tau_M, \tau_{M}+1]$ and hence

\begin{align*} \int_0^{tN+1}\sum_{x}g_{1/N}^4(x)\Bigg(\eta_s^4(x)+\sum_{y\sim x}\eta_s^4(y)\Bigg)\,{\mathrm{d}} s \geq N^2M^2\exp\bigg\{2\min\bigg\{\frac{1}{2\lambda d}-1,0\bigg\}\bigg\}. \end{align*}

Since the state of a vertex jumps to 0 at rate $1/(2\lambda d)$ , applying the strong Markov property of $\{\eta_t\}_{t\geq 0}$ ,

\begin{multline*} \mathbb{P}\Bigg(\int_0^{tN+1}\sum_{x}g_{1/N}^4(x)\Bigg(\eta_s^4(x)+\sum_{y\sim x}\eta_s^4(y)\Bigg)\,{\mathrm{d}} s \geq N^2M^2\exp\bigg\{2\min\bigg\{\frac{1}{2\lambda d}-1,0\bigg\}\bigg\}\Bigg) \\ \geq \mathbb{P}(\tau_M\leq tN)\exp\bigg\{{-}\frac{2d+1}{2\lambda d}\bigg\}. \end{multline*}

Then, by Markov’s inequality, Proposition 2.4, and the fact that

\begin{align*} \sum_x g_{1/N}^4(x)\leq \sum_x g_0^4(x)<+\infty, \end{align*}

we have

(3.20) \begin{align} & \mathbb{P}\bigg(\sup_{0\leq s\leq tN}\frac{1}{N}\sup_{x} \max_{y\sim x}\{g^2_{1/N}(x)\eta^2_s(x), g^2_{1/N}(x)\eta^2_s(y)\}>M\bigg) \notag \\ & \leq \frac{\exp\{({2d+1})/{2\lambda d}\}}{N^2M^2\exp{2\min\{{1}/({2\lambda d})-1,0\}}} \int_0^{tN+1}\sum_xg_0^4(x)(\mathbb{E}_{\lambda,d}(\eta^4(O))(2d+1))\,{\mathrm{d}} s \leq \frac{C_3}{NM^2}, \end{align}

where $C_3<+\infty$ is independent of N and M. Applying the Fubini theorem, for any positive random variable V and any $c>0$ ,

\begin{align*} \mathbb{E}\big(V\textbf{1}_{\{V>c\}}\big) = c\mathbb{P}(V>c) + \int_c^{+\infty}\mathbb{P}(V\geq u)\,{\mathrm{d}} u. \end{align*}

Hence, for any $\varepsilon>0$ ,

\begin{align*} \mathbb{E}\bigg(\sup_{0\leq s\leq tN}\frac{1}{N}(M^{1/N}_{s}-M^{1/N}_{s-})^2\bigg) \leq \varepsilon + \varepsilon\frac{C_3}{N\varepsilon^2} + \int_{\varepsilon}^{+\infty}\frac{C_3}{Nu^2}\,{\mathrm{d}} u = \varepsilon + \frac{2C_3}{N\varepsilon} \end{align*}

according to (3.19) and (3.20). As a result,

\begin{align*} \limsup_{N\rightarrow+\infty}\mathbb{E}\bigg(\sup_{0\leq s\leq tN}\frac{1}{N}(M^{1/N}_{s}-M^{1/N}_{s-})^2\bigg) \leq \varepsilon. \end{align*}

Since $\varepsilon$ is arbitrary, let $\varepsilon\rightarrow 0$ and the proof is complete.