Proof. For ease of use, we define the following functions \[\Phi(y,s\, ;\, \zeta,x,t) := G(x,y,t-s) \mathrm{\boldsymbol{1}}_{[\zeta,t]}(s)\quad \&\quad \Psi(y,s\, ;\, \zeta,t) := \big{\|} \Phi(y,s\, ;\, \zeta,\, \cdot\, ,t) \big{\|}_{L^{\infty}(D)}.\] We note that \(\Psi\) is well defined (one can either see that directly from the series definition of \(G\) or more easily from the estimate (15 ) \(i)\)). Furthermore, it is easy to prove that \(\Phi,\Psi\in L^{2} ( D\times[0,T] )\) thus, according to 1) of Remark 4, it holds \(\Phi,\Psi \in \mathcal{H}\) meaning that the action of the operator \(K_{H}^{*}\) on them is well defined. Lastly, again for ease of use, we define the processes \[M_{\tau}(x\, ;\, \zeta,\, t) := \int_{0}^{\tau} \int_{D} \big[ K^{*}_{H} ( \Phi ) \big] (y,s\, ;\, \zeta,\, x\, ,t) W(dy,ds)\quad \&\quad M^*(x\, ;\, \zeta,\, t) := \sup_{\tau \in [0,T]} |M_{\tau}(x\, ;\, \zeta,\, t)|.\] We note that, due to the preceding remarks on \(\Phi\), for each fixed \(x \in D\), the process \(\tau \mapsto M_{\tau}(x\, ;\, \zeta,\, t)\) is a continuous, centered, square-integrable \(\mathcal{F}_{\tau}\)-martingale with \(M_0(x\, ;\, \zeta,\, t) = 0\) and deterministic quadratic variation \[\langle M(x\, ;\, \zeta,\, t) \rangle_{\tau} = \left \| \big[ K_{H}^{*} ( \Phi ) \big] (\diamond, \bullet\, ;\, \zeta,\, x\, ,t) \right \|^2_{L^{2}(D\times[0,\tau])}.\] For any \(x,x' \in D\) and any \(p \geq 2\), the reverse triangle inequality for suprema yields \[E \Big{(} |M^*(x\, ;\, \zeta,\, t) - M^*(x'\, ;\, \zeta,\, t)|^p \Big{)} \leq E \bigg{(} \sup_{\tau \in [0,T]} |M_{\tau}(x\, ;\, \zeta,\, t) - M_{\tau}(x'\, ;\, \zeta,\, t)|^p \bigg{)}.\] Now, the difference \(M_{\tau}(x\, ;\, \zeta,\, t) - M_{\tau}(x'\, ;\, \zeta,\, t)\) is a continuous martingale vanishing at zero so, applying the Burkholder-Davis-Gundy inequality (cf. [16], Theorem 4.1), we get \[E \bigg{(} \sup_{\tau \in [0,T]} |M_{\tau}(x\, ;\, \zeta,\, t) - M_{\tau}(x'\, ;\, \zeta,\, t)|^p \bigg{)} \leq c(p) \big{\|} \big[ K_{H}^{*} ( \Phi ) \big] (\diamond, \bullet\, ;\, \zeta,\, x\, ,t) - \big[ K_{H}^{*} ( \Phi ) \big] (\diamond, \bullet\, ;\, \zeta,\, x'\, ,t) \big{\|}^p_{L^{2}(D\times[0,T])}.\] Since \(K_{H}^{*}\) is an isometry between \(\mathcal{H}\) and \(L^{2} ( D\times[0,T] )\), we have \[\left \| \big[ K_{H}^{*} ( \Phi ) \big] (\diamond, \bullet\, ;\, \zeta,\, x\, ,t) - \big[ K_{H}^{*} ( \Phi ) \big] (\diamond, \bullet\, ;\, \zeta,\, x'\, ,t) \right \|^p_{L^{2}(D\times[0,T])} = \left \| \Phi(\diamond, \bullet\, ;\, \zeta,\, x\, ,t) - \Phi(\diamond, \bullet\, ;\, \zeta,\, x'\, ,t)\right \|_{\mathcal{H}}^p.\] Due to the aforementioned continuous embedding \(L^{\frac{1}{H}}(D\times[0,T])\hookrightarrow\mathcal{H}\), it holds \[\begin{align} \left \| \Phi(\diamond, \bullet\, ;\, \zeta,\, x\, ,t) - \Phi(\diamond, \bullet\, ;\, \zeta,\, x'\, ,t)\right \|_{\mathcal{H}}^p &\leq c(H,p) \left \| \Phi(\diamond, \bullet\, ;\, \zeta,\, x\, ,t) - \Phi(\diamond, \bullet\, ;\, \zeta,\, x'\, ,t) \right \|_{L^{\frac{1}{H}}(D\times[0,T])}^{p} \\ &= c(H,p) \bigg{(} \int_{\zeta}^{t} \int_{D} | G(x,y,t-s) - G(x',y,t-s)|^{\frac{1}{H}}\, dyds \bigg{)}^{pH} \\ \overset{\text{F.T.C.}}&{=} c(H,p) \bigg{(} \int_{\zeta}^{t} \int_{D} \bigg{|} \int_{x'}^x G_{\xi}(\xi,y,t-s)\, d\xi \bigg{|}^{\frac{1}{H}}\, dyds \bigg{)}^{pH} \\ \overset{\text{Jensen}}&{\leq} c(H,p)\, |x - x'|^{p -pH} \bigg{(} \int_{\zeta}^{t} \int_{D} \int_{x'}^x |G_{\xi}(\xi,y,t-s)|^{\frac{1}{H}}\, d\xi dyds \bigg{)}^{pH} \\ &\overset{\substack{\text{Fubini}\\[0.05cm](\ref{Green39s32function39s32estimates})\;ii)}}{\leq} c(H,p)\, |x - x'|^{p -pH} \bigg{(} \int_{\zeta}^{t} (t-s)^{-\frac{1}{2H}} \int_{x'}^x \int_{D} \exp\Big{(}-C\, |\xi -y|^{\frac{4}{3}}\, (t-s)^{-\frac{1}{3}} \Big{)} dy d\xi ds \bigg{)}^{pH} \\ \overset{(\ref{specific32calculation})}&{\leq} c(H,p)\, |x - x'|^{p - pH} \bigg{(} |x - x'| \int_{\zeta}^{t} (t-s)^{-\frac{1}{2H}} (t-s)^{\frac{1}{4}}\, ds \bigg{)}^{pH} \\ &= c(H,p)\, |x - x'|^p \left ( \int_{\zeta}^{t} (t-s)^{\frac{H-2}{4H}}\, ds \right )^{pH} = c(H,p)\, |x - x'|^p\, (t-\zeta)^{\frac{p(5H-2)}{4}} \end{align}\] (the last integral is finite if and only if \(\frac{H-2}{4H} > -1 \iff H > \frac{2}{5}\) which holds since \(H > \frac{1}{2}\)) and thus \[\label{Kol-Che} E \Big{(} |M^*(x\, ;\, \zeta,\, t) - M^*(x'\, ;\, \zeta,\, t)|^p \Big{)} \leq c(H,p)\, |x - x'|^p\, (t-\zeta)^{\frac{p(5H-2)}{4}}.\tag{17}\] As a consequence, according to the one-dimensional Kolmogorov-Chentsov theorem (cf. [16], Theorem 2.1), the process \(x \mapsto M^*(x\, ;\, \zeta,\, t)\) admits a continuous modification whose paths are locally \(\gamma\)-H\(\ddot{\text{o}}\)lder for every \(\gamma \in \Big{(}0, 1 - \frac{1}{p} \Big{)}\) making \(\sup\limits_{x \in D} M^*(x\, ;\, \zeta,\, t)\) well-defined and measurable. Furthermore, it is easy to verify now that the mapping \((x,\tau) \mapsto M_{\tau}(x\, ;\, \zeta,\, t)\) is jointly measurable, making any future exchange of spatial and temporal suprema valid.
By a simpler, analogous procedure to the one we followed for (17 ), using (15 ) \(i)\) in place of (15 ) \(ii)\), we can also calculate that, for any \(x \in D\) and any \(p \geq 2\) \[\label{Kol-Che322} E \big{(} |M^*(x\, ;\, \zeta,\, t)|^p \big{)} \leq c(H,p) (t-\zeta)^{\frac{p(5H-1)}{4}} = c(H,p)\, (t-\zeta)^{\frac{p}{4}}(t-\zeta)^{\frac{p(5H-2)}{4}} \leq c(H,p,T) (t-\zeta)^{\frac{p(5H-2)}{4}}\tag{18}\] (we note that the bound is independent of \(x\) since, upon applying (15 ) \(i)\), the substitution \(z = y - x\) and relation (16 ) absorb all \(x\)-dependence).
Applying now the Garsia-Rodemich-Rumsey lemma (cf. [17], p. 566) for \(\varPsi(u) = u^p\) and \(\varrho(u) = u^{\alpha}\) with \(\alpha \in \Big{(} \frac{2}{p}, 1 + \frac{1}{p} \Big{)}\) for \(p \geq 2\), we obtain \[\sup_{x \in D} |M^*(x\, ;\, \zeta,\, t)| \leq |M^*(0\, ;\, \zeta,\, t)| + C(p)\, \bigg{(} \int_D \int_D \frac{|M^*(x\, ;\, \zeta,\, t) - M^*(y\, ;\, \zeta,\, t)|^p}{|x - y|^{\alpha p}} dxdy \bigg{)}^{\frac{1}{p}}\] (the finite constant \(C(p)\) is a direct consequence of the lower bound \(\alpha > \frac{2}{p}\)). Taking \(p\)-th power, expectation and then applying Tonelli’s theorem, yields \[\begin{align} E \bigg{(} \sup_{x \in D} |&M^*(x\, ;\, \zeta,\, t)|^p \bigg{)} \leq c(p) \left ( E \big{(} |M^*(0\, ;\, \zeta,\, t)|^p \big{)} + \int_D \int_D \frac{E \left ( |M^*(x\, ;\, \zeta,\, t) - M^*(y\, ;\, \zeta,\, t)|^p \right )}{|x - y|^{\alpha p}}\, dxdy \right ) \\ \overset{\substack{(\ref{Kol-Che})\\(\ref{Kol-Che322})}}&{\leq} c(p)\, (t-\zeta)^{\frac{p(5H-2)}{4}} \left ( c(H,p,T) + c(H,p)\int_D \int_D |x - y|^{p(1 - \alpha)}\, dxdy \right ) = c(H,p,T)\, (t-\zeta)^{\frac{p(5H-2)}{4}} \end{align}\] (the double integral is finite if and only if \(p(1 - \alpha) > -1 \iff \alpha < 1 + \frac{1}{p}\)). Gathering now all the above, we have \[\begin{align} E \bigg{(} \bigg{\|} \int_{\zeta}^{t} \int_{D} G(\cdot,y,t&-s) W_{H}(dy,ds) \bigg{\|}_{L^{\infty}(D)}^{p} \bigg{)} = E \bigg{(} \sup_{x \in D} |M_T(x\, ;\, \zeta,\, t)|^p \bigg{)} \leq E \bigg{(} \sup_{\tau \in [0,T]} \sup_{x \in D} |M_{\tau}(x\, ;\, \zeta,\, t)|^p \bigg{)} \\ &= E \bigg{(} \sup_{x \in D} \sup_{\tau \in [0,T]} |M_{\tau}(x\, ;\, \zeta,\, t)|^p \bigg{)} = E \bigg{(} \sup_{x \in D} |M^*(x\, ;\, \zeta,\, t)|^p \bigg{)} \leq c(H,p,T)\, (t-\zeta)^{\frac{p(5H-2)}{4}}. \end{align}\] which concludes the proof of (?? ). Regarding now the second estimate, we have \[\begin{align} \label{relation32between32Phi44Psi} \begin{aligned} \big{\|} \big[ K_{H}^{*} ( \Phi ) \big] (y,s\, ;\, \zeta,\, \cdot\, ,t) \big{\|}_{L^{\infty}(D)} \overset{(\ref{formula32132for32K95123H1259412342}})}&{=} \Bigg{\|} \int_{s}^{T} \Phi (y,r\, ;\, \zeta,\, \cdot\, ,t) \frac{\partial K_{H}}{\partial r}(r,s)\, dr \Bigg{\|}_{L^{\infty}(D)} \\[0.1cm] &\leq \int_{s}^{T} \big{\|} \Phi (y,r\, ;\, \zeta,\, \cdot\, ,t) \big{\|}_{L^{\infty}(D)} \frac{\partial K_{H}}{\partial r}(r,s)\, dr \\[0.1cm] &= \int_{s}^{T} \Psi(y,r\, ;\, \zeta,t) \frac{\partial K_{H}}{\partial r}(r,s)\, dr \overset{(\ref{formula32132for32K95123H1259412342}})}{=} \big[ K_{H}^{*} ( \Psi ) \big] (y,s\, ;\, \zeta,t) \end{aligned} \end{align}\tag{19}\] thus \[\begin{align} \int_{0}^{T} \int_{D} \big{\|} \big[ K^{*}_{H} \big{(} G(\cdot,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[\zeta,t]}(\bullet) \big{)} \big] (y,s) \big{\|}^{2}_{L^{\infty}(D)}\, dyds &= \int_{0}^{T} \int_{D} \big{\|} \big[ K_{H}^{*} ( \Phi ) \big] (y,s\, ;\, \zeta,\, \cdot\, ,t) \big{\|}^{2}_{L^{\infty}(D)}\, dyds \\[0.1cm] \overset{(\ref{relation32between32Phi44Psi})}&{\leq} \int_{0}^{T} \int_{D} \big{(} \big[ K_{H}^{*} ( \Psi ) \big] (y,s\, ;\, \zeta,t) \big{)}^{2}\, dyds \\[0.1cm] &= \big{\|} \big[ K_{H}^{*} ( \Psi ) \big] (\diamond,\bullet\, ;\, \zeta,t) \big{\|}^{2}_{L^{2}(D\times[0,T])}. \end{align}\] As we already mentioned, \(K_{H}^{*}\) is an isometry between \(\mathcal{H}\) and \(L^{2} ( D\times[0,T] )\), so \[\begin{align} \big{\|} \big[ K_{H}^{*} ( \Psi ) \big] (\diamond,\bullet\, ;\, \zeta,t) \big{\|}^{2}_{L^{2}(D\times[0,T])} = \big{\|} \Psi (\diamond,\bullet\, ;\, \zeta,t) \big{\|}^{2}_{\mathcal{H}} \leq c(H) \big{\|} \Psi (\diamond,\bullet\, ;\, \zeta,t) \big{\|}^{2}_{L^{\frac{1}{H}}(D\times[0,T])}. \end{align}\] Lastly, \[\begin{align} \big{\|} \Psi (\diamond,\bullet\, ;\, \zeta,t) \big{\|}^{2}_{L^{\frac{1}{H}}(D\times[0,T])} &= \bigg{(} \int_{0}^{T} \int_{D} \mathrm{\boldsymbol{1}}_{[\zeta,t]}(s) \left \| G(\cdot,y,t-s) \right \|_{L^{\infty}(D)}^{\frac{1}{H}} dyds \bigg{)}^{2H} \\ &= \bigg{(} \int_{\zeta}^{t} \int_{D} \left \| G(\cdot,y,t-s) \right \|_{L^{\infty}(D)}^{\frac{1}{H}} dyds \bigg{)}^{2H} \\ \overset{(\ref{Green39s32function39s32estimates})\;i)}&{\leq} c(H) \bigg{(} \int_{\zeta}^{t} (t-s)^{-\frac{1}{4H}} \bigg[ \int_{D} \Big{\|} \exp\Big{(}-C\, |\cdot-y|^{\frac{4}{3}}\, (t-s)^{-\frac{1}{3}} \Big{)} \Big{\|}^{\frac{1}{H}}_{L^{\infty}(D)} dy \bigg] ds \bigg{)}^{2H} \\ &\leq c(H) \bigg{(} \int_{\zeta}^{t} (t-s)^{-\frac{1}{4H}} \bigg[ \int_{D} 1\;dy \bigg] ds \bigg{)}^{2H} \\ &= c(H) \bigg{(} \int_{\zeta}^{t} (t-s)^{-\frac{1}{4H}}\, ds \bigg{)}^{2H} = c(H) (t-\zeta)^{\frac{4H-1}{2}}. \end{align}\] Gathering now all the above we finally obtain (?? ).  ◻

Proof. The right-hand side of (12 ) consists of three terms: \[A_{1}(x,t):=\int_{D}G(x,y,t) u_{0}(y)\, dy,\quad A_{2}(x,t;\omega):=\int_{0}^{t}\int_{D} G_{yy}(x,y,t-s) f\big{(}u(y,s)\big{)}\, dyds,\] \[A_{3}(x,t;\omega):=\sigma \int_{0}^{t} \int_{D} G(x,y,t-s) W_{H} (dy,ds).\] Since \(u_{0}\) is continuous, Lemma 2.1 of [4] guarantees the space-time continuity of the term \(A_{1}\). Similarly, Lemma 4.1 of [13] establishes the \(\mu\)-H\(\ddot{\text{o}}\)lder in \(t\) and \(\nu\)-H\(\ddot{\text{o}}\)lder in \(x\) almost sure continuity of the term \(A_{3}\), where \(\mu \in \big[ 0,H-\frac{1}{4} \big{)}\) and \(\nu\in[0,1]\) (our case corresponds to \(d=1\)). Lastly, pages 797-799 of [4] are dedicated to the study of \(A_{2}\). The almost sure continuity of this term can be shown by using a factorization method (applicable here since [13] guarantees \(u\in C([0,T];L^{p}(D))\) a.s. for some suitable constants \(p\geq4\)) to express it as \[\dfrac{\sin(\pi a)}{\pi}\mathcal{F}\big{(}\mathcal{K}(u)\big{)}(x,t)\] for all \((x,t)\in D\times[0,T]\), where \(a \in (0,1)\) is fixed and \[\begin{align} \mathcal{F}(v)(x,t)&:=\int_{0}^{t}\int_{D} G(x,z,t-s)(t-s)^{-a}v(z,s)\, dzds,\\ \mathcal{K}(v)(z,s)&:=\int_{0}^{s}\int_{D} G_{yy}(z,y,s-s')(s-s')^{a-1}f\big{(}v(y,s')\big{)}\, dyds', \end{align}\] and then proving that the operator \(\mathcal{K}\) maps \(L^{\infty}([0,T];L^{q}(D))\) into itself and \(\mathcal{F}\) is H\(\ddot{\text{o}}\)lder continuous if \(v\in L^{\infty}([0,T];L^{q}(D))\). ◻

Proof. We define \[u_{n,0}(x,t):=\int_{D} G(x,y,t) u_{0}(y)\, dy\] and, for any \(k\in\mathbb{N}\), we consider the following Picard iteration scheme \[\begin{align} \label{Picard32scheme} \begin{aligned} u_{n,k+1}(x,t) = u_{n,0}(x,t) &+ \int_{0}^{t}\int_{D} G_{yy}(x,y,t-s) f_{n}\big{(}u_{n,k}(y,s)\big{)}\, dyds \\ &+ \sigma\int_{0}^{t}\int_{D} G(x,y,t-s) W_{H}(dy,ds). \end{aligned} \end{align}\tag{22}\] Relation (22 ) yields for any \(k\in\mathbb{N}_{*}\) \[\begin{align} |u_{n,k+1}(x,t)-u_{n,k}(x,t)| &\leq \int_{0}^{t} \int_{D} |G_{yy}(x,y,t-s)|\, \big{|}f_{n}\big{(}u_{n,k}(y,s)\big{)}-f_{n}\big{(}u_{n,k-1}(s,y)\big{)}\big{|}\, dyds\\ &\leq c(n) \int_{0}^{t} \int_{D} |G_{yy}(x,y,t-s)|\, |u_{n,k}(y,s)-u_{n,k-1}(s,y)|\, dyds, \end{align}\] where for the second inequality we used the fact that \(f_{n}\) is Lipschitz. Taking now supremum for any \(x\in D\), then \(p\) powers for \(p>2\) and finally expectation, we get \[\begin{align} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) &\|_{L^{\infty}(D)}^{p} \Big{)} \leq \\ &\leq c(n,p)\, E \Bigg{(} \bigg{\|} \int_{0}^{t} \int_{D} |G_{yy}(\cdot,y,t-s)|\, |u_{n,k}(y,s) -u_{n,k-1}(s,y)|\, dyds \bigg{\|}_{L^{\infty}(D)}^{p} \Bigg{)}. \end{align}\] In the proof of Lemma 2 we explained that the continuity of \(u_{0}\) implies the space-time continuity of \(u_{n,0}\). Since \(f_{n}\) is Lipschitz, the iteration scheme and similar reasoning as in Lemma 2 imply the almost sure space-time continuity of \(u_{n,k}\) for any \(k\in\mathbb{N}_{*}\). In particular, for any \(k\in\mathbb{N}_{*}\), it holds almost surely \[u_{n,k}-u_{n,k-1}\in L^{1}\big{(}[0,T];L^{\infty}(D)\big{)}.\] Thus, we can invoke inequality (1.12) of [4] for \(v(y,s):=|u_{n,k}(y,s)-u_{n,k-1}(y,s)|\) to obtain \[E \Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq c(n,p)\, E \bigg{(} \bigg{(} \int_{0}^{t} (t-s)^{-\frac{1}{2}} \| u_{n,k}(\cdot,s)-u_{n,k-1}(\cdot,s) \|_{L^{\infty}(D)}\, ds \bigg{)}^{p} \bigg{)}.\] Applying H\(\ddot{\text{o}}\)lder’s inequality, we get \[\begin{align} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) &\|_{L^{\infty}(D)}^{p} \Big{)} \leq \\ &\leq c(n,p)\, E \bigg{(} \bigg{(} \int_{0}^{t} (t-s)^{-\frac{p}{2(p-1)}} ds \bigg{)}^{p-1} \int_{0}^{t} \| u_{n,k}(\cdot,s)-u_{n,k-1}(\cdot,s) \|_{L^{\infty}(D)}^{p}\, ds \bigg{)}. \end{align}\] The first integral is finite only if \(p>2\) which is in accordance with our assumption for \(p\). Thus, \[\begin{align} \label{recursive} \begin{aligned} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} &\leq c(n,p)\, E \bigg{(} t^{\frac{p-2}{2}} \int_{0}^{t} \|u_{n,k}(\cdot,s)-u_{n,k-1}(\cdot,s) \|_{L^{\infty}(D)}^{p}\, ds \bigg{)} \\ &\leq c(n,p,T) \int_{0}^{t} E \Big{(} \| u_{n,k}(\cdot,s)-u_{n,k-1}(\cdot,s) \|_{L^{\infty}(D)}^{p} \Big{)} ds. \end{aligned} \end{align}\tag{23}\]

We now establish the following auxiliary estimate, which will be needed below. In particular, for any \(p>2\), there exists a constant \(\mathcal{C}_{1}=\mathcal{C}_{1}(\sigma,f,H,n,p,T,u_{0})>0\) such that \[\label{sup32bound32of32dif4632of32first32232terms} \sup\limits_{t\in[0,T]}E \Big{(} \| u_{n,1}(\cdot,t)-u_{n,0}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)}\leq \mathcal{C}_{1}.\tag{24}\] Indeed, by (22 ) we have \[u_{n,1}(x,t) - u_{n,0}(x,t) = \int_{0}^{t}\int_{D} G_{yy}(x,y,t-s) f_{n}\big{(}u_{n,0}(y,s)\big{)}\, dyds + \sigma\int_{0}^{t}\int_{D} G(x,y,t-s) W_{H}(dy,ds)\] from which we obtain \[\begin{align} \sup\limits_{t\in[0,T]} E \Big{(} \| u_{n,1}(\cdot,t)-u_{n,0}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} &\leq c(p) \sup\limits_{t\in[0,T]} \bigg{\|} \int_{0}^{t} \int_{D} |G_{yy}(\cdot,y,t-s)|\, \big{|} f_{n}\big{(}u_{n,0}(y,s) \big{)} \big{|}\, dyds \bigg{\|}_{L^{\infty}(D)}^{p} \\ &\phantom{=}+ c(\sigma,p) \sup\limits_{t\in[0,T]} E \Bigg{(} \bigg{\|} \int_{0}^{t}\int_{D} G(\cdot,y,t-s) W_{H}(dy,ds) \bigg{\|}_{L^{\infty}(D)}^{p} \Bigg{)}. \end{align}\] Regarding the first term, since \(f_{n}\) is Lipschitz and \(u_{n,0}\) is space-time continuous, the composition \(f_{n}\circ u_{n,0}\) is also space-time continuous. Thus \(f_{n}\circ u_{n,0}\in C([0,T];L^{\infty}(D))\) so we can invoke once again inequality (1.12) of [4] for \(v(y,s):=|f_{n}\circ u_{n,0}(y,s)|\) to obtain \[\begin{align} \sup\limits_{t\in[0,T]} \bigg{\|} \int_{0}^{t}\int_{D} |G_{yy}(\cdot,y,t-s)|\, \big{|} f_{n}\big{(}u_{n,0}(y,s) \big{)} \big{|}\, &dyds \bigg{\|}_{L^{\infty}(D)}^{p} \leq \\ &\leq c \sup\limits_{t\in[0,T]} \bigg{(} \int_{0}^{t} (t-s)^{-\frac{1}{2}} \left \| f_{n}\big{(}u_{n,0}(\cdot,s) \big{)} \right \|_{L^{\infty}(D)} ds \bigg{)}^{p} \\ \overset{\text{H\ddot{\text{o}}lder}}&{\leq} c(p,T) \sup\limits_{t\in[0,T]} \int_{0}^{t} \big{\|} f_{n}\big{(}u_{n,0}(\cdot,s) \big{)} \big{\|}_{L^{\infty}(D)}^{p}\, ds \\ &\leq c(p,T) \left \| f_{n}\circ u_{n,0} \right \|_{C([0,T];L^{\infty}(D))}^{p}. \end{align}\] Regarding the second term, by (?? ) (for \(\zeta = 0\)), we deduce the existence of a constant \(c(H,p,T)>0\) such that \[\sup\limits_{t\in[0,T]} E \Bigg{(} \bigg{\|} \int_{0}^{t}\int_{D} G(\cdot,y,t-s) W_{H}(dy,ds) \bigg{\|}_{L^{\infty}(D)}^{p} \Bigg{)} \leq c(H,p,T).\] Gathering the above, we get the desired result \[\sup\limits_{t\in[0,T]} E \Big{(} \| u_{n,1}(\cdot,t)-u_{n,0}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq c(\sigma,H,p,T)\, \| f_{n}\circ u_{n,0} \|_{C([0,T];L^{\infty}(D))}^{p}=:\mathcal{C}_{1}.\] Now then, by repeatedly applying inequality (23 ) for the integrand on the right-hand side, we get \[\begin{align} E\Big{(} \| u_{n,k+1}&(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq \\[0.1cm] &\leq c(n,p,T)^{k} \int_{0}^{t} \int_{0}^{s_{k}} \int_{0}^{s_{k-1}} \dots \int_{0}^{s_{2}} E \Big{(} \| u_{n,1}(\cdot,s_{1})-u_{n,0}(\cdot,s_{1}) \|_{L^{\infty}(D)}^{p} \Big{)} ds_{1}\dots ds_{k-2}ds_{k-1}ds_{k} \\[0.1cm] &\leq c(n,p,T)^{k} \int_{0}^{t} \int_{0}^{s_{k}} \int_{0}^{s_{k-1}} \dots \int_{0}^{s_{2}} \sup_{t\in[0,T]}E \Big{(} \| u_{n,1}(\cdot,t)-u_{n,0}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} ds_{1}\dots ds_{k-2}ds_{k-1}ds_{k} \\[0.1cm] &= \sup_{t\in[0,T]}E \Big{(} \| u_{n,1}(\cdot,t)-u_{n,0}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)}\, c(n,p,T)^{k}\, \frac{t^{k}}{k!} \end{align}\] and thus, taking supremum for \(t\in[0,T]\) and using (24 ), we obtain \[\label{sup32bound32of32dif4632of32sequential32terms} \sup_{t \in [0,T]} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq \mathcal{C}_{1}\, \frac{c(n,p,T)^{k}}{k!}.\tag{25}\] An immediate consequence of (25 ) is that \[\label{A46W46321} \sum\limits_{k=0}^{+\infty} \sup_{t \in [0,T]} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq \mathcal{C}_{1} \sum\limits_{k=0}^{+\infty} \frac{c(n,p,T)^{k}}{k!} = \mathcal{C}_{1}\, e^{c(n,p,T)}=:\mathcal{C}_{2}\tag{26}\] which implies \[\sup_{t \in [0,T]} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \to0\quad \text{as}\;k\to+\infty\] and thus \[\begin{align} \| u_{n,k+1}-u_{n,k} \|_{L^{\infty}([0,T];L^{p}(\Omega;L^{\infty}(D)))} :&= \sup_{t \in [0,T]} \bigg[ \left ( E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \right )^{\frac{1}{p}} \bigg] \\ &\leq \bigg{(} \sup_{t \in [0,T]} E\Big{(} \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \bigg{)}^{\frac{1}{p}} \to0\quad \text{as}\;k\to+\infty. \end{align}\] For any \((x,t)\in D\times[0,T]\), we have \[\begin{align} \| u_{n,k+1}(x,t)-u_{n,k}(x,t) \|_{L^{p}(\Omega)} &\leq \| u_{n,k+1}(\cdot,t)-u_{n,k}(\cdot,t) \|_{L^{p}(\Omega;L^{\infty}(D))} \\ &\leq \| u_{n,k+1}-u_{n,k} \|_{L^{\infty}([0,T];L^{p}(\Omega;L^{\infty}(D)))}\to0\quad \text{as}\;k\to+\infty. \end{align}\] This implies that the sequence \(\{u_{n,k}\}_{k\in\mathbb{N}}\) is Cauchy in \(L^{p}(\Omega), L^{p}(\Omega;L^{\infty}(D))\) and \(L^{\infty}\big{(}[0,T];L^{p}(\Omega;L^{\infty}(D))\big{)}\).
Since all of them are Banach spaces, \(\{u_{n,k}\}_{k\in\mathbb{N}}\) converges to some unique \(u_{n}\in L^{p}(\Omega),\;\tilde{u}_{n}\in L^{p}(\Omega;L^{\infty}(D))\)
and \(\hat{u}_{n}\in L^{\infty}\big{(}[0,T];L^{p}(\Omega;L^{\infty}(D))\big{)}\) in their respective spaces. It holds however \[L^{\infty}\big{(}[0,T];L^{p}(\Omega;L^{\infty}(D))\big{)} \subset L^{p}(\Omega;L^{\infty}(D)) \subset L^{p}(\Omega)\] thus \(\tilde{u}_{n},\hat{u}_{n}\in L^{p}(\Omega)\) and so, due to the uniqueness of the limit, we have \(u_{n}=\tilde{u}_{n}=\hat{u}_{n}\) almost surely. Through the scheme (22 ), by a standard argument, where we take limits in the \(L^{p}(\Omega)\) norm, and use the fact that \(f_{n}\) is uniformly continuous (since it is Lipschitz), we conclude that \(u_{n}\) satisfies (21 ) almost surely.

Regarding now the uniqueness of the solution, if we suppose that there exists another solution \(w_{n}\) of (21 ), then by subtracting their respective equations, taking absolute value on both sides and using the Lipschitz property of \(f_{n}\), we get \[\big{|} u_{n}(x,t) - w_{n}(x,t) \big{|} \leq c(n) \int_{0}^{t}\int_{D} |G_{yy}(x,y,t-s)|\, |u_{n}(y,s)-w_{n}(s,y)|\, dyds.\] Taking now supremum for any \(x\in D\), then \(p\) powers for \(p>2\) and finally expectation, we get \[E \Big{(} \| u_{n}(\cdot,t) - w_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq c(n,p) E \Bigg{(} \bigg{\|} \int_{0}^{t}\int_{D} |G_{yy}(\cdot,y,t-s)|\, |u_{n}(y,s)-w_{n}(s,y)|\, dyds \bigg{\|}_{L^{\infty}(D)}^{p} \Bigg{)}.\] Following a similar procedure to the one for deriving (23 ), we obtain \[E \Big{(} \| u_{n}(\cdot,t) - w_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq c(n,p,T) \int_{0}^{t} E \Big{(} \| u_{n}(\cdot,s)-w_{n}(\cdot,s) \|_{L^{\infty}(D)}^{p} \Big{)} ds.\] Hence, by applying Gr\(\ddot{\text{o}}\)nwall’s lemma to the previous inequality, we get \[\begin{align} E \Big{(} \| u_{n}(\cdot,t) - w_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq 0,\;\forall t\in[0,T] &\implies \| u_{n}(\cdot,t)-w_{n}(\cdot,t) \|_{L^{p}(\Omega;L^{\infty}(D))} = 0,\;\forall t \in [0,T] \end{align}\] and thus, due to the relation between the norms of the spaces \(L^{p}(\Omega), L^{p}(\Omega;L^{\infty}(D))\), we obtain \[\| u_{n}(x,t)-w_{n}(x,t) \|_{L^{p}(\Omega)}=0,\;\forall (x,t)\in D\times[0,T].\] This yields \(u_{n}(x,t)=w_{n}(x,t)\) almost surely on \(\Omega\) and \(\Omega_{n}\) (since \(\Omega_{n}\subset\Omega\)) for any \((x,t)\in D\times[0,T]\), i.e., \[\mathbb{P}\big{(} \big{\{} \omega\in\Omega\;(\text{or}\;\Omega_{n})\, :\, u_{n}(x,t;\omega)=w_{n}(x,t;\omega) \big{\}} \big{)}=1,\;\forall (x,t)\in D\times[0,T],\] and so the processes \(u_{n},w_{n}\) are modifications of each other on \(\Omega\) and \(\Omega_{n}\). In addition, \(u_{n},w_{n}\) satisfy equation (12 ) almost surely on \(\Omega_{n}\) (since \(f_{n}(u_{n})=f(u_{n})\) and \(f_{n}(w_{n})=f(w_{n})\) on \(\Omega_{n}\)) and by Lemma 2 we know that the solution of (12 ) is almost surely continuous in space-time. So, \(u_{n},w_{n}\) are almost surely continuous on \(\Omega_{n}\) and thus they are indistinguishable on \(\Omega_{n}\), i.e., \[\mathbb{P}\big{(} \big{\{} \omega\in\Omega_{n}\, :\, u_{n}(x,t;\omega)=w_{n}(x,t;\omega) ,\;\forall (x,t)\in D\times[0,T] \big{\}} \big{)}=1.\] This proves the uniqueness of solution of equation (21 ) with uniquely defined paths almost surely on \(\Omega_{n}\).

Lastly, the convergence \(u_{n,k}\to u_{n}\) in \(L^{\infty}\big{(}[0,T];L^{p}(\Omega;L^{\infty}(D))\big{)}\) for \(p>2\), established previously, implies the existence of a constant \(c = c(n,p,T)>0\) such that \[\label{bound32of32dif32of32u95nk4432u95n} \| u_{n,k}-u_{n} \|_{L^{\infty}([0,T];L^{p}(\Omega;L^{\infty}(D)))} \leq c,\;\forall k\in\mathbb{N},\tag{27}\] and therefore also \[\label{bound32of32dif4632of32u95nk40x44t41-u95n40x44t41} \| u_{n,k}(\cdot,t)-u_{n}(\cdot,t) \|_{L^{p}(\Omega;L^{\infty}(D))} \leq c,\;\forall t\in [0,T],\, \forall k\in\mathbb{N}.\tag{28}\] Thus, for \(p>2\) and any \(t \in [0,T]\), we have \[\begin{align} \label{bound32when32p622} E \Big{(} \| u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} &\leq E \Big{(} \| u_{n,k}(\cdot,t) - u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} + \sum\limits_{i=k}^{+\infty} E \Big{(} \| u_{n,i+1}(\cdot,t) - u_{n,i}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \notag\\ &\leq E \Big{(} \| u_{n,k}(\cdot,t) - u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} + \sum\limits_{i=0}^{+\infty} \sup_{t\in[0,T]} E \Big{(} \| u_{n,i+1}(\cdot,t) - u_{n,i}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \\[-0.1cm] \overset{\substack{(\ref{A46W46321})\\(\ref{bound32of32dif4632of32u95nk40x44t41-u95n40x44t41})}}&{\leq} c+\mathcal{C}_{2}=:\mathcal{C}_{3}, \notag \end{align}\tag{29}\] while for \(p=2\), using H\(\ddot{\text{o}}\)lder’s inequality for the expectation, we get \[E \Big{(} \| u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{2} \Big{)} \leq \left ( E \Big{(} \| u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{4} \Big{)} \right )^{\frac{1}{2}} \overset{(\ref{bound32when32p622})}{\leq} \Big{(} \mathcal{C}_{3}\big{|}_{p=4} \Big{)}^{\frac{1}{2}}=:\mathcal{C}_{4}.\] Hence, for any \(p\geq 2\), we have \[E \Big{(} \| u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq \mathcal{C}:= \begin{cases} \mathcal{C}_{3}, & \text{if}\;p>2,\\[0.1cm] \mathcal{C}_{4}, & \text{if}\;p=2, \end{cases}\;\forall t \in [0,T] \implies \sup_{t\in[0,T]} E \Big{(} \| u_{n}(\cdot,t) \|_{L^{\infty}(D)}^{p} \Big{)} \leq \mathcal{C}\] which proves (?? ) and finally concludes the proof of the lemma. ◻

Proof. Consider the sequence \(\{u_{n,k}\}_{k\in\mathbb{N}}\) defined in the proof of Lemma 3. We will use induction and the Picard iteration scheme to prove that \[\{u_{n,k}(x,t)\}_{k\in\mathbb{N}}\subseteq\mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T].\]

The term \(u_{n,0}(x,t)\) is a deterministic function, thus \(u_{n,0}(x,t)\in\mathbb{D}^{1,2}\) for every \((x,t)\in D\times[0,T]\) with \[\label{mal4632der4632of32u95n0} D_{y,s} u_{n,0}(x,t) = 0,\;\forall (x,t),(y,s) \in D\times[0,T].\tag{31}\]

Fix now an arbitrary \(k\in\mathbb{N}_{*}\). Our induction hypothesis will consist of two parts. Firstly \[u_{n,1}(x,t),\dotsc,u_{n,k}(x,t)\in\mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T]\] and secondly, we assume that there exists a constant \(C=C(\sigma,H,n,T)>0\) (to be specified later) such that \[\label{induction32hyp46322} \sup_{t\in[0,T]}\sup_{i\in\mathbb{N}\cap[0,k]} E \bigg{(} \int_{0}^{T} \int_{D} \left \| D_{y,s} u_{n,i}(\cdot,t) \right \|_{L^{\infty}(D)}^{2} dyds \bigg{)} \leq C.\tag{32}\] We note that the integral over \([0,T]\) coincides with the integral over \([0,t]\), since the Malliavin derivative involved is zero for any \(s>t\). In order to complete the induction we have to prove that \[\begin{cases} u_{n,k+1}(x,t)\in\mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T], \\[0.1cm] \displaystyle\sup_{t\in[0,T]}\sup_{i\in\mathbb{N}\cap[0,k+1]} E \bigg{(} \int_{0}^{T} \int_{D} \left \| D_{y,s} u_{n,i}(\cdot,t) \right \|_{L^{\infty}(D)}^{2} dyds \bigg{)} \leq C. \end{cases}\] To prove the first requirement, we will use the expression (22 ) of \(u_{n,k+1}\) and prove that each term on the right-hand side belongs in \(\mathbb{D}^{1,2}\) for any \((x,t)\in D\times[0,T]\). Since we have already explained that \(u_{n,0}(x,t)\in\mathbb{D}^{1,2}\) for any \((x,t)\in D\times[0,T]\), all that is left is to show that the second and third terms meet our needs.

Now then, since \(f_{n}\) is Lipschitz, with Lipschitz constant \(c(n)>0\), and \(u_{n,k}(x,t)\in\mathbb{D}^{1,2}\) for every \((x,t) \in D\times[0,T]\) (by our induction hypothesis), Proposition 1.2.4 of [11] asserts that \(f_{n} \big{(} u_{n,k}(x,t) \big{)} \in \mathbb{D}^{1,2}\) for every \((x,t) \in D\times[0,T]\) and there exists a stochastic process \(\mathcal{G}_{n,k}(x,t)\) such that \[\label{mal4632der4632of32composition} D_{y,s} \left ( f_{n} \big{(} u_{n,k}(x,t) \big{)} \right ) = \mathcal{G}_{n,k}(x,t) D_{y,s}u_{n,k}(x,t)\;\text{a.s.},\;\forall (x,t),(y,s)\in D\times[0,T]\tag{33}\] with \[\label{bound32of32st46p4632G} \left | \mathcal{G}_{n,k}(x,t) \right | \leq c(n)\;\text{a.s.},\;\forall (x,t)\in D\times[0,T]\implies \sup_{k\in\mathbb{N}}\sup_{(x,t)\in D\times[0,T]} \left | \mathcal{G}_{n,k}(x,t) \right | \leq c(n)\;\text{a.s.}\;.\tag{34}\] In addition, we have \(f_{n}(u_{n,k})\in\mathbb{L}^{1,2}\) since \[\begin{align} \|f_{n}(u_{n,k})\|^{2}_{\mathbb{L}^{1,2}} &= \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + \left \| D_{y,s} \left ( f_{n} \big{(} u_{n,k}(x,t) \big{)} \right ) \right \|^{2}_{L^{2}(\Omega\times (D\times[0,T])^{2})} \\ \overset{(\ref{mal4632der4632of32composition})}&{=} \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + \left \| G_{n,k}(x,t) D_{y,s}u_{n,k}(x,t) \right \|^{2}_{L^{2}(\Omega\times (D\times[0,T])^{2})} \\ &= \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + E \bigg{(} \int_{0}^{T} \int_{D} \int_{0}^{T} \int_{D} |G_{n,k}(x,t)|^{2}\, |D_{y,s}u_{n,k}(x,t)|^{2}\, dydsdxdt \bigg{)} \\ \overset{(\ref{bound32of32st46p4632G})}&{\leq} \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + c(n)\;E \bigg{(} \int_{0}^{T} \int_{D} \int_{0}^{T} \int_{D} \left | D_{y,s}u_{n,k}(x,t) \right |^{2} dydsdxdt \bigg{)} \\ \overset{\text{Fubini}}&{=} \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + c(n) \int_{0}^{T} \int_{D} E \bigg{(} \int_{0}^{T} \int_{D} \left | D_{y,s}u_{n,k}(x,t) \right |^{2} dyds \bigg{)} dxdt \\ &\leq \left \| f_{n} \big{(} u_{n,k}(x,t) \big{)} \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + c(n) \int_{0}^{T} \int_{D}\;\sup_{t\in[0,T]} E \bigg{(} \int_{0}^{T} \int_{D} \left \| D_{y,s} u_{n,k}(\cdot,t) \right \|_{L^{\infty}(D)}^{2} dyds \bigg{)} dxdt \\ \overset{\substack{(\ref{R4632A3})\\(\ref{induction32hyp46322})}}&{\leq} \mathcal{C}+ C\;(<+\infty). \end{align}\] Lastly, using (15 ) \(iii)\) we can see that \(G_{\diamond\diamond}(x,\diamond,t-\bullet)\mathrm{\boldsymbol{1}}_{[0,t]}(\bullet)\in L^{2}(D\times[0,T])\) for any \((x,t)\in D\times[0,T]\). So, by Proposition 3.4.3 of [18], we obtain \[\big{\langle} f_{n} \big{(} u_{n,k}(\diamond,\bullet) \big{)}, G_{\diamond\diamond}(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{\rangle}_{L^{2}(D\times[0,T])} \in \mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T]\] and \[\begin{align} \label{mal4632der4632of32int46321} \begin{aligned} D_{y,s} \bigg{(} \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) f_{n}\big{(}u_{n,k}(z,\tau)\big{)}\, dzd\tau \bigg{)} &= \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) D_{y,s} \left ( f_{n} \big{(} u_{n,k}(z,\tau) \big{)} \right ) dzd\tau \\ \overset{(\ref{mal4632der4632of32composition})}&{=} \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n,k}(z,\tau) D_{y,s}u_{n,k}(z,\tau)\, dzd\tau \end{aligned} \end{align}\tag{35}\] almost surely, for every \((x,t),(y,s)\in D\times[0,T]\).

Regarding now the last term of the right-hand side of (22 ), by (14 ) we know that \[\sigma \int_{0}^{t}\int_{D} G(x,z,t-\tau) W_{H}(dz,d\tau) = \sigma \int_{0}^{T} \int_{D} \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (z,\tau) W(dz,d\tau).\] The last integral is of the form (3 ) for \(n=1,\;f(x):=x,\;h(z,\tau) := \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (z,\tau)\) and
the isonormal Gaussian process \(\{\hat{W}(h)\}_{L^{2}(D\times[0,T])}\) given by (30 ), thus it belongs to \(\mathcal{S}\subseteq\mathbb{D}^{1,2}\) with \[\label{mal4632der4632of32int46322} D_{y,s} \bigg{(} \int_{0}^{T} \int_{D} \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (z,\tau) W(dz,d\tau) \bigg{)} = \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s)\tag{36}\] almost surely, for every \((x,t),(y,s)\in D\times[0,T]\).

Now that we have shown that every term on the right-hand side belongs to \(\mathbb{D}^{1,2}\) for any \((x,t)\in D\times[0,T]\), the linearity of the space \(\mathbb{D}^{1,2}\) yields the desired result.

To prove the second requirement, we apply the Malliavin derivative to both sides of (22 ) and use the linearity of the operator to get \[\begin{align} D_{y,s}u_{n,k+1}(x,t) = D_{y,s}u_{n,0}(x,t) &+ D_{y,s} \bigg{(} \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) f_{n}\big{(}u_{n,k}(z,\tau)\big{)} \, dzd\tau \bigg{)} \\ &+ \sigma D_{y,s} \bigg{(} \int_{0}^{t}\int_{D} G(x,z,t-\tau) W_{H}(dz,d\tau) \bigg{)}. \end{align}\] Using relations ([formula32of32fr4632stoch4632int46]), (31 ), (35 ) and (36 ), we obtain \[\begin{align} \label{mal4632der4632of32recursive} \begin{aligned} D_{y,s}u_{n,k+1}(x,t) = \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n,k}(z,\tau) D_{y,s}u_{n,k}(z,&\tau)\, dzd\tau\;+ \\[0.1cm] &+\sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \end{aligned} \end{align}\tag{37}\] almost surely, for every \((x,t),(y,s)\in D\times[0,T]\).
We take now absolute value and use triangle inequality to get \[\begin{align} |D_{y,s}u_{n,k+1}(x,t)| &\leq \int_{s}^{t} \int_{D} \left | G_{zz}(x,z,t-\tau) \right | \left | \mathcal{G}_{n,k}(z,\tau) \right | \left | D_{y,s}u_{n,k}(z,\tau) \right | dzd\tau\;+ \\ &+|\sigma| \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \right | \\ \overset{(\ref{bound32of32st46p4632G})}&{\leq} c(n) \int_{s}^{t} \int_{D} \left | G_{zz}(x,z,t-\tau) \right | \left | D_{y,s}u_{n,k}(z,\tau) \right | dzd\tau\;+ \\ &+|\sigma| \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \right |. \end{align}\] We then take supremum for any \(x\in D\), raise to the second power, integrate for \((y,s)\in D\times[0,t]\) and finally take expectation to obtain \[\begin{align} \label{R46R461} \begin{aligned} E \bigg{(} \int_{0}^{t} \int_{D} \| D_{y,s}&u_{n,k+1}(\cdot,t) \|^{2}_{L^{\infty}(D)}\, dyds \bigg{)} \leq \\ &\leq c(n)\, E \Bigg{(} \int_{0}^{t} \int_{D} \bigg{\|} \int_{s}^{t} \int_{D} \left | G_{zz}(\cdot,z,t-\tau) \right | \left | D_{y,s}u_{n,k}(z,\tau) \right | dzd\tau \bigg{\|}^{2}_{L^{\infty}(D)} dyds \Bigg{)} \\ &\phantom{\leq.}+ c(\sigma) \int_{0}^{t} \int_{D} \big{\|} \big[ K^{*}_{H} \big{(} G(\cdot,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{\|}^{2}_{L^{\infty}(D)}\, dyds. \end{aligned} \end{align}\tag{38}\]

Regarding the first term on the right-hand side of (38 ), we can invoke relations (15 ) \(iii)\), (16 ) (or, alternatively, inequality (1.12) of [4] for \(v(z,\tau):=|D_{y,s}u_{n,k}(z,\tau)|\)) to get \[\begin{align} \label{R46R462} E \bigg{(} \int_{0}^{t} \int_{D} \bigg{\|} \int_{s}^{t} &\int_{D} \left | G_{zz}(\cdot,z,t-\tau) \right | \left | D_{y,s}u_{n,k}(z,\tau) \right | dzd\tau \bigg{\|}^{2}_{L^{\infty}(D)} dyds \bigg{)} \leq \notag\\ &\leq c\, E \bigg{(} \int_{0}^{t} \int_{D} \bigg{(} \int_{s}^{t} (t-\tau)^{-\frac{1}{2}} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|_{L^{\infty}(D)} d\tau \bigg{)}^{2} dyds\bigg{)} \notag\\ &= c\, E \bigg{(} \int_{0}^{t} \int_{D} \bigg{(} \int_{s}^{t} (t-\tau)^{-\frac{3}{8}}(t-\tau)^{-\frac{1}{8}} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|_{L^{\infty}(D)} d\tau \bigg{)}^{2} dyds\bigg{)} \notag\\ \overset{\text{H\ddot{\text{o}}lder}}&{\leq} c\, E \bigg{(} \int_{0}^{t} \int_{D} \bigg{(} \int_{s}^{t} (t-\tau)^{-\frac{3}{4}} d\tau \bigg{)} \bigg{(} \int_{s}^{t} (t-\tau)^{-\frac{1}{4}} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|_{L^{\infty}(D)}^{2} d\tau \bigg{)} dyds\bigg{)} \\ &\leq c(T)\, E \bigg{(} \int_{0}^{t} \int_{D} \int_{s}^{t} (t-\tau)^{-\frac{1}{4}} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|_{L^{\infty}(D)}^{2} d\tau dyds\bigg{)} \notag\\ &= c(T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} E \bigg{(} \int_{0}^{t} \int_{D} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} d\tau \notag\\ &= c(T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} E \bigg{(} \int_{0}^{\tau} \int_{D} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} d\tau \notag, \end{align}\tag{39}\] where, for the second-to-last equality, we used Fubini’s theorem (which is permitted by (32 )) and, for the last equality, we used the fact that the Malliavin derivative is zero when \(s>\tau\).

Regarding the second term on the right-hand side of (38 ), by (?? ) (for \(\zeta=0\)), we know that there exists a constant \(c(H)>0\) such that \[\int_{0}^{t} \int_{D} \big{\|} \big[ K^{*}_{H} \big{(} G(\cdot,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{\|}^{2}_{L^{\infty}(D)}\, dyds \leq c(H)\, t^{\frac{4H-1}{2}},\;\forall t\in[0,T].\]

By (38 ), (39 ) and the above inequality, we obtain for any \(t\in[0,T]\) \[\begin{align} \label{R46R463} \begin{aligned} E \bigg{(} \int_{0}^{t} \int_{D} \| D_{y,s}&u_{n,k+1}(\cdot,t) \|^{2}_{L^{\infty}(D)}\, dyds \bigg{)} \leq \\ &\leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} E \bigg{(} \int_{0}^{\tau} \int_{D} \left \| D_{y,s}u_{n,k}(\cdot,\tau) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} d\tau + c(\sigma,H)\, t^{\frac{4H-1}{2}}. \end{aligned} \end{align}\tag{40}\] Taking now supremum for \(i\in\mathbb{N}\cap[0,k]\) yields \[\begin{align} \begin{aligned} \sup_{i\in\mathbb{N}\cap[0,k]} E \bigg{(} &\int_{0}^{t} \int_{D} \left \| D_{y,s}u_{n,i+1}(\cdot,t) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} \leq \\ &\leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{i\in\mathbb{N}\cap[0,k]} E \bigg{(} \int_{0}^{\tau} \int_{D} \left \| D_{y,s}u_{n,i}(\cdot,\tau) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} d\tau + c(\sigma,H)\, t^{\frac{4H-1}{2}} \\ &\leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{i\in\mathbb{N}\cap[0,k]} E \bigg{(} \int_{0}^{\tau} \int_{D} \left \| D_{y,s}u_{n,i+1}(\cdot,\tau) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)} d\tau + c(\sigma,H)\, t^{\frac{4H-1}{2}}. \end{aligned} \end{align}\] If we define \[A_{n,k+1}(t) := \sup_{i\in\mathbb{N}\cap[0,k+1]} E \bigg{(} \int_{0}^{t} \int_{D} \left \| D_{y,s}u_{n,i}(\cdot,t) \right \|^{2}_{L^{\infty}(D)} dyds \bigg{)},\;t\in[0,T]\] then the above relation (along with (31 )) gives \[\label{Gronwall32for32mal4632der46} A_{n,k+1}(t) \leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} A_{n,k+1}(\tau)\, d\tau + c(\sigma,H)\, t^{\frac{4H-1}{2}},\;\forall t\in[0,T].\tag{41}\] Thus, by the singular Gr\(\ddot{\text{o}}\)nwall inequality (cf. [19], Lemma 7.7.1), we get \[\begin{align} A_{n,k+1}(t) &\leq c(\sigma,H)\, \Gamma \Big{(} \frac{4H+1}{2} \Big{)}\, t^{\frac{4H-1}{2}}\, \text{E}_{\frac{3}{4}, \frac{4H+1}{2}} \bigg{(} c(n,T)\, \Gamma \Big{(} \frac{3}{4} \Big{)}\, t^{\frac{3}{4}} \bigg{)},\;\forall t\in[0,T] \end{align}\] where \(\Gamma\) is the gamma function and E is the two-parameter Mittag-Leffler function \(\text{E}_{a,b}(z) := \sum\limits_{n = 0}^{+\infty} \frac{z^n}{\Gamma(an + b)}\).
From the known exponential upper bound of the Mittag-Leffler functions (cf. [20], p. 71), for any \(\tilde{c} > 1\), there exists \(c > 0\) such that \[\text{E}_{\frac{3}{4}, \frac{4H+1}{2}} \bigg{(} c(n,T)\, \Gamma \Big{(} \frac{3}{4} \Big{)}\, t^{\frac{3}{4}} \bigg{)} \leq c\, \exp \Bigg[ \tilde{c}\, \bigg{(} c(n,T)\, \Gamma \Big{(} \frac{3}{4} \Big{)}\, t^{\frac{3}{4}} \bigg{)}^{\frac{4}{3}} \Bigg] = c \exp \big{(} c(n,T)\, t \big{)}.\] Gathering the above we obtain \[A_{n,k+1}(t) \leq c(\sigma,H)\, t^{\frac{4H-1}{2}}\, \exp \big{(} c(n,T)\, t \big{)},\;\forall t \in [0,T] \implies \sup_{t\in[0,T]} A_{n,k+1}(t) \leq c(\sigma,H)\, T^{\frac{4H-1}{2}} \exp \big{(} c(n,T)\, T \big{)}\] which finally establishes the second requirement of our induction for \[C := c(\sigma,H)\, T^{\frac{4H-1}{2}} \exp \big{(} c(n,T)\, T \big{)}.\]

Finally, as a consequence of relation (32 ), for every \((x,t)\in D\times[0,T]\), it holds \[\label{sup32bound32of32mal4632der4632of32u95nk} \sup_{k\in\mathbb{N}} \left \| D_{\diamond,\bullet}u_{n,k}(x,t) \right \|_{L^{2}(\Omega;L^{2}(D\times[0,T]))} = \sup_{k\in\mathbb{N}} \left \| D_{\diamond,\bullet}u_{n,k}(x,t) \right \|_{L^{2}(\Omega\times D\times[0,T])}\leq C.\tag{42}\] Since, for any \((x,t)\in D\times[0,T]\), we have proven that \(\{u_{n,k}(x,t)\}_{k\in\mathbb{N}}\subseteq\mathbb{D}^{1,2}\) and we already know (from Lemma 3) that \(u_{n,k}(x,t)\to u_{n}(x,t)\) in \(L^{p}(\Omega)\) for \(p>2\) (which implies convergence in \(L^{2}(\Omega)\) as well), Lemma 1.2.3 of [11] now asserts that \(\{u_{n}(x,t)\}_{n\in\mathbb{N}_{*}}\subseteq\mathbb{D}^{1,2}\) for every \((x,t) \in D\times[0,T]\) and \[\label{weak32conv4632of32mal4632der46} D_{\diamond,\bullet}u_{n,k}(x,t)\rightharpoonup D_{\diamond,\bullet}u_{n}(x,t)\quad \text{in}\;L^{2} \left ( \Omega; L^{2} (D\times[0,T]) \right )\quad \text{as}\;k\to+\infty.\tag{43}\]

What is left to do for 1) is to identify which stochastic partial differential equation the Malliavin derivative of \(u_{n}(x,t)\) satisfies, and prove that it does so uniquely. By Corollary 1.2.1 of [11], it follows immediately that \(D_{y,s}u_{n}(x,t)=0\) almost surely if \((y,s)\in D\times(t,T]\), so the focus will be on proving (?? ).

Since \(f_{n}\) is Lipschitz (with Lipschitz constant \(c(n)>0\)) and \(u_{n}(x,t)\in\mathbb{D}^{1,2}\) for any \((x,t)\in D\times[0,T]\), Proposition 1.2.4 of [11] asserts that \(f_{n} \big{(} u_{n}(x,t) \big{)}\in\mathbb{D}^{1,2}\) for any \((x,t)\in D\times[0,T]\) and there exists a stochastic process \(\mathcal{G}_{n}(x,t)\) such that \[\label{mal4632der4632of32composition322} D_{y,s} \big{(} f_{n} \big{(} u_{n}(x,t) \big{)} \big{)} = \mathcal{G}_{n}(x,t) D_{y,s}u_{n}(x,t)\;\text{a.s.},\;\forall (x,t),(y,s)\in D\times[0,T]\tag{44}\] with \[\label{bound32of32st46p4632G322} \left | \mathcal{G}_{n}(x,t) \right | \leq c(n)\;\text{a.s.},\;\forall (x,t)\in D\times[0,T]\implies \sup_{(x,t)\in D\times[0,T]} \left | \mathcal{G}_{n}(x,t) \right | \leq c(n)\;\text{a.s.}\;.\tag{45}\] Furthermore, by the weak convergence stated in (43 ), we get \[\begin{align} \begin{aligned} E \bigg{(} \int_{0}^{T} \int_{D} \left | D_{y,s} u_{n}(x,t) \right |^{2} dyds \bigg{)} &= \left \| D_{\diamond,\bullet}u_{n}(x,t) \right \|^{2}_{L^{2}(\Omega;L^{2}(D\times[0,T]))} \\[0.1cm] &\leq \liminf\limits_{k\to+\infty} \left ( \left \| D_{\diamond,\bullet}u_{n,k}(x,t) \right \|^{2}_{L^{2}(\Omega;L^{2}(D\times[0,T]))} \right ) \\[0.1cm] &= \lim\limits_{k\to+\infty} \left ( \inf_{m\geq k} \left \| D_{\diamond,\bullet}u_{n,m}(x,t) \right \|^{2}_{L^{2}(\Omega;L^{2}(D\times[0,T]))} \right ) \\[0.1cm] &\leq \lim\limits_{k\to+\infty} \left ( \sup_{m\geq k} \left \| D_{\diamond,\bullet}u_{n,m}(x,t) \right \|^{2}_{L^{2}(\Omega;L^{2}(D\times[0,T]))} \right ) \\[0.1cm] &\leq \sup_{k\in\mathbb{N}} \left \| D_{\diamond,\bullet}u_{n,k}(x,t) \right \|^{2}_{L^{2}(\Omega\times D\times[0,T])} \overset{(\ref{sup32bound32of32mal4632der4632of32u95nk})}{\leq} C \end{aligned} \end{align}\] for every \((x,t) \in D\times[0,T]\) and thus \[\label{R46R464} \sup_{(x,t)\in D\times[0,T]} E \bigg{(} \int_{0}^{T} \int_{D} \left | D_{y,s} u_{n}(x,t) \right |^{2} dyds \bigg{)} \leq C.\tag{46}\] Using now (46 ) and similar argumentation to that used for deriving (35 ), we conclude that \[\big{\langle} f_{n} \big{(} u_{n}(\diamond,\bullet) \big{)}, G_{\diamond\diamond}(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{\rangle}_{L^{2}(D\times[0,T])} \in \mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T]\] and \[\begin{align} \label{R46R465} \begin{aligned} D_{y,s} \bigg{(} \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) f_{n}\big{(}u_{n}(z,\tau)\big{)}\, dz&d\tau \bigg{)} = \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) D_{y,s} \big{(} f_{n} \big{(} u_{n}(z,\tau) \big{)} \big{)}\, dzd\tau \\ &\overset{(\ref{mal4632der4632of32composition322})}{=} \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau \end{aligned} \end{align}\tag{47}\] almost surely, for every \((x,t),(y,s)\in D\times[0,T]\).
Lastly, we have already explained that the last term on the right-hand side of (21 ) belongs in \(\mathbb{D}^{1,2}\) and we have calculated its Malliavin derivative in (36 ). Thus, applying Malliavin derivative to both sides of (21 ) and using (31 ), (36 ) and (47 ), we finally obtain (?? ).

Now that we have shown that the Malliavin derivative of \(u_{n}\) satisfies (?? ), to complete 1), we have to justify that it does so uniquely. To demonstrate this, all we have to do is prove the uniqueness of the solution for the following stochastic partial differential equation: \[\begin{align} \label{equation32of32V} \begin{aligned} V(x,y,t,s\, ;\, n) = \int_{0}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) V(z,y,\tau,s\, &;\, n)\, dzd\tau\;+ \\ &+\sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s). \end{aligned} \end{align}\tag{48}\] a.s. for \((x,t) \in D \times [0,T]\) and \((y,s) \in D \times [0,t]\) while \(V(x,y,t,s\, ;\, n) = 0\) when \((y,s) \in D \times (t,T]\), with \[\sup_{(x,t)\in D\times[0,T]} E \bigg{(} \int_{0}^{T} \int_{D} \left | V(x,y,t,s\, ;\, n) \right |^{2} dyds \bigg{)} < +\infty.\] Suppose there exist two solutions \(V,U\) of (48 ). By subtracting the equations they respectively satisfy and taking absolute value on both sides, we get \[\begin{align} |V(x,y,t,s\, ;\, n) - U(x,y,t,s\, ;\, n)| &= \bigg{|} \int_{0}^{t}\int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) (V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) )\, dzd\tau \bigg{|} \\ \overset{(\ref{bound32of32st46p4632G322})}&{\leq} c(n) \int_{0}^{t}\int_{D} \left | G_{zz}(x,z,t-\tau) \right | \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right | dzd\tau. \end{align}\] We then raise to the second power, integrate for \((y,s)\in D\times[0,t]\), take expectation and finally take supremum for any \(x\in D\) to obtain \[\begin{align} \label{R46R463213} \begin{aligned} \sup_{x\in D} E &\bigg{(} \int_{0}^{t} \int_{D} \left | V(x,y,t,s\, ;\, n) - U(x,y,t,s\, ;\, n) \right |^{2} dyds \bigg{)} \leq \\ &\leq c(n)\, \sup_{x \in D} E \bigg{(} \int_{0}^{t} \int_{D} \left ( \int_{0}^{t} \int_{D} \left | G_{zz}(x,z,t-\tau) \right | \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right | dzd\tau \right )^{2} dyds \bigg{)} \\ &\leq c(n)\, \sup_{x \in D} \left ( \int_{0}^{t} \int_{D} |G_{zz}(x,z,t-\tau)| \left ( E \left ( \int_{0}^{t} \int_{D} \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right |^{2} dyds \right ) \right )^{\frac{1}{2}} dzd\tau \right )^{2} \\ &\leq c(n)\, \sup_{x \in D} \left ( \int_{0}^{t} \left [ \int_{D} |G_{zz}(x,z,t-\tau)| dz \right ] \left ( \sup_{z \in D} E \left ( \int_{0}^{t} \int_{D} \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right |^{2} dyds \right ) \right )^{\frac{1}{2}} d\tau \right )^{2} \\ &\leq c(n) \left ( \int_{0}^{t} (t-\tau)^{-\frac{1}{2}} \left ( \sup_{z \in D} E \left ( \int_{0}^{t} \int_{D} \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right |^{2} dyds \right ) \right )^{\frac{1}{2}} d\tau \right )^{2} \\ &\leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{z \in D} E \left ( \int_{0}^{\tau} \int_{D} \left | V(z,y,\tau,s\, ;\, n) - U(z,y,\tau,s\, ;\, n) \right |^{2} dyds \right ) d\tau \end{aligned} \end{align}\tag{49}\] where for the second inequality we used Minkowski’s integral inequality and for the remaining calculations we followed the same approach as in (39 ).
If we define now \[B_{n}(t) := \sup_{x \in D} E \bigg{(} \int_{0}^{t} \int_{D} |V(x,y,t,s\, ;\, n) - U(x,y,t,s\, ;\, n)|^{2}\, dyds \bigg{)},\;t\in[0,T],\] then (49 ) is written equivalently as \[B_{n}(t) \leq c(n,T) \int_{0}^{t} (t-\tau)^{-\frac{1}{4}} B_{n}(\tau)\, d\tau,\;\forall t\in[0,T].\] Thus, by the singular Gr\(\ddot{\text{o}}\)nwall inequality and the non-negativity of \(B_{n}\), we obtain \(B_{n}(t)=0\) for all \(t\in[0,T]\) which in turn yields the almost sure equality of \(V,U\) for all \((x,t),(y,s)\in D\times[0,T]\).

To complete the lemma, we need to prove 2), i.e., \(\{u_{n}\}_{n\in\mathbb{N}_{*}} \subset \mathbb{L}^{1,2}\). We have \[\begin{align} \|u_{n}\|^{2}_{\mathbb{L}^{1,2}} &= \left \| u_{n}(x,t) \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + \left \| D_{y,s}u_{n}(x,t) \right \|^{2}_{L^{2}(\Omega\times (D\times[0,T])^{2})} \\ &= \left \| u_{n}(x,t) \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + E \bigg{(} \int_{0}^{T} \int_{D} \int_{0}^{T} \int_{D} \left | D_{y,s}u_{n}(x,t) \right |^{2} dydsdxdt \bigg{)} \\ \overset{\text{Fubini}}&{=} \left \| u_{n}(x,t) \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + \int_{0}^{T} \int_{D} E \bigg{(} \int_{0}^{T} \int_{D} \left | D_{y,s}u_{n}(x,t) \right |^{2} dyds \bigg{)} dxdt \\ &\leq \left \| u_{n}(x,t) \right \|_{L^{2}(\Omega\times D\times[0,T])}^{2} + \int_{0}^{T} \int_{D}\;\sup_{(x,t)\in D\times[0,T]} E \bigg{(} \int_{0}^{T} \int_{D} \left | D_{y,s} u_{n}(x,t) \right |^{2} dyds \bigg{)} dxdt \overset{\substack{(\ref{important32bound321})\\(\ref{R46R464})}}{\leq} \mathcal{C}+ C, \end{align}\] which directly confirms our claim and concludes the proof of the lemma. ◻

Proof. We start from equation (?? ), i.e., \[D_{y,s}u_{n}(x,t) = \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau\;+ \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s)\] which holds almost surely for any \((x,t)\in D\times[0,T]\) and any \((y,s)\in D\times[0,t]\). We take absolute value on both sides and use triangle inequality to get \[\begin{align} \big{|} D_{y,s}u_{n}(x,t) \big{|} &\leq \int_{s}^{t} \int_{D} |G_{zz}(x,z,t-\tau)|\, |\mathcal{G}_{n}(z,\tau)|\, |D_{y,s}u_{n}(z,\tau)|\, dzd\tau\;+ \\ &+|\sigma| \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \right | \\ \overset{(\ref{bound32of32st46p4632G322})}&{\leq} c(n) \int_{s}^{t} \int_{D} |G_{zz}(x,z,t-\tau)|\, |D_{y,s}u_{n}(z,\tau)|\, dzd\tau\;+ \\ &+|\sigma| \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \right |. \end{align}\] We then raise to the second power, integrate for \((y,s)\in D\times[\zeta,t]\), where \(0\leq\zeta\leq t\), take expectation and finally take supremum for any \(x\in D\) to obtain \[\begin{align} \label{R46R467} \begin{aligned} \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} | D_{y,s}&u_{n}(x,t) |^{2}\, dyds \bigg{)} \leq \\ &\leq c(n)\, \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} \bigg{(} \int_{s}^{t} \int_{D} |G_{zz}(x,z,t-\tau)|\, |D_{y,s}u_{n}(z,\tau)|\, dzd\tau \bigg{)}^{2} dyds \bigg{)} \\ &\phantom{\leq.}+ c(\sigma) \sup_{x \in D} \int_{\zeta}^{t} \int_{D} \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \right |^{2} dyds. \end{aligned} \end{align}\tag{50}\] Regarding the first term on the right-hand side of (50 ), following similar calculations as those done for (49 ), we obtain \[\begin{align} \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} \bigg{(} \int_{s}^{t} \int_{D} |G_{zz}(x,z,t-\tau)|\, &|D_{y,s}u_{n}(z,\tau)|\, dzd\tau \bigg{)}^{2} dyds \bigg{)} \leq \\ &\leq c(T) \int_{\zeta}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{z \in D} E \bigg{(} \int_{\zeta}^{\tau} \int_{D} | D_{y,s}u_{n}(z,\tau) |^{2}\, dyds \bigg{)} d\tau. \end{align}\] Regarding now the second term, it is easy to verify that, for any \(\zeta\leq s\), we have \[\begin{align} \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) = \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[\zeta,t]}(\bullet) \big{)} \big] (y,s) \end{align}\] and thus \[\begin{align} \sup_{x \in D} \int_{\zeta}^{t} \int_{D} \big{ |} \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{|}^{2}\, dyds &= \sup_{x \in D} \int_{\zeta}^{t} \int_{D} \left | \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[\zeta,t]}(\bullet) \big{)} \big] (y,s) \right |^{2} dyds \\ &\leq \int_{0}^{T} \int_{D} \big{\|} \big[ K^{*}_{H} \big{(} G(\cdot,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[\zeta,t]}(\bullet) \big{)} \big] (y,s) \big{\|}^{2}_{L^{\infty}(D)}\, dyds \\ \overset{(\ref{second32estimate})}&{\leq} c(H)\, (t-\zeta)^{\frac{4H-1}{2}}. \end{align}\] Gathering the above estimates, we end up with \[\begin{align} \begin{aligned} \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} &| D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)} \leq \\ &\leq c(n,T) \int_{\zeta}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{z \in D} E \bigg{(} \int_{\zeta}^{\tau} \int_{D} | D_{y,s}u_{n}(z,\tau) |^{2}\, dyds \bigg{)} d\tau + c(\sigma,H)\, (t-\zeta)^{\frac{4H-1}{2}}. \end{aligned} \end{align}\] If we define \[L_{n}(t) := \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} | D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)},\;t\in[\zeta,T],\] then the above relation becomes \[L_{n}(t) \leq c(n,T) \int_{\zeta}^{t} (t-\tau)^{-\frac{1}{4}} L_{n}(\tau)\, d\tau + c(\sigma,H)\, (t-\zeta)^{\frac{4H-1}{2}},\;\forall t\in[\zeta,T].\] Thus, by an argument similar to that used for (41 ), we obtain \[\begin{align} L_{n}(t) &\leq c(\sigma,H)\, (t-\zeta)^{\frac{4H-1}{2}} \exp \big[ c(n,T) (t - \zeta) \big] \\ \iff \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} | D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)} &\leq c(\sigma,H)\, (t-\zeta)^{\frac{4H-1}{2}} \exp \big[ c(n,T) (t - \zeta) \big],\;\forall t\in[\zeta,T]. \end{align}\] Consider now any \(\hat{s} \in (0,T]\), \(\zeta\in[0,\hat{s}]\). Then, for every \(t \in [\zeta,\hat{s}]\), we have (since \(D_{y,s}u_{n}(x,t)=0\) when \(s>t\)) \[\begin{align} \sup_{x \in D} E \bigg{(} \int_{\zeta}^{\hat{s}} \int_{D} | D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)} &= \sup_{x \in D} E \bigg{(} \int_{\zeta}^{t} \int_{D} | D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)} \\[0.1cm] &\leq c(\sigma,H)\, (t - \zeta)^{\frac{4H-1}{2}} \exp \big[ c(n,T) (t - \zeta) \big] \\[0.1cm] &\leq c(\sigma,H)\, (\hat{s} - \zeta)^{\frac{4H-1}{2}} \exp \big[ c(n,T) (\hat{s} - \zeta) \big]. \end{align}\] Choosing now \(\zeta = \hat{s} - \epsilon\) for any \(\epsilon \in (0,\hat{s}]\), we obtain for every \(t \in [\hat{s}- \epsilon,\hat{s}]\) \[\sup_{x \in D} E \bigg{(} \int_{\hat{s}-\epsilon}^{\hat{s}} \int_{D} | D_{y,s}u_{n}(x,t) |^{2}\, dyds \bigg{)} \leq c(\sigma,H)\, \epsilon^{\frac{4H-1}{2}} \exp \big{(} c(n,T)\, \epsilon \big{)}.\] Taking finally supremum for \(t\in[\hat{s}-\epsilon,\hat{s}]\) yields the desired result. ◻

Proof. In order to prove our claim, we invoke Theorem 2.1.3 of [11] which requires us to show that \[\Big{(} u_{n}(x,t) \in \mathbb{D}^{1,1}_{\mathrm{\text{loc}}}\quad \&\quad \left \| D_{\diamond,\bullet}u_{n}(x,t) \right \|_{L^{2}(D\times[0,T])}>0\;\text{a.s.} \Big{)},\;\forall (x,t)\in D\times(0,T].\]

The first requirement is immediate since we have already shown in Lemma 4 that \[\{u_{n}(x,t)\}_{n\in\mathbb{N}_{*}}\subset\mathbb{D}^{1,2},\;\forall (x,t)\in D\times[0,T]\] and we know from 1) of Remark 3 that \(\mathbb{D}^{1,2}\subseteq\mathbb{D}^{1,2}_{\mathrm{\text{loc}}}\subseteq\mathbb{D}^{1,1}_{\mathrm{\text{loc}}}\).
We focus now on the second requirement. Starting again from equation (?? ), i.e., \[D_{y,s}u_{n}(x,t) = \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau\;+ \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s)\] which holds almost surely for any \((x,t)\in D\times[0,T]\) and any \((y,s)\in D\times[0,t]\), we take absolute value on both sides and square to get \[\left | D_{y,s}u_n(x,t) \right |^{2} = \bigg{|} \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau\;+ \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \bigg{|}^{2}.\] We make now use of the relation \[(a+b)^{2}\geq \dfrac{a^{2}}{2}-b^{2},\;\forall a,b\in\mathbb{R}\quad \bigg{(}\iff \Big{(}\dfrac{\sqrt{2}}{2}a+\sqrt{2}b\Big{)}^{2}\geq0,\;\forall a,b\in\mathbb{R}\bigg{)}\] for \[\begin{align} a= \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s)\quad \&\quad b= \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau \end{align}\] to obtain \[\begin{align} \label{R46R468} \begin{aligned} \left | D_{y,s}u_{n}(x,t) \right |^{2} &\geq \dfrac{\big{|} \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{|}^{2}}{2}\;- \\ &- \bigg{|} \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau \bigg{|}^{2}. \end{aligned} \end{align}\tag{51}\] Let now \((x,t)\in D\times(0,T]\) be arbitrary and consider any \(\epsilon\in(0,t]\). We will now prove some important estimates that we are going to need later on. More specifically, we will show that \[\begin{align} E \bigg{(} \int_{t-\epsilon}^{t} \int_{D} \bigg{|} \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau \bigg{|}^{2} dyds \bigg{)} &\leq \mathcal{C}_{1}\, \epsilon^{2H+\frac{1}{4}}\, \exp (\mathcal{C}_2\, \epsilon), \tag{52} \\[0.1cm] \int_{t-\epsilon}^{t} \int_{D} \big{|} \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{|}^{2}\, dyds &\geq \mathcal{C}_{3}\, \Lambda(H,t,\epsilon) \tag{53}, \end{align}\] where \(\mathcal{C}_{1} = \mathcal{C}_{1}(\sigma,H,n,T),\;\mathcal{C}_{2} = \mathcal{C}_{2}(n,T),\, \mathcal{C}_3 = \mathcal{C}_3(\sigma,H)\) are positive constants and \[\Lambda(H,t,\epsilon):=\dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}} \left [ \dfrac{t^{4-2H}-(t-\epsilon)^{4-2H}}{(4-2H)(3-2H)} - \dfrac{(t-\epsilon)^{2-2H} \big[ t^{2}-(t-\epsilon)^{2}\big]}{2}+ \dfrac{(2-2H)(t-\epsilon)^{3-2H}\, \epsilon}{3-2H} \right ].\] We note that \(\Lambda(H,t,\epsilon) \geq 0\), with equality holding if and only if \(\epsilon = t\), as will become clear from the derivation of (53 ). Following similar calculations as those done for (49 ), we obtain \[\begin{align} E \bigg{(} \int_{t-\epsilon}^{t} \int_{D} \bigg{|} \int_{s}^{t} \int_{D} G_{zz}(x,&z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau \bigg{|}^{2} dyds \bigg{)} \\[0.1cm] &\leq c(n,T) \int_{t-\epsilon}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{z \in D} E \bigg{(} \int_{t-\epsilon}^{\tau} \int_{D} \left | D_{y,s}u_{n}(z,\tau) \right |^2 dyds \bigg{)} d\tau \\[0.1cm] &\leq c(n,T) \int_{t-\epsilon}^{t} (t-\tau)^{-\frac{1}{4}} \sup_{\tau\in[t-\epsilon,t]} \sup_{z \in D} E \bigg{(} \int_{t-\epsilon}^{t} \int_{D} \left | D_{y,s}u_{n}(z,\tau) \right |^2 dyds \bigg{)} d\tau \\[0.1cm] \overset{(\ref{R46R466})}&{\leq} c(\sigma,H,n,T)\, \epsilon^{\frac{4H-1}{2}} \exp \big{(} c(n,T)\, \epsilon \big{)} \int_{t-\epsilon}^{t} (t-\tau)^{-\frac{1}{4}}\, d\tau \\[0.1cm] &= c(\sigma,H,n,T)\, \epsilon^{\frac{4H-1}{2}} \exp \big{(} c(n,T)\, \epsilon \big{)}\, \frac{4}{3}\epsilon^{\frac{3}{4}} = c(\sigma,H,n,T)\, \epsilon^{2H+\frac{1}{4}} \exp \big{(} c(n,T)\, \epsilon \big{)}, \end{align}\] which establishes (52 ). Before proving (53 ), we mention one important property regarding the Green’s function defined in (13 ). More specifically, for any \(x,y \in D\) and any \(t_{1},t_{2} > 0\), it holds \[\label{int4632property32of32Green39s32function} \int_{D} G(x,z,t_{1}) G(z,y,t_{2})\, dz = G(x,y,t_{1}+t_{2}).\tag{54}\] Having said that, we can now move on with the proof of (53 ): \[\begin{align} \int_{t-\epsilon}^{t} \int_{D} \big{|} \sigma \big[ K^{*}_{H} \big{(} &G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s) \big{|}^{2}\, dyds = \sigma^{2} \int_{t-\epsilon}^{t} \int_{D} \bigg{(} \int_{s}^{t} G(x,y,t-r)\, \frac{\partial K_{H}(r,s)}{\partial r}\, dr \bigg{)}^{2} dyds \\[0.3cm] &= \sigma^{2} \int_{t-\epsilon}^{t} \int_{s}^{t} \int_{s}^{t} \int_{D} \frac{\partial K_{H}(r,s)}{\partial r}\, \frac{\partial K_{H}(r',s)}{\partial r'}\, G(x,y,t-r)\, G(x,y,t-r')\, dy dr dr' ds \\[0.3cm] \overset{(\ref{int4632property32of32Green39s32function})}&{=} \sigma^{2} \int_{t-\epsilon}^{t} \int_{s}^{t} \int_{s}^{t} \frac{\partial K_{H}(r,s)}{\partial r}\, \frac{\partial K_{H}(r',s)}{\partial r'}\, G(x,x,2t-r-r')\, dr dr' ds \\[0.3cm] \overset{\text{Fubini}}&{=} \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{r'} \int_{s}^{t} \frac{\partial K_{H}(r,s)}{\partial r}\, \frac{\partial K_{H}(r',s)}{\partial r'}\, G(x,x,2t-r-r')\, dr ds dr' \\[0.3cm] \overset{\text{Fubini}}&{=} \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{\min\{r,r'\}} \frac{\partial K_{H}(r,s)}{\partial r}\, \frac{\partial K_{H}(r',s)}{\partial r'}\, G(x,x,2t-r-r')\, ds dr dr' \\[0.3cm] &= \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} G(x,x,2t-r-r')\, \bigg{(} \int_{t-\epsilon}^{\min\{r,r'\}} \frac{\partial K_{H}(r,s)}{\partial r}\, \frac{\partial K_{H}(r',s)}{\partial r'}\, ds \bigg{)} dr dr' \\[0.3cm] \overset{(\ref{par4632der4632of32K95H})}&{=} \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} G(x,x,2t-r-r')\, \bigg{(} \int_{t-\epsilon}^{\min\{r,r'\}} \dfrac{s^{1-2H}\, (r-s)^{H-\frac{3}{2}}\, (r'-s)^{H-\frac{3}{2}}}{(rr')^{\frac{1}{2}-H}}\, ds \bigg{)} dr dr' \\[0.3cm] &= \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} G(x,x,2t-r-r')\, \bigg{(} \int_{t-\epsilon}^{\min\{r,r'\}} \dfrac{(rr')^{H-\frac{1}{2}}}{s^{2H-1}\, (r-s)^{\frac{3}{2}-H}\, (r'-s)^{\frac{3}{2}-H}}\, ds \bigg{)} dr dr' \\[0.3cm] \overset{s\geq t-\epsilon}&{\geq} \sigma^{2} \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} \dfrac{G(x,x,2t-r-r')\, (rr')^{H-\frac{1}{2}}}{(r-t+\epsilon)^{\frac{3}{2}-H}\, (r'-t+\epsilon)^{\frac{3}{2}-H}}\, \bigg{(} \int_{t-\epsilon}^{\min\{r,r'\}} \dfrac{1}{s^{2H-1}}\, ds \bigg{)} dr dr' \\[0.3cm] &= c(\sigma,H) \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} \dfrac{G(x,x,2t-r-r')\, (rr')^{H-\frac{1}{2}}}{(r-t+\epsilon)^{\frac{3}{2}-H}\, (r'-t+\epsilon)^{\frac{3}{2}-H}} \left [ \min\{r,r'\}^{2-2H}-(t-\epsilon)^{2-2H} \right ] dr dr'. \end{align}\] We pause for a moment our calculations to give a lower bound for the term \(G(x,x,2t-r-r')\). It holds \[\label{R46R469} G(x,x,2t-r-r') \overset{(\ref{Green39s32function})}{=} \sum\limits_{k=0}^{+\infty} a_{k}^{2}(x) e^{-k^{4}(2t-r-r')} = \frac{1}{\pi} + \frac{2}{\pi} \sum\limits_{k=1}^{+\infty} \cos^{2}(kx) e^{-k^{4}(2t-r-r')} \geq \frac{1}{\pi}\tag{55}\] (since \(r\leq t\) & \(r'\leq t\), we have \(2t-r-r'\geq0\), thus the last series is not only convergent but converges to a non-negative value). We can resume now with the calculations were we left \[\begin{align} c(\sigma,H) \int_{t-\epsilon}^{t} &\int_{t-\epsilon}^{t} \dfrac{G(x,x,2t-r-r')\, (rr')^{H-\frac{1}{2}}}{(r-t+\epsilon)^{\frac{3}{2}-H}\, (r'-t+\epsilon)^{\frac{3}{2}-H}} \left [ \min\{r,r'\}^{2-2H}-(t-\epsilon)^{2-2H} \right ] dr dr' \\[0.3cm] \overset{(\ref{R46R469})}&{\geq} c(\sigma,H) \int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} \dfrac{(rr')^{H-\frac{1}{2}}}{(r-t+\epsilon)^{\frac{3}{2}-H}\, (r'-t+\epsilon)^{\frac{3}{2}-H}} \left [ \min\{r,r'\}^{2-2H}-(t-\epsilon)^{2-2H} \right ] dr dr' \\[0.3cm] \overset{t-\epsilon\leq r,r'\leq t}&{\geq} c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}}\;\int_{t-\epsilon}^{t} \int_{t-\epsilon}^{t} \min\{r,r'\}^{2(1-H)}-(t-\epsilon)^{2(1-H)}\, dr dr' \\[0.3cm] &\geq c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}}\;\int_{t-\epsilon}^{t} \int_{t-\epsilon}^{r'} r^{2(1-H)}-(t-\epsilon)^{2(1-H)}\, dr dr' \\[0.3cm] &= c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}} \int_{t-\epsilon}^{t} \dfrac{(r')^{3-2H}- (t-\epsilon)^{3-2H}}{3-2H} -(t-\epsilon)^{2(1-H)}(r'-t+\epsilon)\, dr' \\[0.3cm] &= c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}}\, \left ( \int_{t-\epsilon}^{t} \dfrac{(r')^{3-2H}}{3-2H} -(t-\epsilon)^{2(1-H)}r'\, dr' -\dfrac{(t-\epsilon)^{3-2H}\;\epsilon}{3-2H}+(t-\epsilon)^{3-2H}\, \epsilon \right ) \\[0.3cm] &= c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}}\, \left ( \int_{t-\epsilon}^{t} \dfrac{(r')^{3-2H}}{3-2H} -(t-\epsilon)^{2(1-H)}r'\, dr' + \dfrac{(2-2H)(t-\epsilon)^{3-2H}\, \epsilon}{3-2H} \right ) \\[0.3cm] &= c(\sigma,H)\, \dfrac{(t-\epsilon)^{2H-1}}{\epsilon^{3-2H}} \left [ \dfrac{t^{4-2H}-(t-\epsilon)^{4-2H}}{(4-2H)(3-2H)} - \dfrac{(t-\epsilon)^{2-2H} \left [ t^{2}-(t-\epsilon)^{2} \right ]}{2}+ \dfrac{(2-2H)(t-\epsilon)^{3-2H}\, \epsilon}{3-2H} \right ], \end{align}\] which establishes (53 ). We are now in position to conclude the proof of the lemma.

Let \((x,t) \in D\times(0,T]\) be arbitrary and consider any \(\epsilon \in (0,t)\). Then, using again the notation \[a= \sigma \big[ K^{*}_{H} \big{(} G(x,\diamond,t-\bullet) \mathrm{\boldsymbol{1}}_{[0,t]}(\bullet) \big{)} \big] (y,s)\quad \&\quad b= \int_{s}^{t} \int_{D} G_{zz}(x,z,t-\tau) \mathcal{G}_{n}(z,\tau) D_{y,s}u_{n}(z,\tau)\, dzd\tau\] we have \[\begin{align} \label{R46R4610} \begin{aligned} 1\geq \mathbb{P}\bigg{(}\int_{0}^{T}\int_{D} |D_{y,s}u_{n}(x,t)|^{2}\, dyds>0\bigg{)} &\geq \mathbb{P}\bigg{(}\int_{t-\epsilon}^{t}\int_{D} |D_{y,s}u_{n}(x,t)|^{2}\, dyds>0\bigg{)} \\[0.1cm] \overset{(\ref{R46R468})}&{\geq} \mathbb{P}\bigg{(} \dfrac{1}{2} \int_{t-\epsilon}^{t} \int_{D} a^{2}\, dyds -\int_{t-\epsilon}^{t} \int_{D} b^{2}\, dyds > 0\bigg{)} \\[0.1cm] \overset{(\ref{lower32bound})}&{\geq} \mathbb{P}\bigg{(}\int_{t-\epsilon}^{t} \int_{D} b^{2}\, dyds < \frac{\mathcal{C}_{3}\, \Lambda(H,t,\epsilon)}{2} \bigg{)} \\[0.1cm] &= 1- \mathbb{P}\bigg{(}\int_{t-\epsilon}^{t} \int_{D} b^{2}\, dyds \geq \frac{\mathcal{C}_{3}\, \Lambda(H,t,\epsilon)}{2} \bigg{)} \\[0.1cm] \overset{\text{Markov}}&{\geq} 1-E\bigg{(} \int_{t-\epsilon}^{t} \int_{D} b^{2}\, dyds\bigg{)}\, \dfrac{2}{\mathcal{C}_{3}\, \Lambda(H,t,\epsilon)} \\[0.1cm] \overset{(\ref{upper32bound})}&{\geq} 1- \frac{2\mathcal{C}_{1}}{\mathcal{C}_{3}}\, \dfrac{\epsilon^{2H+\frac{1}{4}}}{\Lambda(H,t,\epsilon)}\, \exp (\mathcal{C}_2\, \epsilon). \end{aligned} \end{align}\tag{56}\] We want to pass now to the limit as \(\epsilon\to0^{+}\). We define for ease of use \[c_{1}:=2H-1,\quad c_{2}:=2-2H,\quad c_{3}:=3-2H,\quad c_{4}:=4-2H\] \(\big{(}\)we recall that we are working in the case \(H \in \big{(} \frac{1}{2},1 \big{)}\) thus \(c_{1},c_{2},c_{3},c_{4}>0 \big{)}\) and then calculate \[\begin{align} \lim\limits_{\epsilon\to0^{+}} \bigg{(} \dfrac{\epsilon^{2H+\frac{1}{4}}}{\Lambda(H,t,\epsilon)} \bigg{)} &= \lim\limits_{\epsilon\to0^{+}} \left ( \frac{\epsilon^{2H + \frac{1}{4}}}{\dfrac{(t-\epsilon)^{c_{1}}}{\epsilon^{c_{3}}}\, \bigg[ \dfrac{t^{c_{4}}-(t-\epsilon)^{c_{4}}}{c_{4}c_{3}} - \dfrac{(t-\epsilon)^{c_{2}} \big[ t^{2}-(t-\epsilon)^{2}\big]}{2}+ \dfrac{c_{2}(t-\epsilon)^{c_{3}}\, \epsilon}{c_{3}} \bigg]} \right ) \\[0.3cm] &= \lim\limits_{\epsilon\to0^{+}} \left ( \frac{2c_{3}c_{4}}{(t-\epsilon)^{c_{1}}}\cdot\frac{\epsilon^{\frac{13}{4}}}{2\big[t^{c_{4}}-(t-\epsilon)^{c_{4}}\big] - c_{3}c_{4}(t-\epsilon)^{c_{2}} \big[ t^{2}-(t-\epsilon)^{2}\big] +2c_{2}c_{4} (t-\epsilon)^{c_{3}}\, \epsilon} \right )\\[0.3cm] \end{align}\]
with \[\begin{align} &\lim\limits_{\epsilon\to0^{+}} \Bigg{(} \frac{ \epsilon^{\frac{13}{4}}}{2\big[t^{c_{4}}-(t-\epsilon)^{c_{4}}\big]-c_{3}c_{4}(t-\epsilon)^{c_{2}} \big[t^{2}-(t-\epsilon)^{2}\big] + 2c_{2} c_{4} (t-\epsilon)^{c_{3}}\, \epsilon} \Bigg{)} =\\[0.1cm] \overset{(\frac{0}{0})}{=} &\lim\limits_{\epsilon\to0^{+}} \bigg{(} \dfrac{\frac{13}{4}\epsilon^{\frac{9}{4}}}{2c_{4}(t-\epsilon)^{c_{4}-1} +c_{2}c_{3}c_{4}(t-\epsilon)^{c_{2}-1}\big[t^{2}-(t-\epsilon)^{2}\big]-2c_{3}c_{4}(t-\epsilon)^{c_{2}+1}} \\[0.1cm] &\phantom{\overset{(\frac{0}{0})}{=} \lim\limits_{\epsilon\to0^{+}} \bigg{(}}\frac{}{-2c_{2}c_{3}c_{4}(t-\epsilon)^{c_{3}-1}\, \epsilon+2c_{2}c_{4}(t-\epsilon)^{c_{3}}} \bigg{)}, \end{align}\] where we applied L’Hôspital’s rule. Notice that \(c_{4}-1=c_{3}=c_{2}+1\) thus the first, third and last terms of the denominator cancel out giving us \[\begin{align} \lim\limits_{\epsilon\to0^{+}} &\Bigg{(} \frac{ \epsilon^{\frac{13}{4}}}{2\big[t^{c_{4}}-(t-\epsilon)^{c_{4}}\big]-c_{3}c_{4}(t-\epsilon)^{c_{2}} \big[t^{2}-(t-\epsilon)^{2}\big]+2c_{2}c_{4}(t-\epsilon)^{c_{3}}\, \epsilon} \Bigg{)} =\\[0.1cm] &= \lim\limits_{\epsilon\to0^{+}} \left ( \dfrac{\frac{13}{4}\epsilon^{\frac{9}{4}}}{c_{2}c_{3}c_{4}(t-\epsilon)^{c_{2}-1}\big[t^{2}-(t-\epsilon)^{2}\big]-2c_{2}c_{3}c_{4}(t-\epsilon)^{c_{3}-1}\, \epsilon} \right ) = \lim\limits_{\epsilon\to0^{+}} \left ( \dfrac{\frac{13}{4}\epsilon^{\frac{9}{4}}}{c_{2}c_{3}c_{4}(t-\epsilon)^{c_{2}-1}\, \epsilon^{2}} \right ) = 0. \end{align}\] Gathering the above, we conclude that \[\lim\limits_{\epsilon\to0^{+}} \left ( \dfrac{\epsilon^{2H+\frac{1}{4}}}{\Lambda(H,t,\epsilon)} \right ) = 0\] and so, taking finally limits as \(\epsilon\to0^{+}\) in relation (56 ) yields \[\mathbb{P}\bigg{(}\int_{0}^{T}\int_{D} |D_{y,s}u_{n}(x,t)|^{2}\, dyds>0\bigg{)}=1 \iff \left \| D_{\diamond,\bullet}u_{n}(x,t) \right \|_{L^{2}(D\times[0,T])}>0\;\text{a.s.}\] for any \((x,t)\in D\times(0,T]\) which is exactly the second requirement as mentioned in the beginning. ◻

Proof. As we did for the analogous result in Lemma 6, we will invoke Theorem 2.1.3 of [11] which requires us to show that \[\Big{(} u(x,t) \in \mathbb{D}^{1,1}_{\mathrm{\text{loc}}}\quad \&\quad \left \| D_{\diamond,\bullet}u(x,t) \right \|_{L^{2}(D\times[0,T])}>0\;\text{a.s.} \Big{)},\;\forall (x,t)\in D\times(0,T].\]

The first requirement is immediate since we have already explained at the end of Section 3 that \(u(x,t)\in\mathbb{D}^{1,2}_{\mathrm{\text{loc}}}\) for all \((x,t) \in D\times[0,T]\) and we know from 1) of Remark 3 that \(\mathbb{D}^{1,2}_{\mathrm{\text{loc}}}\subseteq\mathbb{D}^{1,1}_{\mathrm{\text{loc}}}\).
Regarding the second requirement, instead of following an arduous process like the one for deriving the analogous result for \(u_{n}(x,t)\) in Lemma 6, we will use that result efficiently to obtain what we need. In detail, we fix an arbitrary \(n \in \mathbb{N}_{*}\) and define the following sets \[\begin{align} {2} A_{1} &:= \Big{\{} \omega\in\Omega\, :\, \left \| D_{\diamond,\bullet}u(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}>0 \Big{\}},\quad &&B_{1} := \Big{\{} \omega\in\Omega\, :\, \left \| D_{\diamond,\bullet}u_{n}(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}>0 \Big{\}}, \\[0.1cm] A_{2} &:= \Big{\{} \omega\in\Omega_{n}\, :\, \left \| D_{\diamond,\bullet}u(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}>0 \Big{\}},\quad &&B_{2} := \Big{\{} \omega\in\Omega_{n}\, :\, \left \| D_{\diamond,\bullet}u_{n}(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}>0 \Big{\}}, \end{align}\] where \(\Omega_{n}\) is given by (20 ). We explained at the end of Subsection 3.3 that \(D_{y,s}u(x,t) = D_{y,s}u_{n}(x,t)\) a.s. on \(\Omega_{n}\) for all \((x,t),(y,s)\in D\times[0,T]\); therefore it holds that \(\mathbb{P}(B_{2}) = \mathbb{P}(A_{2})\).
Additionally, it was established in Lemma 6 that \(\mathbb{P}(B_{1})=1\), thus \(\mathbb{P}(B_{1}^{c})=0\). Notice however that \[\Omega_{n} - B_{2} = \left \{ \omega\in\Omega_{n}\, :\, \left \| D_{\diamond,\bullet}u_{n}(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}=0 \right \} \subseteq \left \{ \omega\in\Omega\, :\, \left \| D_{\diamond,\bullet}u_{n}(x,t;\omega) \right \|_{L^{2}(D\times[0,T])}=0 \right \} = B_{1}^{c}\] so we get \(\mathbb{P}(\Omega_{n} - B_{2}) \leq \mathbb{P}(B_{1}^{c})\) which yields \[\mathbb{P}(\Omega_{n} - B_{2}) = 0 \iff \mathbb{P}(\Omega_{n}) = \mathbb{P}(B_{2}) \implies \mathbb{P}(\Omega_{n}) = \mathbb{P}(A_{2}) \overset{A_{2} \subseteq A_{1}}{\implies} \mathbb{P}(\Omega_{n}) \leq \mathbb{P}(A_{1}).\] The last inequality holds for every \(n \in \mathbb{N}_{*}\) so, passing to the limit as \(n \to +\infty\), we obtain \(\mathbb{P}(A_{1})=1\) which is exactly the second requirement. ◻