Appendix A: A Technical Lemma
Lemma A.1
If \(H:[0,\infty ) \rightarrow [0,\infty )\) is a non-decreasing function such that
$$\begin{aligned} \gamma :=\inf \left\{ \beta >0:\, \int _0^\infty \mathrm {e}^{-\beta t} H(t)\mathrm {d}t<\infty \right\} <\infty , \end{aligned}$$
then
$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{1}{t}\log H(t)\le \gamma . \end{aligned}$$
Proof
We will prove this lemma by contradiction. Suppose that \(\limsup _{t\rightarrow \infty }t^{-1}\log H(t)>\gamma \). Then there exist \(\epsilon _0>0\) and a non-decreasing sequence \(\{t_n\}_{n\ge 1}\) such that \(0\le t_n\uparrow \infty \) as \(n\rightarrow \infty \) and
$$\begin{aligned} \log H(t_n)\ge t_n(\gamma +\epsilon _0),\quad \text {for all }n\ge 1. \end{aligned}$$
Moreover, we assume that \(t_n \ge s_{n-1}^*\) for a certain sequence \((s_n^*)_{n \ge 1}\) which will be constructed below. By definition of \(\gamma \), we have that
$$\begin{aligned} \int _{0}^\infty \mathrm {e}^{-(\gamma +\epsilon ) t}H(t)\mathrm {d}t<\infty ,\quad \text {for all }\epsilon >0. \end{aligned}$$
(A.1)
We claim that there exits \(s_1>t_1\) such that \(H(s_1)\le \mathrm {e}^{(\gamma +\epsilon _0/2)s_1}\). If this is not true, then
$$\begin{aligned} H(t)1_{\{t> t_1\}}> \mathrm {e}^{(\gamma +\epsilon _0/2) t}1_{\{t>t_1\}}, \end{aligned}$$
which leads to the following contradiction with (A.1):
$$\begin{aligned} \int _{t_1}^\infty \mathrm {e}^{-(\gamma +\epsilon _0/4) t}H(t)\mathrm {d}t \ge \int _{t_1}^\infty \mathrm {e}^{-(\gamma +\epsilon _0/4) t}\mathrm {e}^{(\gamma +\epsilon _0/2) t}\mathrm {d}t = \infty . \end{aligned}$$
Let \(r_1=\inf \{s_1>t_1; H(s_1) \le e^{(\gamma +\epsilon _0/2)s_1} \}\). Then \(H(t) > e^{(\gamma +\epsilon _0/2)t}\) for any \(t \in [t_1,r_1)\). Since H(t) is non-decreasing and \(H(t_1) \ge e^{(\gamma +\epsilon _0)t_1}\), the smallest possible value for \(r_1\) is obtained in the case when the function H(t) is constant with value equal to \(\mathrm {e}^{(\gamma +\epsilon _0)t_1}\) starting from \(t_1\) until it crosses the function \(\mathrm {e}^{(\gamma +\epsilon _0/2)t}\). In this case, \(r_1=s_1^*\) where \(\mathrm {e}^{(\gamma +\epsilon _0/2)s_1^*} = \mathrm {e}^{(\gamma +\epsilon _0)t_1}\). For a general non-decreasing function \(H, r_1 \ge s_1^*\). Hence,
$$\begin{aligned} H(t) 1_{\{t\in [t_1,s_1^*]\}} \ge \mathrm {e}^{(\gamma +\epsilon _0/2)t_1} 1_{\{t\in [t_1,s_1^*]\}}, \quad \text {with }s_1^*=\left( 1+\frac{\epsilon _0}{2\gamma +\epsilon _0}\right) t_1. \end{aligned}$$
We now select \(t_2\) such that \(t_2>s_1^*\) and \(t_2 \ge t_1\). In the same way, we have that
$$\begin{aligned} H(t) 1_{\{t\in [t_2,s_2^*]\}} \ge \mathrm {e}^{(\gamma +\epsilon _0/2)t_2} 1_{\{t\in [t_2,s_2^*]\}}, \quad \text {with } s_2=\left( 1+\frac{\epsilon _0}{2\gamma +\epsilon _0}\right) t_2. \end{aligned}$$
In this way, we can find a sequence of disjoint nonempty intervals \(\{[t_n,s_n^*] \}_{n\ge 1}\) such that
$$\begin{aligned} H(t) 1_{\{t\in [t_n,s_n^*]\}} \ge \mathrm {e}^{(\gamma +\epsilon _0/2)t_n} 1_{\{t\in [t_n,s_n^*]\}}, \quad \text {with } s_n^*=\left( 1+\frac{\epsilon _0}{2\gamma +\epsilon _0}\right) t_n, \end{aligned}$$
for all \(n\ge 1\). Now we have that
$$\begin{aligned} \int _{0}^\infty \mathrm {e}^{-(\gamma +\epsilon _0/2) t}H(t)\mathrm {d}t&\ge \sum _{n=1}^\infty \int _{t_n}^{s_n^*} \mathrm {e}^{-(\gamma +\epsilon _0/2) t}\mathrm {e}^{(\gamma +\epsilon _0/2) t_n}\mathrm {d}t\\&= \sum _{n=1}^\infty \frac{1}{\gamma +\epsilon _0/2}\left( 1-\mathrm {e}^{-(\gamma +\epsilon _0/2)(s_n^*-t_n)} \right) \\&= \sum _{n=1}^\infty \frac{1}{\gamma +\epsilon _0/2}\left( 1-\mathrm {e}^{-(\gamma +\epsilon _0/2)\frac{\epsilon _0 t_n}{2\gamma +\epsilon _0}}\right) \\&\ge \sum _{n=1}^\infty \frac{2}{2\gamma +\epsilon _0}\left( 1-\mathrm {e}^{-(\gamma +\epsilon _0/2)\frac{\epsilon _0 t_1}{2\gamma +\epsilon _0}}\right) =\infty , \end{aligned}$$
which contradicts with (A.1). This proves Lemma A.1. \(\square \)
Appendix B: Continuity of \(J_n\) in \(L^p(\Omega )\)
The following result is an extension of Proposition A.3 of [6] to higher dimensions d.
Proposition B.1
Fix \((t,x)\in (0,\infty )\times \mathbb {R}^d\). Set
$$\begin{aligned} B_{t,x}:=\left\{ \left( t',x'\right) \in (0,\infty ) \times \mathbb {R}^d:\, 0< t'\le t+\frac{1}{2}\,,\,\, \left| x'-x\right| \le 1 \right\} \end{aligned}$$
Then there exists \(a=a_{t,x}>0\) such that for all \(\left( t',x'\right) \in B_{t,x}\) and all \(s\in [0,t']\) and \(y \in \mathbb {R}^d\) with \(|y|\ge a\),
$$\begin{aligned} G(t'-s,x'-y) \le G(t+1-s,x-y)\,. \end{aligned}$$
(B.1)
Proof
By direct calculation, we see that inequality (B.1) is equivalent to
$$\begin{aligned} \sum _{i=1}^d \left( -\frac{(x'_i-y_i)^2}{t'-s} +\frac{(x_i-y_i)^2}{t+1-s}\right) \le d \, \log \left( \frac{t'-s}{t+1-s}\right) , \end{aligned}$$
(B.2)
where \(x=(x_1,\ldots ,x_d), x'=(x_1',\ldots ,x_d')\) and \(y=(y_1,\ldots ,y_d)\).
We fix (t, x). In order to find \(a=a_{t,x}\), we will freeze \(d-1\) coordinates. Because
$$\begin{aligned} -\frac{(x'_i-y_i)^2}{t'-s} +\frac{(x_i-y_i)^2}{t+1-s}&=-\frac{1+t-t'}{(1+t-s)(t'-s)}\\&\quad \left( y-\frac{x'(1+t-s)-x(t'-s)}{1+t-t'}\right) ^2 +\frac{(x_i-x_i')^2}{1+t-t'} \\&\quad \le \frac{(x_i-x_i')^2}{1+t-t'} \le 2(x_i-x_i')^2\le 2, \end{aligned}$$
we have
$$\begin{aligned}&\sum _{i=1}^d \left( -\frac{(x'_i-y_i)^2}{t'-s} +\frac{(x_i-y_i)^2}{t+1-s}\right) \\&\quad \le 2(d-1) +\left( -\frac{(x'_j-y_j)^2}{t'-s} +\frac{(x_j-y_j)^2}{t+1-s}\right) . \end{aligned}$$
for any index \(j=1, \ldots ,d\). Hence, inequality (B.2) holds, provided that there exists an index \(j=1, \ldots ,d\) such that
$$\begin{aligned} -\frac{(x'_j-y_j)^2}{t'-s} +\frac{(x_j-y_j)^2}{t+1-s} \le d \, \log \left( \frac{t'-s}{t+1-s}\right) - 2(d-1)\;. \end{aligned}$$
(B.3)
This shows that condition (B.2) holds, if for some index \(j=1,\ldots ,d\), we have:
$$\begin{aligned} -\frac{(x'_j-y_j)^2}{t'-s} +\frac{(x_j-y_j)^2}{t+1-s} \le 2 d \, \log \left( \frac{t'-s}{t+1-s}\right) \;, \end{aligned}$$
(B.4)
and
$$\begin{aligned} -\frac{(x'_j-y_j)^2}{t'-s} +\frac{(x_j-y_j)^2}{t+1-s}\le -4(d-1). \end{aligned}$$
(B.5)
By Proposition A.3 of [6], there exists a constant \(a_1=a_{1,t,x}>0\) such that (B.4) and (B.5) hold for any \((t',x_j')\) with \(0<t' \le t+1/2\) and \(|x_j'-x_j| \le 1\), and for any \(y_j \in \mathbb {R}\) with \(|y_j|>a_1\).
Let \(a:=a_1 \sqrt{d}\). Note that \(\{y\in \mathbb {R}^d: |y|\ge a \} \subset \bigcup _{j=1}^d B_j\), where
$$\begin{aligned} B_j=\left\{ y=(y_1,\ldots ,y_d) \in \mathbb {R}^d: |y_j|\ge a_1 \right\} , \quad j=1,\ldots ,d. \end{aligned}$$
Therefore, for any \(y\in \mathbb {R}^d\) with \(|y|\ge a\), there exists an index \(j=1,\ldots ,d\) such that \(|y_j|\ge a_1\). As we have shown above, this means that condition (B.2) holds for this y, for any \((t',x') \in B_{t,x}\). \(\square \)
Lemma B.2
\(J_0\) is continuous on \((0,\infty ) \times \mathbb {R}^d\).
Proof
Fix \(t>0\) and \(x \in \mathbb {R}^d\). By the definition of \(J_0\), we have:
$$\begin{aligned} |J_0(t,x)-J_0(t',x')| \le \int _{\mathbb {R}^d}|G(t,x-y)-G(t',x'-y)|\,|u_0|(\mathrm {d}y)=:L(t,t',x,x'). \end{aligned}$$
We claim that:
$$\begin{aligned} \lim _{(t',x') \rightarrow (t,x)}L(t,t',x,x')=0. \end{aligned}$$
(B.6)
To see this, we write \(L(t,t',x,x')=L_1(t,t',x,x')+L_2(t,t',x,x')\) where
$$\begin{aligned} L_1(t,t',x,x')&= \int _{|y| \ge a}|G(t,x-y)-G(t',x'-y)|\,|u_0|(\mathrm {d}y), \quad \text {and}\\ L_2(t,t',x,x')&= \int _{|y|<a}|G(t,x-y)-G(t',x'-y)|\,|u_0|(\mathrm {d}y), \end{aligned}$$
and \(a=a_{t,x}\) is the constant given by Proposition B.1. By enlarging a if necessary, we may assume that \(t>1/a\). By the dominated convergence theorem and the continuity of the function G, we see that \(L_i(t,t',x,x') \rightarrow 0\) when \((t',x') \rightarrow (t,x)\), for \(i=1,2\). To justify the application of this theorem, we argue as follows. For \(L_1(t,t',x,x')\), we use Proposition B.1 to infer that for any \((t',x') \in B_{t,x}\) and for any \(y \in \mathbb {R}^d\) with \(|y| \ge a\), we have:
$$\begin{aligned} |G(t,x-y)-G(t',x'-y)| \le 2 G(t+1,x-y). \end{aligned}$$
For \(L_2(t,t',x,x')\), we use the fact that for any \(t'>1/a, x' \in \mathbb {R}^d\) and \(y\in \mathbb {R}^d\) with \(|y| \le a\),
$$\begin{aligned} \frac{G(t',x'-y)}{G(t,x-y)}= & {} \frac{\sqrt{t}}{\sqrt{t'}}\exp \left( -\frac{(x'-y)^2}{2t'}+\frac{(x-y)^2}{2t} \right) \\\le & {} \frac{\sqrt{t}}{\sqrt{1/a}}\exp \left( \frac{|x|^2+|a|^2}{t}\right) =:C_{t,x}, \end{aligned}$$
and hence \(|G(t',x'-y)-G(t,x-y)| \le (C_{t,x}+1)G(t,x-y)\). \(\square \)
Lemma B.3
For any \(p \ge 2\) and \(n \ge 1, J_n\) is \(L^p(\Omega )\)-continuous on \((0,\infty ) \times \mathbb {R}^d\).
Proof
We proceed as in the proof of Lemma 3.6 of [2]. We divide the proof in three steps.
Step 1. (right-continuity in time) We will prove that for any \(t>0\) and \(a>0\),
$$\begin{aligned} \lim _{h \downarrow 0}\Vert J_n(t+h,x)-J_n(t,x)\Vert _p=0 \quad \text{ uniformly } \text{ in } \ x \in [-a,a]^d. \end{aligned}$$
(B.7)
For any \(h>0\), we have:
$$\begin{aligned} \Vert J_n(t+h,x)-J_n(t,x)\Vert _p^2\le & {} (p-1)^n \Vert J_n(t+h,x)-J_n(t,x)\Vert _2^2 \nonumber \\= & {} (p-1)^n n! \, \Vert \widetilde{f}_n(\cdot ,t+h,x)-\widetilde{f}_n(\cdot ,t,x)\Vert _{\mathcal {H}^{\otimes n}}^2 \nonumber \\\le & {} \frac{2}{n!} \left( A_n(t,x,h)+B_n(t,x,h) \right) , \end{aligned}$$
(B.8)
where
$$\begin{aligned} A_n(t,x,h)&= (n!)^2 \Vert \widetilde{f}_n(\cdot ,t+h,x)1_{[0,t]^{n}}-\widetilde{f}_n(\cdot ,t,x) \Vert _{\mathcal {H}^{\otimes n}}^{2}, \end{aligned}$$
(B.9)
$$\begin{aligned} B_n(t,x,h)&= (n!)^2 \Vert \widetilde{f}_n(\cdot ,t+h,x)1_{[0,t+h]^{n}\setminus [0,t]^n} \Vert _{\mathcal {H}^{\otimes n}}^{2}. \end{aligned}$$
(B.10)
We evaluate \(A_n(t,h,x)\) first. We have:
$$\begin{aligned} A_n(t,h,x) = \int _{[0,t]^{2n}} \prod _{j=1}^{n}\gamma (t_j-s_j) \psi _{t,h,x}^{(n)}(\mathbf{t}, \mathbf{s})\mathrm {d}\mathbf{t} \mathrm {d}\mathbf{s}, \end{aligned}$$
where
$$\begin{aligned} \psi _{t,h,x}^{(n)}(\mathbf{t},\mathbf{s})&=\frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}} \mathcal {F}(g_{\mathbf{t},t+h,x}^{(n)}-g_{\mathbf{t},t,x}^{(n)}) (\xi _1,\ldots ,\xi _n) \\&\quad \times \overline{\mathcal {F}(g_{\mathbf{s},t+h,x}^{(n)}-g_{\mathbf{s},t,x}^{(n)}) (\xi _1, \ldots ,\xi _n)} \,\, \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n). \end{aligned}$$
Similarly to (3.4), we have:
$$\begin{aligned} A_n(t,h,x) \le \Gamma _t^n \int _{[0,t]^n}\psi _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}=\Gamma _t^n \sum _{\rho \in S_n}\int _{0<t_{\rho (1)}<\cdots<t_{\rho (n)}<t}\psi _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}. \end{aligned}$$
(B.11)
If \(t_{\rho (1)}<\cdots<t_{\rho (n)}<t=:t_{\rho (n+1)}\), then by (3.5),
$$\begin{aligned}&|\mathcal {F}(g_{\mathbf{t},t+h,x}^{(n)}-g_{\mathbf{t},t,x}^{(n)}) (\xi _1,\ldots ,\xi _n) |^2 \\&\quad \le \lambda ^{2n} J_+^2(t,x) \prod _{k=1}^{n-1} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \\&\quad \left| \exp \left( -\frac{1}{2}\frac{t+h-t_{\rho (n)}}{t_{\rho (n)}(t+h)} \left| \sum _{j=1}^{n}t_{j} \xi _{j} \right| ^2 \right) -\exp \left( -\frac{1}{2}\frac{t-t_{\rho (n)}}{t_{\rho (n)}t} \left| \sum _{j=1}^{n}t_{j} \xi _{j} \right| ^2 \right) \right| ^2 \\&\quad = \lambda ^{2n} J_+^2(t,x) \prod _{k=1}^{n} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \\&\quad \left[ 1-\exp \left( -\frac{h}{2t(t+h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2, \end{aligned}$$
and hence
$$\begin{aligned} \psi _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})&\le \Gamma _t^n J_+^2(t,x) \frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}} \prod _{k=1}^{n} \exp \nonumber \\&\quad \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \nonumber \\&\quad \times \left[ 1-\exp \left( -\frac{h}{2t(t+h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2 \mu (\mathrm {d}\xi _1)\ldots \mu _n(\mathrm {d}\xi _n). \end{aligned}$$
(B.12)
Using (B.11) and (B.12), it follows that
$$\begin{aligned} A_n(t,h,x)&\le \Gamma _t^n \lambda ^{2n}J_+^2(t,x) n! \frac{1}{(2\pi )^{nd}} \int _{0<t_1<\cdots<t_n<t} \int _{\mathbb {R}^{nd}} \prod _{k=1}^{n}\exp \nonumber \\&\quad \left( -\frac{t_{k+1}-t_k}{t_k t_{k+1}} \left| \sum _{j=1}^{k}t_j \xi _j \right| ^2\right) \nonumber \\&\quad \times \left[ 1-\exp \left( -\frac{h}{2t(t+h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2 \nonumber \\&\quad \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n) \mathrm {d}t_1 \ldots \mathrm {d}t_n, \end{aligned}$$
(B.13)
with the convention \(t_{n+1}=t\). By the dominated convergence theorem and (3.20), we conclude that
$$\begin{aligned} \lim _{h \downarrow 0}A_n(t,h,x)= 0 \quad \text{ uniformly } \text{ in } x \in [-a,a]^d. \end{aligned}$$
(B.14)
As for \(B_n(t,h,x)\), note that
$$\begin{aligned} B_n(t,h,x) =\int _{[0,t+h]^{2n}} \prod _{j=1}^{n}\gamma (t_j-s_j) \gamma _{t,h,x}^{(n)}(\mathbf{t}, \mathbf{s})1_{D_{t,h}}(\mathbf{t}) 1_{D_{t,h}}(\mathbf{s})\mathrm {d}\mathbf{t} d\mathbf{s}, \end{aligned}$$
where
and
$$\begin{aligned} \gamma _{t,h,x}^{(n)}(\mathbf{t},\mathbf{s})= & {} \frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}} \mathcal {F}g_{\mathbf{t},t+h,x}^{(n)}(\xi _1, \ldots ,\xi _n)\\&\overline{\mathcal {F}g_{\mathbf{s},t+h,x}^{(n)}(\xi _1, \ldots ,\xi _n)} \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n). \end{aligned}$$
Similarly to (46) of [2], it can be proved that
$$\begin{aligned} B_n(t,h,x) \le \Gamma _{t+h}^n \int _{[0,t+h]^n} \gamma _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})1_{D_{t,h}}(\mathbf{t})\mathrm {d}\mathbf{t}. \end{aligned}$$
(B.15)
If \(t_{\rho (1)}<\cdots<t_{\rho (n)}<t+h\), then by (3.5),
$$\begin{aligned}&|\mathcal {F}g_{\mathbf{t},t+h,x}(\xi _1,\ldots ,\xi _n)|^2 \\&\quad \le \lambda ^{2n} J_+^2(t,x) \prod _{k=1}^{n-1}\exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| \sum _{j=1}^k t_{\rho (j)}\xi _{\rho (j)} \right| ^2\right) \\&\quad \quad \times \exp \left( -\frac{t+h-t_{\rho (n)}}{(t+h)t_{\rho (n)}} \left| \sum _{j=1}^n t_{j}\xi _{j} \right| ^2\right) , \end{aligned}$$
and hence, by Lemma 3.4
$$\begin{aligned} \gamma _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})&\le \lambda ^{2n}J_+^2(t,x) \frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}}\prod _{k=1}^{n-1}\exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| \sum _{j=1}^k t_{\rho (j)}\xi _{\rho (j)} \right| ^2\right) \nonumber \\&\quad \times \exp \left( -\frac{t+h-t_{\rho (n)}}{(t+h)t_{\rho (n)}} \left| \sum _{j=1}^n t_{j}\xi _{j} \right| ^2\right) \mu (\mathrm {d}\xi _1) \ldots \mu _n(\mathrm {d}\xi _n) \nonumber \\&\le \lambda ^{2n} J_+^2(t,x) \frac{1}{(2\pi )^{nd}} \prod _{k=1}^{n-1} \int _{\mathbb {R}^d} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| t_{\rho (k)}\xi _{k} \right| ^2\right) \mu (\mathrm {d}\xi _k) \nonumber \\&\quad \times \int _{\mathbb {R}^d}\exp \left( -\frac{t+h-t_{\rho (n)}}{(t+h)t_{\rho (n)}} \left| t_{\rho (n)}\xi _{n} \right| ^2\right) \mu (\mathrm {d}\xi _n). \end{aligned}$$
(B.16)
Using relations (B.15) and (B.16), and the fact that
$$\begin{aligned} D_{t,h}=\bigcup _{\rho \in S_n}\{(t_1,\ldots ,t_n);0<t_{\rho (1)}<\cdots<t_{\rho (n)}<t+h, \, t_{\rho (n)}>t\}, \end{aligned}$$
we obtain that
$$\begin{aligned} B_n(t,h,x)&\le \Gamma _{t+h}^n \sum _{\rho \in S_n} \int _{t}^{t+h} \int _{0<t_{\rho (1)}<\cdots<t_{\rho (n-1)}<t_{\rho (n)}}\nonumber \\&\quad \gamma _{t,h,x}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}t_{\rho (1)} \ldots \mathrm {d}t_{\rho (n-1)} \mathrm {d}t_{\rho (n)} \nonumber \\&\le \Gamma _{t+h}^n \lambda ^{2n} J_+^2(t,x) n! \, \frac{1}{(2\pi )^{nd}} \int _{t}^{t+h} \int _{0<t_{1}<\cdots<t_{n-1}<t_{n}} \nonumber \\&\quad \prod _{k=1}^{n-1} \int _{\mathbb {R}^d} \exp \left( -\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}} \left| t_{k}\xi _{k} \right| ^2\right) \mu (\mathrm {d}\xi _k)\nonumber \\&\quad \times \int _{\mathbb {R}^d}\exp \left( -\frac{t+h-t_{n}}{(t+h)t_{n}} \left| t_{n}\xi _{n} \right| ^2\right) \mu (\mathrm {d}\xi _n) \mathrm {d}t_{1} \ldots \mathrm {d}t_{n-1} \mathrm {d}t_{n} \nonumber \\&= \Gamma _{t+h}^n \lambda ^{2n} J_+^2(t,x) n! \, \int _{t}^{t+h} \int _{0<t_{1}<\cdots<t_{n-1}<t_{n}} J_{t_n}^{(n-1)}(t_1,\ldots ,t_{n-1}) k\nonumber \\&\quad \left( \frac{2(t+h-t_n)t_n}{t+h}\right) \mathrm {d}t_n \nonumber \\&\le \Gamma _{t+h}^n \lambda ^{2n} J_+^2(t,x) n! \, 2^{n-1}\int _{t}^{t+h}h_{n-1}(t_n) k\left( \frac{2(t+h-t_n)t_n}{t+h}\right) \mathrm {d}t_n \nonumber \\&= \Gamma _{t+h}^n \lambda ^{2n}J_+^2(t,x) n! \, 2^{n-1}\int _{0}^{h}h_{n-1}(t+s) k\left( \frac{2(h-s)(t+s)}{t+h}\right) \mathrm {d}s \nonumber \\&\le \Gamma _{t+h}^n \lambda ^{2n} J_+^2(t,x) n! \, 2^{n-1} h_{n-1}(t+h) \int _0^h k \left( \frac{2(h-s)t}{t+h} \right) \mathrm {d}s \end{aligned}$$
(B.17)
where the second last inequality is due to Lemma 3.6, and for the last inequality we used the fact that \(h_{n-1}\) is non-decreasing and k is non-increasing. By the dominated convergence theorem and (3.20), we infer that
$$\begin{aligned} \lim _{h \downarrow 0}B_n(t,h,x) = 0 \quad \text{ uniformly } \text{ in } x \in [-a,a]^d. \end{aligned}$$
(B.18)
Relation (B.7) follows from (B.8), (B.14) and (B.18).
Step 2. (left-continuity in time) We will prove that for any \(t>0\) and \(a>0\),
$$\begin{aligned} \lim _{h \downarrow 0}\Vert J_n(t-h,x)-J_n(t,x)\Vert _p=0 \quad \text{ uniformly } \text{ in } \ x \in [-a,a]^d. \end{aligned}$$
(B.19)
For any \(h>0\), we have:
$$\begin{aligned} \Vert J_n(t-h,x)-J_n(t,x)\Vert _p^2\le & {} (p-1)^n \Vert J_n(t-h,x)-J_n(t,x)\Vert _2^2 \nonumber \\= & {} (p-1)^n n! \, \Vert \widetilde{f}_n(\cdot ,t-h,x)-\widetilde{f}_n(\cdot ,t,x)\Vert _{\mathcal {H}^{\otimes n}}^2 \nonumber \\\le & {} \frac{2}{n!} \left( A_n'(t,x,h)+B_n'(t,x,h) \right) , \end{aligned}$$
(B.20)
where
$$\begin{aligned} A_n'(t,x,h)= (n!)^2 \Vert \widetilde{f}_n(\cdot ,t-h,x)-\widetilde{f}_n(\cdot ,t,x) 1_{[0,t-h]^n}\Vert _{\mathcal {H}^{\otimes n}}^{2}, \end{aligned}$$
(B.21)
$$\begin{aligned} B_n'(t,x,h)= (n!)^2 \Vert \widetilde{f}_n(\cdot ,t,x)1_{[0,t]^{n}\setminus [0,t-h]^n} \Vert _{\mathcal {H}^{\otimes n}}^{2}. \end{aligned}$$
(B.22)
We evaluate \(A_n'(t,h,x)\) first. We have:
$$\begin{aligned} A_n'(t,h,x) = \int _{[0,t-h]^{2n}} \prod _{j=1}^{n}\gamma (t_j-s_j) \psi _{t,h,x}^{(n)'}(\mathbf{t}, \mathbf{s})\mathrm {d}\mathbf{t} \mathrm {d}\mathbf{s}, \end{aligned}$$
where
$$\begin{aligned} \psi _{t,h,x}^{(n)}(\mathbf{t},\mathbf{s})'&=\frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}} \mathcal {F}(g_{\mathbf{t},t,x}^{(n)}-g_{\mathbf{t},t-h,x}^{(n)}) (\xi _1,\ldots ,\xi _n) \\&\quad \times \overline{\mathcal {F}(g_{\mathbf{s},t,x}^{(n)}-g_{\mathbf{s},t-h,x}^{(n)}) (\xi _1, \ldots ,\xi _n)}\mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n). \end{aligned}$$
Similarly to (3.4), we have:
$$\begin{aligned} A_n'(t,h,x)\le & {} \Gamma _{t-h}^n \int _{[0,t-h]^n}\psi _{t,h,x}^{(n)'}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}\nonumber \\= & {} \Gamma _{t-h}^n \sum _{\rho \in S_n}\int _{0<t_{\rho (1)}<\cdots<t_{\rho (n)}<t-h}\psi _{t,h,x}^{(n)'}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}. \end{aligned}$$
(B.23)
If \(t_{\rho (1)}<\cdots<t_{\rho (n)}<t-h\), then by (3.5),
$$\begin{aligned}&|\mathcal {F}(g_{\mathbf{t},t,x}^{(n)}-g_{\mathbf{t},t-h,x}^{(n)}) (\xi _1,\ldots ,\xi _n) |^2 \\&\quad \le \lambda ^{2n} J_+^2(t,x) \prod _{k=1}^{n-1} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \\&\quad \quad \times \left| \exp \left( -\frac{1}{2}\frac{t-t_{\rho (n)}}{tt_{\rho (n)}} \left| \sum _{j=1}^{n}t_{j} \xi _{j} \right| ^2 \right) -\exp \left( -\frac{1}{2}\frac{t-h-t_{\rho (n)}}{(t-h)t_{\rho (n)}} \left| \sum _{j=1}^{n}t_{j} \xi _{j} \right| ^2 \right) \right| ^2 \\&\quad = \lambda ^{2n} J_+^2(t,x) \prod _{k=1}^{n} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \\&\quad \quad \times \exp \left( -\frac{t-h-t_{\rho (n)}}{(t-h)t_{\rho (n)}}\left| \sum _{j=1}^n t_j \xi _j \right| ^2 \right) \left[ 1-\exp \left( -\frac{h}{2t(t-h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2, \end{aligned}$$
and hence
$$\begin{aligned} \psi _{t,h,x}^{(n)'}(\mathbf{t},\mathbf{t})&\le \Gamma _t^n J_+^2(t,x) \frac{1}{(2\pi )^{nd}}\int _{\mathbb {R}^{nd}} \prod _{k=1}^{n-1} \exp \left( -\frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)} t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)} \xi _{\rho (j)} \right| ^2 \right) \nonumber \\&\quad \times \exp \left( -\frac{t-h-t_{\rho (n)}}{(t-h)t_{\rho (n)}}\left| \sum _{j=1}^n t_j \xi _j \right| ^2 \right) \nonumber \\&\quad \times \left[ 1-\exp \left( -\frac{h}{2t(t-h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2 \mu (\mathrm {d}\xi _1)\ldots \mu _n(\mathrm {d}\xi _n). \end{aligned}$$
(B.24)
It follows that
$$\begin{aligned} A_n'(t,h,x)&\le \Gamma _t^n \lambda ^{2n}J_+^2(t,x) n! \frac{1}{(2\pi )^{nd}} \int _{0<t_1<\cdots<t_n<t-h} \int _{\mathbb {R}^{nd}}\nonumber \\&\quad \prod _{k=1}^{n-1}\exp \left( -\frac{t_{k+1}-t_k}{t_k t_{k+1}} \left| \sum _{j=1}^{k}t_j \xi _j \right| ^2\right) \nonumber \\&\quad \times \exp \left( -\frac{t-h-t_{n}}{(t-h)t_{n}}\left| \sum _{j=1}^n t_j \xi _j \right| ^2 \right) \nonumber \\&\quad \times \left[ 1-\exp \left( -\frac{h}{2t(t-h)}\left| \sum _{j=1}^{n}t_j \xi _j\right| ^2 \right) \right] ^2 \nonumber \\&\quad \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n) \mathrm {d}t_1 \ldots \mathrm {d}t_n, \end{aligned}$$
(B.25)
We will now prove that
$$\begin{aligned} \lim _{h \downarrow 0}A_n'(t,h,x)= 0 \quad \text{ uniformly } \text{ in } x \in [-a,a]^d. \end{aligned}$$
(B.26)
For this, we assume that \(h\in [0,t/2]\). Notice that:
$$\begin{aligned}&\exp \left( -\frac{t-h-t_n}{(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \left[ 1-\exp \left( -\frac{h}{2t(t-h)}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \right] ^{2}\\&\quad \le \exp \left( -\frac{t-h-t_n}{(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \min \left( \frac{h}{t^{2}}\left| \sum _{j=1}^n t_j \xi _j\right| ^{2},1\right) . \end{aligned}$$
For this, we used the fact that \((1-e^{-x})^2 \le 1-e^{-x} \le \min (x,1)\) for \(x>0\). Now we move the exponential inside \(\min (\ldots )\) and consider the two competing terms separately. For \(A>0\) and \(x\ge 0\), we see that
$$\begin{aligned} \exp \left( -\frac{A}{2} x^2\right) x^{2} =\exp \left( -\frac{A}{2} x^2 + 2 \log x\right) \le (2/e) A^{-1}. \end{aligned}$$
(B.27)
This can be seen by noticing that the function \(f(x)=-\frac{A}{2} x^2 +2 \log x,x>0\) attains its maximum at \(x_0=\sqrt{2/A}\). Hence, inequality (B.27) implies that
$$\begin{aligned}&\exp \left( -\frac{t-h-t_n}{(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \frac{h}{t^2}\left| \sum _{j=1}^n t_j \xi _j\right| ^{2} \\&\quad \le \frac{2h}{e t^2} \frac{(t-h)t_n}{t-h-t_n} \exp \left( -\frac{t-h-t_n}{2(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \\&\quad \le \frac{2h}{e(t-h-t_n)} \exp \left( -\frac{t-h-t_n}{2(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) . \end{aligned}$$
The second term is bounded by
$$\begin{aligned} \exp \left( -\frac{t-h-t_n}{(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \le \exp \left( -\frac{t-h-t_n}{2(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) . \end{aligned}$$
Therefore,
$$\begin{aligned} \exp&\left( -\frac{t-h-t_n}{(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \min \left( \frac{h}{t^{2}}\left| \sum _{j=1}^n t_j \xi _j\right| ^{2},1\right) \\&\le \exp \left( -\frac{t-h-t_n}{2(t-h)t_n}\left| \sum _{j=1}^n t_j \xi _j\right| ^2\right) \min \left( \frac{2h}{e(t-h-t_n)},1\right) . \end{aligned}$$
Putting the above bounds back into the expression of \(A_n'(t,h,x)\), we see that
$$\begin{aligned} A_n'(t,h,x)&\le \Gamma _t^n \lambda ^{2n} J_+^2(t,x) n! \frac{1}{(2\pi )^{nd}} \int _{0<t_1<\cdots<t_n<t-h} \mathrm {d}t_1 \ldots \mathrm {d}t_n \\&\quad \int _{\mathbb {R}^{nd}} \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n)\\&\quad \times \prod _{k=1}^{n-1}\exp \left( -\frac{t_{k+1}-t_k}{2t_k t_{k+1}} \left| \sum _{j=1}^{k}t_j \xi _j \right| ^2\right) \\&\quad \times \exp \left( -\frac{t-h-t_n}{2(t-h)t_n} \left| \sum _{j=1}^{n}t_j \xi _j \right| ^2\right) \min \left( \frac{2h}{e(t-h-t_n)},1\right) \\&=: \Gamma _t^n \lambda ^{2n} J_+^2(t,x) n!\, A''_n(t,h). \end{aligned}$$
Relation (B.26) will follow from (3.20), once we prove that:
$$\begin{aligned} \lim _{h\rightarrow 0} A_n''(t,h) =0 . \end{aligned}$$
(B.28)
We will use the fact that
$$\begin{aligned} \frac{1}{2}\frac{t_{k+1}-t_k}{t_{k}t_{k+1}} \left| \sum _{j=1}^{k}t_j \xi _j\right| ^2= \frac{\frac{t_{k+1}}{2}-\frac{t_k}{2}}{\frac{t_{k+1}}{2}\frac{t_k}{2}} \left| \sum _{j=1}^{k}\frac{t_j}{2}\xi _j\right| ^2 \end{aligned}$$
(B.29)
for any \(k=1, \ldots ,n\), with \(t_{n+1}=t-h\). Using the change of variables \(t_k'=t_k/2\) for \(k=1,\ldots ,n\), and recalling the definition of the integral \(I_{t}^{(n)}(t_1,\ldots ,t_n)\) given in Lemma 3.3, we see that
$$\begin{aligned} A_n''(t,h)&= 2^n \int _{0<t_1<\cdots<t_{n}<\frac{t-h}{2}}\min \left( \frac{h}{e(\frac{t-h}{2}-t_n)},1 \right) I_{\frac{t-h}{2}}^{(n)}(t_1,\ldots ,t_n)\mathrm {d}t_1 \ldots \mathrm {d}t_n\\&\le 2^n \int _{0<t_1<\cdots<t_{n}<\frac{t-h}{2}}\min \left( \frac{h}{e(\frac{t-h}{2}-t_n)},1 \right) J_{\frac{t-h}{2}}^{(n)}(t_1,\ldots ,t_n)\mathrm {d}t_1 \ldots \mathrm {d}t_n \\&= 2^n \int _{0}^{\frac{t-h}{2}} \min \left( \frac{h}{e(\frac{t-h}{2}-t_n)},1 \right) k\left( \frac{2(\frac{t-h}{2}-t_n)t_n}{\frac{t-h}{2}}\right) \\&\quad \times \left( \int _{0<t_1<\cdots<t_{n-1}<t_{n}}J_{t_n}^{(n-1)}(t_1,\ldots ,t_{n-1})\mathrm {d}t_1 \ldots \mathrm {d}t_{n-1}\right) \mathrm {d}t_n\\&\le 2^{2n-1} \int _{0}^{\frac{t-h}{2}} \min \left( \frac{h}{e(\frac{t-h}{2}-s)},1 \right) k\left( \frac{2(\frac{t-h}{2}-s)s}{\frac{t-h}{2}}\right) h_{n-1}(s) \mathrm {d}s\\&\le 2^{2n} \int _{0}^{\frac{t-h}{2}} \min \left( \frac{h}{es},1 \right) k(s)h_{n-1}(t-s) \mathrm {d}s, \end{aligned}$$
where the first inequality is due to Lemma 3.5, the second last inequality is due to Lemma 3.6, and the last inequality can be proved similarly to (3.12). By the dominated convergence theorem, the last integral converges to 0 as \(h \rightarrow 0\), because \(\int _0^{t} k(s) h_{n-1}(t-s) \mathrm {d}s = h_n(t)<\infty \). This concludes the proof of (B.28).
As for \(B_n'(t,h,x)\), note that
$$\begin{aligned} B_n'(t,h,x) =\int _{[0,t]^{2n}} \prod _{j=1}^{n}\gamma (t_j-s_j) \psi _{t,x}^{(n)}(\mathbf{t}, \mathbf{s})1_{D_{t,h}'}(\mathbf{t}) 1_{D_{t,h}'}(\mathbf{s})\mathrm {d}\mathbf{t} \mathrm {d}\mathbf{s}, \end{aligned}$$
where
and \(\psi _{t,x}^{(n)}(\mathbf{t},\mathbf{s})\) is given by (3.2).
Similarly to (B.15), we have:
$$\begin{aligned} B_n'(t,h,x) \le \Gamma _{t}^n \int _{[0,t]^n} \psi _{t,x}^{(n)}(\mathbf{t},\mathbf{t})1_{D_{t,h}'}(\mathbf{t})\mathrm {d}\mathbf{t}. \end{aligned}$$
(B.30)
Using Lemmas 3.2, 3.4 and 3.6, and the fact that
$$\begin{aligned} D_{t,h}'=\bigcup _{\rho \in S_n}\{(t_1,\ldots ,t_n);0<t_{\rho (1)}<\cdots<t_{\rho (n)}<t, \, t_{\rho (n)}>t-h\}, \end{aligned}$$
we obtain that
$$\begin{aligned} B_n'(t,h,x)&\le \Gamma _{t}^n \sum _{\rho \in S_n} \int _{t-h}^{t} \int _{0<t_{\rho (1)}<\cdots<t_{\rho (n-1)}<t_{\rho (n)}} \psi _{t,x}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}t_{\rho (1)} \ldots \mathrm {d}t_{\rho (n-1)} \mathrm {d}t_{\rho (n)} \nonumber \\&\le \Gamma _{t}^n \lambda ^{2n} J_+^2(t,x) n! \, \frac{1}{(2\pi )^{(n-1)d}} \int _{t-h}^{t} \int _{0<t_{1}<\cdots<t_{n-1}<t_{n}} \mathrm {d}t_{1} \ldots \mathrm {d}t_{n-1} \mathrm {d}t_{n}\nonumber \\&\quad \times \prod _{k=1}^{n-1} \int _{\mathbb {R}^d} \exp \left( -\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}} \left| t_{k}\xi _{k} \right| ^2\right) \mu (\mathrm {d}\xi _k)\nonumber \\&\quad \times \left( \frac{1}{(2\pi )^d}\int _{\mathbb {R}^d} \exp \left( -\frac{t-t_{n}}{t_nt} \left| t_n\xi _{n} \right| ^2\right) \mu (\mathrm {d}\xi _n) \right) \nonumber \\&= \Gamma _{t}^n \lambda ^{2n} J_+^2(t,x) n! \, \int _{t-h}^{t} \mathrm {d}t_n \, k\left( \frac{2(t-t_n)t_n}{t}\right) \nonumber \\&\quad \times \int _{0<t_{1}<\cdots<t_{n-1}<t_{n}} J_{t_n}^{(n-1)}(t_1,\ldots ,t_{n-1})\mathrm {d}t_1 \ldots \mathrm {d}t_{n-1} \nonumber \\&\le \Gamma _{t}^n \lambda ^{2n} J_+^2(t,x) n! \, 2^{n-1}\int _{t-h}^{t}h_{n-1}(t_n) k\left( \frac{2(t-t_n)t_n}{t}\right) \mathrm {d}t_n \nonumber \\&\le \Gamma _{t}^n \lambda ^{2n} J_+^2(t,x) n! \, 2^{n-1}h_{n-1}(t) \int _{0}^{h} k\left( \frac{2s(t-s)}{t}\right) \mathrm {d}s. \end{aligned}$$
(B.31)
By the dominated convergence theorem and (3.20), it follows that
$$\begin{aligned} \lim _{h \downarrow 0}B_n'(t,h,x)= 0 \quad \text{ uniformly } \text{ in } x \in [-a,a]^d. \end{aligned}$$
(B.32)
Relation (B.19) follows from (B.20), (B.26) and (B.32).
Step 3. (continuity in space) We will prove that for any \(t>0\) and \(x \in \mathbb {R}^d\),
$$\begin{aligned} \lim _{|z| \rightarrow 0}\Vert J_n(t,x+z)-J_n(t,x)\Vert _p=0. \end{aligned}$$
(B.33)
For any \(z \in \mathbb {R}^d\), we have
$$\begin{aligned} \Vert J_n(t,x+z)-J_n(t,x)\Vert _p\le & {} (p-1)^{n} \Vert J_n(t,x+z)-J_n(t,x)\Vert _2^2 \nonumber \\= & {} (p-1)^n \frac{1}{n!} \, C_n(t,x,z), \end{aligned}$$
(B.34)
where
$$\begin{aligned} C_n(t,x,z)= & {} (n!)^2 \Vert \widetilde{f}_n(\cdot ,t,x+z)-\widetilde{f}_n(\cdot ,t,x)\Vert _{\mathcal {H}^{\otimes n}}^2 \nonumber \\= & {} \int _{[0,t]^{2n}} \prod _{j=1}^{n}\gamma (t_j-s_j) \psi _{t,x,z}^{(n)}(\mathbf{t},\mathbf{s})\mathrm {d}\mathbf{t}\mathrm {d}\mathbf{s} \end{aligned}$$
(B.35)
and
$$\begin{aligned} \psi _{t,x,z}^{(n)}(\mathbf{t},\mathbf{s})&= \frac{1}{(2\pi )^{nd}} \int _{\mathbb {R}^d} \mathcal {F}(g_{\mathbf{t},t,x+z}^{(n)}-g_{\mathbf{t},t,x}^{(n)})(\xi _1,\ldots ,\xi _n)\\&\quad \times \overline{\mathcal {F}(g_{\mathbf{s},t,x+z}^{(n)}-g_{\mathbf{s},t,x}^{(n)})(\xi _1,\ldots ,\xi _n)} \mu (\mathrm {d}\xi _1) \ldots \mu (\mathrm {d}\xi _n). \end{aligned}$$
Similarly to the previous estimates, we have:
$$\begin{aligned} C_n(t,x,z)\le \Gamma _t^n \int _{[0,t]^n}\psi _{t,x,z}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}=\Gamma _t^n \sum _{\rho \in S_n} \int _{t_{\rho (1)}<\cdots <t_{\rho (n)}}\psi _{t,x,z}^{(n)}(\mathbf{t},\mathbf{t})\mathrm {d}\mathbf{t}. \end{aligned}$$
(B.36)
If \(t_{\rho (1)}<\cdots<t_{\rho (n)}<t=t_{\rho (n+1)}\), then by (3.5),
$$\begin{aligned}&|\mathcal {F}(g_{\mathbf{t},t,x+z}^{(n)}-g_{\mathbf{t},t,x}^{(n)})(\xi _1, \ldots ,\xi _n)|^2\\&\quad =\lambda ^{2n} \prod _{k=1}^{n} \exp \left( - \frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)}\xi _{\rho (j)}\right| ^2 \right) \\&\quad \times \left| \exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot (x+z) \right] \right. \\&\left. \quad \int _{\mathbb {R}^d} \exp \left\{ -i \left[ \sum _{k=1}^{n}\left( 1-\frac{t_k}{t}\right) \xi _k\right] \cdot x_0\right\} G(t,x+z-x_0) u_0(\mathrm {d}x_0)\right. \\&\quad \left. -\exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot x \right] \right. \\&\left. \quad \int _{\mathbb {R}^d} \exp \left\{ -i \left[ \sum _{k=1}^{n}\left( 1-\frac{t_k}{t}\right) \xi _k\right] \cdot x_0\right\} G(t,x-x_0) u_0(\mathrm {d}x_0) \right| ^2. \end{aligned}$$
Inside the squared modulus above, we add and subtract the term
$$\begin{aligned}&\exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot x \right] \int _{\mathbb {R}^d} \exp \left\{ -i \left[ \sum _{k=1}^{n}\left( 1-\frac{t_k}{t}\right) \xi _k\right] \cdot x_0\right\} \\&\quad G(t,x+z-x_0) u_0(\mathrm {d}x_0). \end{aligned}$$
We obtain that
$$\begin{aligned}&|\mathcal {F}(g_{\mathbf{t},t,x+z}^{(n)}-g_{\mathbf{t},t,x}^{(n)})(\xi _1, \ldots ,\xi _n)|^2 \nonumber \\&\quad \le 2 \lambda ^{2n} \prod _{k=1}^{n} \exp \left( - \frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)}\xi _{\rho (j)}\right| ^2 \right) \\&\quad \times \left\{ \left| \exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot z \right] -1\right| ^2 J_0^2(t,x+z)+F^2(t,x,z)\right\} , \end{aligned}$$
where
$$\begin{aligned} F(t,x,z)= & {} L(t,t,x,x+z)=\int _{\mathbb {R}^d}|G(t,x+z-x_0)\nonumber \\&-G(t,x-x_0)|\,|u_0|(\mathrm {d}x_0). \end{aligned}$$
(B.37)
Hence,
$$\begin{aligned} \psi _{t,x,z}^{(n)}(\mathbf{t},\mathbf{t})&\le 2\lambda ^{2n} \frac{1}{(2\pi )^{nd}} \int _{\mathbb {R}^{nd}}\mu (\mathrm {d}\xi _1)\ldots \mu (\mathrm {d}\xi _n) \prod _{k=1}^{n} \exp \nonumber \\&\quad \left( - \frac{t_{\rho (k+1)}-t_{\rho (k)}}{t_{\rho (k)}t_{\rho (k+1)}} \left| \sum _{j=1}^{k}t_{\rho (j)}\xi _{\rho (j)}\right| ^2 \right) \\&\quad \times \left\{ \left| \exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot z \right] -1\right| ^2 J_0^2(t,x+z)+F^2(t,x,z)\right\} . \end{aligned}$$
Using (B.36), it follows that
$$\begin{aligned} C_n(t,x,z)\le 2 \left( C_n^{(1)}(t,x,z)+C_n^{(2)}(t,x,z) \right) , \end{aligned}$$
(B.38)
where
$$\begin{aligned} C_n^{(1)}(t,x,z)&= \Gamma _t^n \lambda ^{2n} J_+^2(t,x+z) n! \frac{1}{(2\pi )^{nd}}\nonumber \\&\quad \int _{0<t_1<\cdots<t_n<t} \int _{\mathbb {R}^{nd}} \prod _{k=1}^{n} \exp \left( - \frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\left| \sum _{j=1}^{k}t_{j}\xi _{j}\right| ^2 \right) \nonumber \\&\quad \times \left| \exp \left[ -\frac{i}{t}\left( \sum _{k=1}^{n}t_k \xi _k\right) \cdot z \right] -1\right| ^2 \mu (\mathrm {d}\xi _1)\ldots \mu (\mathrm {d}\xi _n) \mathrm {d}t_1 \ldots \mathrm {d}t_n, \end{aligned}$$
(B.39)
and
$$\begin{aligned} C_n^{(2)}(t,x,z)&= \Gamma _t^n \lambda ^{2n} F^2(t,x,z) n! \frac{1}{(2\pi )^{nd}}\int _{0<t_1<\cdots<t_n<t} \int _{\mathbb {R}^{nd}} \exp \nonumber \\&\quad \left( - \frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\left| \sum _{j=1}^{k}t_{j}\xi _{j}\right| ^2 \right) \nonumber \\&\quad \times \mu (\mathrm {d}\xi _1)\ldots \mu (\mathrm {d}\xi _n) \mathrm {d}t_1 \ldots \mathrm {d}t_n. \end{aligned}$$
(B.40)
By relation (B.6), \(\lim _{|z|\rightarrow 0}J_+(t,x+z)=J_+(t,x)\). By the dominated convergence theorem, \(\lim _{|z|\rightarrow 0}C_{n}^{(1)}(t,x,z)=0\). By (B.6), \(\lim _{|z| \rightarrow 0}F(t,x,z)=0\), and hence \(\lim _{|z|\rightarrow 0}C_{n}^{(2)}(t,x,z)=0\). Relation (B.33) follows from (B.34) and (B.38). \(\square \)
