Proof of Lemma 4 (Wong–Zakai type approximation)
We make use of a usual trick of avoiding the degeneracy of the porous medium operator, see for instance [19, Proof of Theorem 5.5]. Let \(\varepsilon > 0\) be fixed and choose a bounded smooth function \(a_\varepsilon :\mathbf {R}\rightarrow [m(\frac{\varepsilon }{2})^{m-1},\infty )\) such that it holds \(a_\varepsilon (r)=mr^{m-1}\) for all \(r\in [\varepsilon ,\varepsilon {+}\Vert u_0\Vert _{L^\infty (D)}]\). By the choice of \(a_\varepsilon \) and since \(B^\varepsilon _t\) as defined in (7) is smooth on \([0,T^*]\) almost surely, we can make use of standard quasilinear theory [17] to solve the PDE
$$\begin{aligned} \partial _t u_\varepsilon = \nabla \cdot \big (a_\varepsilon (u_\varepsilon )\nabla u_\varepsilon \big ) + \nu \Big (\frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot \nabla \Big )u_\varepsilon \quad \text {on } \bar{D}\times [0,T^*] \end{aligned}$$
in a classical sense, i.e., we obtain with probability one a classical solution \(u_\varepsilon \in C^{2,1}_{x,t}(\bar{D}{\times }[0,T^*])\cap C^\infty (D{\times }(0,T^*))\). It is then immediate from the choice of \(a_\varepsilon \) and the regularity of \(u_\varepsilon \) that the equations (12)–(14) are satisfied pointwise everywhere. We can infer from the maximum principle that \(\varepsilon \le u_\varepsilon (x,t)\le \varepsilon +\Vert u_0\Vert _{L^\infty (D)}\) holds true for all \((x,t)\in \bar{D}{\times }[0,T^*]\) as it is asserted in (15). The derivation of the energy estimate (16) is standard: multiply the equation with \(u_\varepsilon \), integrate over D and use the regularity of \(u_\varepsilon \) to integrate by parts in the spatial differential operators. Note in this respect that as a consequence of (15) and (14) it holds \(\mathbf {n}_{\partial D}\cdot \nabla u_\varepsilon ^m\le 0\) on \(\partial D\), where \(\mathbf {n}_{\partial D}\) is the exterior unit normal vector field along the \(C^2\) manifold \(\partial D\). This is the only reason for the inequality sign in (16) as we may compute for the second term
$$\begin{aligned} \int _D \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot u_\varepsilon \nabla u_\varepsilon \,\mathrm {d}x&= \int _D \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot u_\varepsilon \nabla (u_\varepsilon {-}\varepsilon ) \,\mathrm {d}x\\&= -\int _D \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot (u_\varepsilon {-}\varepsilon )\nabla u_\varepsilon = -\int _D \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot \frac{1}{2}\nabla |u_\varepsilon {-}\varepsilon |^2 = 0. \end{aligned}$$
We proceed with the bound (17) for the time derivative. Multiplying the equation (12) with \(\partial _t u_\varepsilon ^m\), integrating over D, performing an integration by parts in the term with the porous medium operator and estimating the transport term by Hölder’s and Young’s inequality yields for all \(t\in (0,T^*)\) the estimate
$$\begin{aligned}&\int _D mu_\varepsilon ^{m-1}(t)|\partial _t u_\varepsilon |^2(t) \,\mathrm {d}x+ \frac{\mathrm {d}}{\mathrm {d}t}\int _D \frac{1}{2} |\nabla u_\varepsilon ^m|^2(t) \,\mathrm {d}x\\&\quad \le \frac{1}{2}\int _D mu_\varepsilon ^{m-1}(t)|\partial _t u_\varepsilon |^2(t) \,\mathrm {d}x+\frac{\nu ^2}{2}\int _D \Big |\frac{\mathrm {d}}{\mathrm {d}t} B^\varepsilon _t\Big |^2 mu_\varepsilon ^{m-1}(t)|\nabla u_\varepsilon |^2(t) \,\mathrm {d}x. \end{aligned}$$
Multiplying this bound with t and integrating the resulting estimate over (0, T) we may infer using also (16) and (15)
$$\begin{aligned}&\int _0^T\int _D \frac{t}{2}mu_\varepsilon ^{m-1}|\partial _t u_\varepsilon |^2 \,\mathrm {d}x\,\mathrm {d}t+ \frac{T}{2}\int _D |\nabla u_\varepsilon ^m|^2(T) \,\mathrm {d}x\\&\quad \le \frac{1}{2}\int _0^T\int _D |\nabla u_\varepsilon ^m|^2 \,\mathrm {d}x\,\mathrm {d}t+\sup _{0\le t \le T^*} \Big |\frac{\mathrm {d}}{\mathrm {d}t} B^\varepsilon _t\Big |^2 \frac{T^*\nu ^2}{2}\int _0^T\int _D mu_\varepsilon ^{m-1}|\nabla u_\varepsilon |^2 \,\mathrm {d}x\,\mathrm {d}t\\&\quad \le \Big (\frac{m}{2}(\varepsilon {+}\Vert u_0\Vert _{L^\infty })^{m-1} {+}\frac{T^*\nu ^2}{2}\sup _{0\le t \le T^*} \Big |\frac{\mathrm {d}}{\mathrm {d}t} B^\varepsilon _t\Big |^2\Big ) \int _D\frac{1}{2}|u_\varepsilon |^2(0) \,\mathrm {d}x\end{aligned}$$
for all \(T\in (0,T^*)\). Moreover, it follows from (7) and Doob’s maximal inequality that \(\mathbf {E}\sup _{0\le t \le T^*}|\frac{\mathrm {d}}{\mathrm {d}t} B^\varepsilon _t|^2\le C\varepsilon ^{-\beta }T^*\mathbf {E}|B_{T^*}|^2\) for some absolute constant \(C>0\). This establishes the estimate (17). The bound (18) is now a consequence of plugging in the equation (12), then using the triangle inequality, estimating the term with the time derivative by means of (17) and bounding the transport term similarly as at the end of the proof of (17). This concludes the proof of Lemma 4. \(\square \)
Proof of Proposition 5 (Contraction principle up to time-dependent shift for truncated Wong–Zakai type approximations)
Fix \(\kappa >0\) and let \(\zeta _\delta \) denote the smooth and convex approximation to the positive part truncation \(r\mapsto r_+:=r\vee 0\) on scale \(\delta >0\) such that (19)–(21) hold true. Define \(\zeta ^\delta _\kappa (r):=\kappa + \zeta _\delta (r{-}\kappa )\) which is a smooth and convex approximation to the truncation \(r\mapsto r\vee \kappa \). Finally, fix \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\) and abbreviate for what follows \(v_\varepsilon :=\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon }^\leftarrow \) resp. \(v_{{\hat{\varepsilon }}}:=\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}}^\leftarrow \), where \(q>1\) will be a large exponent to be specified later on in the proof. See (22) for the definition of the shifted densities. Finally, fix a compact set \(K\subset D\).
We aim to derive an estimate for \(\sup _{T\in [0,T^*]}\mathbf {E}\Vert v_\varepsilon {-}v_{{\hat{\varepsilon }}}\Vert _{L^1(K)}\) of the same type as the asserted bound (23). The proof of this proceeds in several steps. For the sake of better readability, let us occasionally break the proof into intermediate results.
Lemma 8
Let the assumptions and notation of Sect. 4.2 be in place. We next choose a smooth, even and convex map \(\eta :\mathbf {R}\rightarrow [0,\infty )\) such that \(\eta (r)=|r|-1\) for \(|r|\ge 2\) and \(\eta (r)\le |r|\) for all \(r\in \mathbf {R}\). Define \(\eta _\delta (r):=\delta \eta (\frac{r}{\delta })\). There is a constant \(C>0\) independent of \(\delta \) such that
$$\begin{aligned}&\sup _{r\in \mathbf {R}} |(\eta _\delta )'(r)| + |r|(\eta _\delta )''(r) \le C, \end{aligned}$$
(30)
$$\begin{aligned}&(\eta _\delta )'(r) \rightarrow \mathrm {sign}(r) \text { and } \eta _\delta (r) \nearrow |r| \text { as } \delta \rightarrow 0, \end{aligned}$$
(31)
$$\begin{aligned}&|\eta _\delta (r){-}|r||\le C\delta \text { for all } r\in \mathbf {R},\text { and } \eta _\delta (r) = |r|-\delta \text { for all } |r|\ge 2\delta . \end{aligned}$$
(32)
Then the following “entropy estimate” holds true
$$\begin{aligned}&-\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon (x,t){-}\hat{\mathcal {Z}}\big ) \partial _t\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \le -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}|\nabla v_\varepsilon ^m(x,t)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}\hat{\mathcal {Z}}^m\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \quad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\eta _\delta \big (v_\varepsilon ^m(x,t){-}\hat{\mathcal {Z}}^m\big ) \Delta \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \quad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2|\nabla v_\varepsilon (x,t)|^2 (\eta _\delta )''\big (v_\varepsilon (x,t){-}\hat{\mathcal {Z}}\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \quad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \eta _\delta \big (v_\varepsilon (x,t){-}\hat{\mathcal {Z}}\big ) \Delta \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \quad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t\bigg |_{{\hat{z}} = \hat{\mathcal {Z}}}\nonumber \\&\qquad \quad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \Delta v_\varepsilon ^m(x,t)\xi (x,t) \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}\hat{\mathcal {Z}}\big ) -(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}\hat{\mathcal {Z}}^m\big ) \big \}\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad \quad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) \nabla v_\varepsilon (x,t)\cdot (\eta _\delta )'\big (v_\varepsilon (x,t){-}\hat{\mathcal {Z}}\big ) \nabla \xi (x,t)\,\mathrm {d}x\,\mathrm {d}t\end{aligned}$$
(33)
for all \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*);[0,\infty ))\) and all bounded random variables \(\hat{\mathcal {Z}}\in L^\infty (\Omega )\). An analogous estimate holds true for \(v_{{\hat{\varepsilon }}}\), see (49) below for the precise statement.
Proof
Step 1 (Equation for \(u_{\varepsilon }^\leftarrow \) and \(u_{{\hat{\varepsilon }}}^\leftarrow \)): The first step is to derive the equation for the shifted densities \(u_{\varepsilon }^\leftarrow \) and \(u_{{\hat{\varepsilon }}}^\leftarrow \), respectively. To this end, we aim to apply Itô’s formula with respect to \(\int _{\mathbf {R}^d} u_\varepsilon (x,t)\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t)\,\mathrm {d}x\), where \(\eta \) is an \(\mathbf {F}\)-adapted random test function \(\eta \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*))\). Note that \(\partial _t u_\varepsilon \equiv 0\) on the lateral boundary \(\partial D\times (0,T^*)\). Hence, it holds \(\partial _tu_\varepsilon \in C(\mathbf {R}^d\times (0,T^*))\) and we thus obtain from an application of Itô’s formula for each \(\mathbf {F}\)-adapted random test function \(\eta \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*))\) with probability one
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\partial _t\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad = \int _0^{T^*}\int _{\mathbf {R}^d} \partial _tu_\varepsilon (x,t)\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\nu \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot \nabla \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\frac{1}{2}\nu ^2 \Delta \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\nu \nabla \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}B_t. \end{aligned}$$
(34)
Since \(u_\varepsilon \) solves the equation (12) classically in \(D\times (0,T^*)\), and is by definition constant outside of it, we may compute
$$\begin{aligned} \begin{aligned}&\int _0^{T^*}\int _{\mathbf {R}^d} \partial _tu_\varepsilon (x,t)\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad =\int _0^{T^*}\int _{\mathbf {R}^d\setminus \partial D} \Delta u_\varepsilon ^m(x,t)\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \nu \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot \nabla u_\varepsilon (x,t) \,\mathrm {d}x\,\mathrm {d}t. \end{aligned} \end{aligned}$$
(35)
(Despite \(\partial D\) having Lebesgue measure zero, the domain of integration in the first right hand side term is \(\mathbf {R}^d\setminus \partial D\) since \(\nabla u_\varepsilon ^m\) may jump across the domain boundary \(\partial D\) for all \(i\in \{1,\ldots ,d\}\), and hence \(\partial _i \nabla u_\varepsilon ^m\) may not exist in the sense of weak derivatives in \(\mathbf {R}^d\times (0,T^*)\) for all \(i\in \{1,\ldots ,d\}\).) Integrating by parts in the second term on the right hand side of the latter identity as well as performing a change of variables \(x\mapsto x+\nu (B_t{-}B^\varepsilon _t)\) yields
$$\begin{aligned} \begin{aligned}&\int _0^{T^*}\int _{\mathbf {R}^d} \partial _tu_\varepsilon (x,t)\eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t) \nu \frac{\mathrm {d}}{\mathrm {d}t}B^\varepsilon _t\cdot \nabla \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad =\int _0^{T^*}\int _{\mathbf {R}^d\setminus ({-}\nu (B_t{-}B^\varepsilon _t)+\partial D)} \Delta (u_{\varepsilon }^\leftarrow )^m(x,t)\eta (x,t) \,\mathrm {d}x\,\mathrm {d}t. \end{aligned} \end{aligned}$$
(36)
Note that there is no boundary integral appearing from the integration by parts in the second term on the right hand side of (35) since \(u_\varepsilon \in C(\mathbf {R}^d\times (0,T^*))\). We compute analogously
$$\begin{aligned} \begin{aligned}&\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\frac{1}{2}\nu ^2 \Delta \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad =-\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2 \nabla u_{\varepsilon }^\leftarrow (x,t)\cdot \nabla \eta (x,t) \,\mathrm {d}x\,\mathrm {d}t\end{aligned} \end{aligned}$$
(37)
as well as
$$\begin{aligned} \begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\nu \nabla \eta (x{-}\nu (B_t{-}B^\varepsilon _t),t) \,\mathrm {d}x\,\mathrm {d}B_t.\\&\quad =\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla u_{\varepsilon }^\leftarrow (x,t) \eta (x,t) \,\mathrm {d}x\,\mathrm {d}B_t. \end{aligned} \end{aligned}$$
(38)
From (34), (36), (37) and (38) we infer that for each \(\mathbf {F}\)-adapted random test function \(\eta \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*))\) it holds with probability one
$$\begin{aligned} \begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} u_{\varepsilon }^\leftarrow (x,t)\partial _t\eta (x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad =\int _0^{T^*}\int _{\mathbf {R}^d\setminus ({-}\nu (B_t{-}B^\varepsilon _t)+\partial D)} \Delta (u_{\varepsilon }^\leftarrow )^m(x,t)\eta (x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \nabla u_{\varepsilon }^\leftarrow (x,t)\cdot \nabla \eta (x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla u_{\varepsilon }^\leftarrow (x,t)\eta (x,t) \,\mathrm {d}x\,\mathrm {d}B_t. \end{aligned} \end{aligned}$$
(39)
Analogously one derives the equation for \(u_{{\hat{\varepsilon }}}^\leftarrow \).
Step 2 (Convex approximation to \(r\mapsto r\vee \kappa \) as test function): In the next step we aim to derive the equation for \(v_\varepsilon =\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow \) based on the equation for the shifted density derived in (39). The idea is to test the equation (39) with the test function \(\eta :=\big ((\zeta _\kappa ^{\varepsilon ^q})'\circ u_\varepsilon ^\leftarrow \big )\xi \), where \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*);[0,\infty ))\) is arbitrary. However, since the shifted density \(u_\varepsilon ^\leftarrow \) is only Hölder continuous in the time variable we have to regularize first. To this end, we make use of the Steklov average \(\eta _h(x,t):=\frac{1}{h}\int _{t-h}^{t}\eta (x,s)\,\mathrm {d}s\) which is an admissible test function for (39) for all sufficiently small \(h>0\) (depending only on the support of \(\xi \)). Since \(\partial _t\eta _h(x,t)=\frac{\eta (x,t){-}\eta (x,t{-}h)}{h}\) we obtain by a simple change of variables
$$\begin{aligned} \begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon ^\leftarrow (x,t)\partial _t\eta _h(x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad =-\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{h} \big (u_\varepsilon ^\leftarrow (x,t{+}h){-}u_\varepsilon ^\leftarrow (x,t)\big ) (\zeta ^{\varepsilon ^q}_\kappa )'(u_\varepsilon ^\leftarrow (x,t))\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\end{aligned} \end{aligned}$$
(40)
for each \(h>0\) and each \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times } (0,T^*);[0,\infty ))\) almost surely. We then deduce from the smoothness and convexity of \(\zeta _\kappa ^{\varepsilon ^q}\) as well as by reverting the change of variables the bound
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon ^\leftarrow (x,t)\partial _t\eta _h(x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad \ge -\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{h} \big (\zeta ^{\varepsilon ^q}_\kappa (u_\varepsilon ^\leftarrow (x,t{+}h)) -\zeta ^{\varepsilon ^q}_\kappa (u_\varepsilon ^\leftarrow (x,t))\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\\&\quad = -\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{h} \big (\xi (x,t{-}h){-}\xi (x,t)\big ) \zeta ^{\varepsilon ^q}_\kappa (u_\varepsilon ^\leftarrow (x,t)) \,\mathrm {d}x\,\mathrm {d}t\end{aligned}$$
for each \(h>0\) and each \(\xi \in C^\infty _{\mathrm {cpt}}(D{\times } (0,T^*);[0,\infty ))\) almost surely. Hence, we may infer from this, (39) and standard properties of the Steklov average after letting \(h\rightarrow 0\) the estimate
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} \zeta ^{\varepsilon ^q}_\kappa (u_\varepsilon ^\leftarrow (x,t))\partial _t\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad \le \int _0^{T^*}\int _{\mathbf {R}^d\setminus ({-}\nu (B_t{-}B^\varepsilon _t)+\partial D)} \Delta (u_\varepsilon ^\leftarrow )^m(x,t)(\zeta ^{\varepsilon ^q}_\kappa )' (u_\varepsilon ^\leftarrow (x,t))\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla u_\varepsilon ^\leftarrow (x,t)\cdot \nabla \big ((\zeta ^{\varepsilon ^q}_\kappa )'(u_\varepsilon ^\leftarrow (x,t))\xi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla u_\varepsilon ^\leftarrow (x,t) (\zeta ^{\varepsilon ^q}_\kappa )'(u_\varepsilon ^\leftarrow (x,t))\xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t, \end{aligned}$$
(41)
which is valid for each \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times } (0,T^*);[0,\infty ))\) on a set with probability one. Note that on \(\mathbf {R}^d\setminus ({-}\nu (B_t{-}B^\varepsilon _t)+\partial D)\) we may apply the chain rule to compute that \(\nabla u_\varepsilon ^\leftarrow (x,t) (\zeta ^{\varepsilon ^q}_\kappa )'(u_\varepsilon ^\leftarrow (x,t)) =\nabla (\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow )(x,t)\). Furthermore, note that \((\zeta ^{\varepsilon ^q}_\kappa )'(r)=0\) holds true for all \(r\le \kappa \). In particular, because of \(u_\varepsilon ^\leftarrow \in C(\mathbf {R}^d{\times }[0,T^*])\) and the choice \(\varepsilon \le \frac{\kappa }{2}\) we have for all \(t\in [0,T^*]\) that \((\zeta ^{\varepsilon ^q}_\kappa )'\circ u_\varepsilon ^\leftarrow (\cdot ,t) \equiv 0\) in a neighborhood of the interface \(({-}\nu (B_t{-}B^\varepsilon _t)+\partial D)\). This in turn means that we can integrate by parts in the first term on the right hand side of (41) without producing an additional surface integral. Taking all of these information together yields the bound
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} (\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow )\partial _t\xi \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad \le -\int _0^{T^*}\int _{\mathbf {R}^d}m(u_\varepsilon ^\leftarrow )^{m-1}|\nabla u_\varepsilon ^\leftarrow |^2 \big ((\zeta ^{\varepsilon ^q}_\kappa )''\circ u_\varepsilon ^\leftarrow \big )\xi \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m(u_\varepsilon ^\leftarrow )^{m-1} \nabla (\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow ) \cdot \nabla \xi \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2|\nabla u_\varepsilon ^\leftarrow |^2 \big ((\zeta ^{\varepsilon ^q}_\kappa )''\circ u_\varepsilon ^\leftarrow \big )\xi \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla (\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow ) \cdot \nabla \xi \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla (\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow ) \xi \,\mathrm {d}x\,\mathrm {d}B_t \end{aligned}$$
(42)
for each \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times } (0,T^*);[0,\infty ))\) almost surely. Exploiting the sign \((\zeta ^{\varepsilon ^q}_\kappa )''\ge 0\) and making use of the abbreviation \(v_\varepsilon =\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow \) we arrive at the estimate
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} v_\varepsilon (x,t)\partial _t\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad \le -\int _0^{T^*}\int _{\mathbf {R}^d}\nabla v_\varepsilon ^m(x,t) \cdot \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla v_\varepsilon (x,t) \cdot \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla v_\varepsilon (x,t)\xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) \nabla v_\varepsilon (x,t)\cdot \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\end{aligned}$$
(43)
which is valid for all \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times } (0,T^*);[0,\infty ))\) almost surely. An analogous estimate also holds true for the pair \((u_{{\hat{\varepsilon }}}^\leftarrow , v_{{\hat{\varepsilon }}}=\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}}^\leftarrow )\), i.e.,
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} v_{{\hat{\varepsilon }}}(y,s)\partial _t{\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\quad \le -\int _0^{T^*}\int _{\mathbf {R}^d}\nabla v_{{\hat{\varepsilon }}}^m(y,s) \cdot \nabla {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla v_{{\hat{\varepsilon }}}(y,s) \cdot \nabla {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \nu \nabla v_{{\hat{\varepsilon }}}(y,s){\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}B_s\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_{{\hat{\varepsilon }}}^\leftarrow )^{m-1}{-}v_{{\hat{\varepsilon }}}^{m-1}\big )(y,s) \nabla v_{{\hat{\varepsilon }}}(y,s)\cdot \nabla {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\end{aligned}$$
(44)
for all \({\tilde{\xi }}\in C^\infty _{\mathrm {cpt}}(D{\times } (0,T^*);[0,\infty ))\) almost surely.
Step 3 (Convex approximation \(\eta _\delta \) to \(r\mapsto |r|\) as test function): We proceed by testing the inequality (43) with \((\eta _\delta )'(v_\varepsilon (x,t){-}{\hat{z}})\xi (x,t)\) where the test function \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*);[0,\infty ))\) and \({\hat{z}}\in \mathbf {R}\) are arbitrary. This again incorporates several integration by parts which we do not want to produce any additional surface integrals. We reiterate that this will indeed not be the case since neither the Dirichlet data nor the Neumann data for \(v_\varepsilon \) jump across the interfaces \({-}\nu (B_t{-}B^\varepsilon _t)+\partial D\) for all \(t\in [0,T^*]\). Hence, arguing similar to the previous step using in particular the Steklov average and the convexity of \(\eta _\delta \) we obtain the estimate
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \partial _t\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad \le -\int _0^{T^*}\int _{\mathbf {R}^d}\nabla v_\varepsilon ^m(x,t) \cdot \nabla \big ((\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big )\xi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla v_\varepsilon (x,t) \cdot \nabla \big ((\eta _\delta )'(v_\varepsilon (x,t){-}{\hat{z}})\xi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) |\nabla v_\varepsilon (x,t)|^2(\eta _\delta )'' \big (v_\varepsilon (x,t){-}{\hat{z}}\big )\xi (x,t)\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) \nabla v_\varepsilon (x,t)\cdot (\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t)\,\mathrm {d}x\,\mathrm {d}t\end{aligned}$$
(45)
for all \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*);[0,\infty ))\) and all \({\hat{z}}\in \mathbf {R}\). As a preparation for what follows, we post-process the right hand side of the latter inequality. Integrating by parts, adding zero and using the chain rule we may rewrite the non-linear diffusion term as follows
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d}\nabla v_\varepsilon ^m(x,t) \cdot \nabla \big ((\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big )\xi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad =-\int _0^{T^*}\int _{\mathbf {R}^d}|\nabla v_\varepsilon ^m(x,t)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d}\eta _\delta \big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \Delta \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \Delta v_\varepsilon ^m(x,t)\xi (x,t) \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) -(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \big \} \,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(46)
Analogously one obtains for the Stratonovich correction term
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2\nabla v_\varepsilon (x,t) \cdot \nabla \big ((\eta _\delta )'(v_\varepsilon (x,t){-}{\hat{z}})\xi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad =-\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2|\nabla v_\varepsilon (x,t)|^2 (\eta _\delta )''\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \Delta \xi (x,t)\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(47)
Since \(\chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon }v_\varepsilon \le \chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon }u_\varepsilon ^\leftarrow \) by the definition of \(v_\varepsilon =\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow \) and the truncation \(\zeta ^{\varepsilon ^q}_\kappa \) we observe that the penultimate term in (45) has a favorable sign. Together with the two identities (46) and (47) the bound (45) thus yields for all \({\hat{z}}\in \mathbf {R}\) and all \(\xi \in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times } (0,T^*);[0,\infty ))\) almost surely the estimate
$$\begin{aligned}&-\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \partial _t\xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad \le -\int _0^{T^*}\int _{\mathbf {R}^d}|\nabla v_\varepsilon ^m(x,t)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d}\eta _\delta \big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \Delta \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2|\nabla v_\varepsilon (x,t)|^2 (\eta _\delta )''\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \Delta \xi (x,t) \,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t\nonumber \\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \Delta v_\varepsilon ^m(x,t)\xi (x,t) \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) -(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}{\hat{z}}^m\big ) \big \}\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) \nabla v_\varepsilon (x,t)\cdot (\eta _\delta )'\big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t)\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(48)
Step 4 (Substituting \({\hat{z}} \rightsquigarrow \hat{\mathcal {Z}}\) and taking expectation): We define for every \(\lambda >0\) a smooth and compactly supported cut-off \({\bar{\rho }}_\lambda :=\frac{1}{\lambda }{\bar{\rho }}(\frac{\cdot }{\lambda })\) by rescaling a standard even cut-off function \({\bar{\rho }}\in C^\infty _{\mathrm {cpt}}((-1,1);[0,\infty ))\) such that \(\int _\mathbf {R}{\bar{\rho }}(r)\,\mathrm {d}r=1\). We may then multiply the inequality (48) with the non-negative random variable \(\rho _{\lambda }({\hat{z}}{-}\hat{\mathcal {Z}})\), take the expected value of the resulting almost sure estimate, and finally integrate over \({\hat{z}}\in \mathbf {R}\). We claim that taking the limit \(\lambda \rightarrow 0\) produces the desired estimate (33).
To this end, we focus on the parameter dependent stochastic integral term \(\mathbf {R}\ni {\hat{z}} \mapsto X({\hat{z}}) := \int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla \xi (x,t) \,\mathrm {d}x\,\mathrm {d}B_t\). Note that because of the maximum principle bound (15), the definition \(v_\varepsilon =\zeta ^{\varepsilon ^q}_\kappa \circ u_\varepsilon ^\leftarrow \), as well as the Burkholder–Davis–Gundy inequality it holds \(\mathbf {E}|X'(0)| + \mathbf {E}|X''(0)| < \infty \). Hence, \(\mathbf {E}\big |X({\hat{z}})|_{{\hat{z}} = \hat{\mathcal {Z}}}\big | \le \Vert \hat{\mathcal {Z}}\Vert _{L^\infty }\mathbf {E}|X'(0)|<\infty \), \(\mathbf {E}\big |X'({\hat{z}})|_{{\hat{z}} = \hat{\mathcal {Z}}}\big | \le \Vert \hat{\mathcal {Z}}\Vert _{L^\infty }\mathbf {E}|X''(0)|<\infty \), and
$$\begin{aligned} \mathbf {E}\bigg |X({\hat{z}})\big |_{{\hat{z}} = \hat{\mathcal {Z}}} -\int _{\mathbf {R}} \rho _{\lambda }({\hat{z}}{-}\hat{\mathcal {Z}}) X({\hat{z}})\,\mathrm {d}{\hat{z}}\bigg | \le \int _{\mathbf {R}} {\hat{z}} \rho _{\lambda }({\hat{z}}) \,\mathrm {d}{\hat{z}} \,\mathbf {E}\big |X'({\hat{z}})|_{{\hat{z}} = \hat{\mathcal {Z}}}\big | \rightarrow 0 \text { as } \lambda \rightarrow 0. \end{aligned}$$
This yields the claim for the noise term appearing in (48). We observe that all the other terms can be dealt with based on the estimates (15)–(18) from Lemma 4, which in turn concludes the proof of (33).
Finally, by taking (44) instead of (43) as a starting point for the previous two steps we obtain along the same lines that
$$\begin{aligned}&-\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_{{\hat{\varepsilon }}}(y,s){-}\mathcal {Z}\big ) \partial _t{\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\quad \le -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}|\nabla v_{{\hat{\varepsilon }}}^m(y,s)|^2 (\eta _\delta )''\big (v_{{\hat{\varepsilon }}}^m(y,s){-}\mathcal {Z}^m\big ) {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\eta _\delta \big (v_{{\hat{\varepsilon }}}^m(y,s){-}\mathcal {Z}^m\big ) \Delta {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2|\nabla v_{{\hat{\varepsilon }}}(y,s)|^2 (\eta _\delta )''\big (v_{{\hat{\varepsilon }}}(y,s){-}\mathcal {Z}\big ) {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \eta _\delta \big (v_{{\hat{\varepsilon }}}(y,s){-}\mathcal {Z}\big ) \Delta {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_{{\hat{\varepsilon }}}(y,s){-}z\big ) \nabla {\tilde{\xi }}(y,s) \,\mathrm {d}y\,\mathrm {d}B_s\bigg |_{z=\mathcal {Z}}\nonumber \\&\qquad +\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \Delta v_{{\hat{\varepsilon }}}^m(y,s){\tilde{\xi }}(y,s) \big \{(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,s){-}\mathcal {Z}\big ) -(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}^m(y,s){-}\mathcal {Z}^m\big ) \big \} \,\mathrm {d}y\,\mathrm {d}s\nonumber \\&\qquad -\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}m\big ((u_{{\hat{\varepsilon }}}^\leftarrow )^{m-1}{-}v_{{\hat{\varepsilon }}}^{m-1}\big )(y,s) \nabla v_{{\hat{\varepsilon }}}(y,s)\cdot (\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,s){-}\mathcal {Z}\big ) \nabla {\tilde{\xi }}(y,s)\,\mathrm {d}y\,\mathrm {d}s, \end{aligned}$$
(49)
for all test functions \({\tilde{\xi }}\in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d\times (0,T^*);[0,\infty ))\), as well as all bounded random variables \(\mathcal {Z}\). This concludes the proof of Lemma 8. \(\square \)
We continue with the proof of Proposition 5. The next step takes care of the proper choice of test functions \(\xi (x,t)\) resp. \({\tilde{\xi }}(y,s)\) in the latter two estimates. After that, we start merging them by i) substituting \(\hat{\mathcal {Z}}=v_{{\hat{\varepsilon }}}(y,s)\) in (33) resp. \(\mathcal {Z}=v_{\varepsilon }(x,t)\) in (49), ii) integrating over the respective independent variables (y, s) resp. (x, t), and iii) summing the two resulting inequalities.
Consider the mollifier \(\rho \in C^\infty _{\mathrm {cpt}}((0,1);[0,\infty ))\) with \(\int _\mathbf {R}\rho (r)\,\mathrm {d}r\) already used in (7), and define for \(\tau >0\) the scaled kernel \(\rho _\tau :=\frac{1}{\tau }\rho (\frac{\cdot }{\tau })\). Let \(\varphi \in C^\infty _{\mathrm {cpt}}((0,T^*);[0,1])\) and fix another even mollifier \(\gamma \in C^\infty _{\mathrm {cpt}}(B_1;[0,\infty ))\) such that \(\int _{B_1}\gamma (x)\,\mathrm {d}x=1\). For \(\theta >0\) let \(\gamma _\theta :=\frac{1}{\theta ^d}\gamma (\frac{\cdot }{\theta })\). Now, since \(K\subset D\) is compact we can find a scale \(s_c\in (0,1)\) such that K is contained in \(D_{s_c}:=\{x\in D:\mathrm {dist}(x,\partial D)>s_c\}\). Moreover, because D has a \(C^2\) boundary \(\partial D\) there exists (cf. [1, Lemma 5.4]) a sequence \(({\bar{\xi }}_h)_h\) and a constant \(C=C(D)\) such that
-
(i)
\({\bar{\xi }}_h\in H^1_0(D;[0,1])\), \({\bar{\xi }}_h=\chi _{D}\) on \(\{x\in D:\mathrm {dist}(x,\partial D)\ge h\}\),
-
(ii)
it holds \(\int _{D}\nabla \phi \cdot \nabla {\bar{\xi }}_h\,\mathrm {d}x\ge 0\) for all \(\phi \in H^1_0(D;[0,\infty ))\),
-
(iii)
\({{\,\mathrm{supp}\,}}\nabla {\bar{\xi }}_h\subset \{x\in D:\mathrm {dist}(x,\partial D)< h\}\), and we have the bounds
$$\begin{aligned} C^{-1}\le \int _{D}|\nabla {\bar{\xi }}_h|\,\mathrm {d}x\le C,\quad \int _{D}|\nabla {\bar{\xi }}_h|^2\,\mathrm {d}x\le Ch^{-1}. \end{aligned}$$
(50)
We then fix once and for all a scale \(h\in (0,s_c)\), and set \({\bar{\xi }}:={\bar{\xi }}_h\). Note that \({\bar{\xi }} \equiv 1\) on K by the choice of \(s_c\) and h. For purely technical reasons, we actually consider in the following a mollified version of \({\bar{\xi }}\). Let \({\bar{\xi }}_l:=\gamma _{l}*{\bar{\xi }}\) for \(l>0\).
Let now \(y\in \mathbf {R}^d\) and \(s\in (0,T^*)\) be fixed. We then define the test function
$$\begin{aligned} \xi (x,t,y,s) := \rho _\tau (t{-}s)\varphi \Big (\frac{t{+}s}{2}\Big ) \gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ), \quad (x,t)\in \mathbf {R}^d{\times }(0,T^*). \end{aligned}$$
(51)
For this to be an admissible choice in (48) we need to restrict the range of the various parameters. Assuming that
$$\begin{aligned} 2\tau<\min \{\inf {{\,\mathrm{supp}\,}}\varphi ,T^*{-}\sup {{\,\mathrm{supp}\,}}\varphi \} \text { and } \theta \vee l<\frac{s_c}{4} \end{aligned}$$
(52)
we observe \(\xi (\cdot ,\cdot ,y,s)\in C^\infty _{\mathrm {cpt}}(\mathbf {R}^d{\times }(0,T^*);[0,\infty ))\) and is thus admissible for (48). Moreover, for every \(x\in \mathbf {R}^d\) and \(t\in (0,T^*)\) the test function
$$\begin{aligned} {\tilde{\xi }}(y,s,x,t) := \xi (x,t,y,s),\quad (y,s)\in \mathbf {R}^d{\times }(0,T^*) \end{aligned}$$
(53)
then also represents an admissible choice for (33) under the same restrictions (52) on the parameters. We have everything in place to merge (48) and (33).
Testing (33) with the admissible test functions \(\xi (\cdot ,\cdot ,y,s)\) from (51) for every \((y,s)\in \mathbf {R}^d{\times } (0,T^*)\), integrating over \((y,s)\in \mathbf {R}^d{\times } (0,T^*)\), then repeating everything with (49) based on the admissible test functions \({\tilde{\xi }}(\cdot ,\cdot ,x,t)\) from (53) for every \((x,t)\in \mathbf {R}^d{\times } (0,T^*)\), and finally summing the two resulting inequalities (using in particular that \(\eta _\delta \) is even) yields an estimate of the form
$$\begin{aligned}&\mathbf {E}R_{\mathrm {dt}} \le \mathbf {E}R_{\mathrm {porMed}} + \mathbf {E}R_{\mathrm {corr}} + \mathbf {E}R_{\mathrm {noise}} + \mathbf {E}R_{\mathrm {error}} \end{aligned}$$
(54)
for all \((\tau ,\theta ,l)\) subject to (52), all \(\delta >0\) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). Here, we introduced for convenience the abbreviations
$$\begin{aligned} R_{\mathrm {dt}} :=&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) (\partial _t{+}\partial _s)\xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {porMed}}:=&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} |\nabla v_\varepsilon ^m(x,t)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big )\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&+\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) \Delta _x\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} |\nabla v_{{\hat{\varepsilon }}}^m(y,s)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big )\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t\\&+\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) \Delta _y\xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {corr}} :=&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2|\nabla v_\varepsilon (x,t)|^2 (\eta _\delta )''\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big )\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&+\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) \Delta _x\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2|\nabla v_{{\hat{\varepsilon }}}(y,s)|^2 (\eta _\delta )''\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big )\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t\\&+\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) \Delta _y\xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {noise}} :=&-\int _0^{T^*}\int _{\mathbf {R}^d} \int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla _x\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}B_t\bigg |_{{\hat{z}}=v_{{\hat{\varepsilon }}}(y,s)} \,\mathrm {d}y\,\mathrm {d}s\\&-\int _0^{T^*}\int _{\mathbf {R}^d} \int _0^{T^*}\int _{\mathbf {R}^d} \nu \eta _\delta \big (z{-}v_{{\hat{\varepsilon }}}(y,s)\big ) \nabla _y\xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}B_s\bigg |_{z=v_\varepsilon (x,t)} \,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
as well as
$$\begin{aligned} R_{\mathrm {error}}&:=\int _0^{T^*}\int _{\mathbf {R}^d} \int _0^{T^*}\int _{\mathbf {R}^d} \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) {-}(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big )\big \}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times (\Delta _x v_\varepsilon ^m)(x,t)\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&\qquad +\int _0^{T^*}\int _{\mathbf {R}^d} \int _0^{T^*}\int _{\mathbf {R}^d} \big \{(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,s){-}v_{\varepsilon }(x,t)\big ) -(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}^m(y,s){-}v_{\varepsilon }^m(x,t)\big )\big \}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times (\Delta _y v_{{\hat{\varepsilon }}}^m)(y,s){\tilde{\xi }}(y,s,x,t) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t)\nabla _x\xi (x,t,y,s)\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \nabla v_\varepsilon (x,t) (\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&\qquad -\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} m\big ((u_{{\hat{\varepsilon }}}^\leftarrow )^{m-1}{-}v_{{\hat{\varepsilon }}}^{m-1}\big )(y,s)\nabla _y{\tilde{\xi }}(y,s,x,t)\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \nabla v_{{\hat{\varepsilon }}}(y,s) (\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,s){-}v_{\varepsilon }(x,t)\big ) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
Before we move on with removing the doubling in the time variable by studying the limit \(\tau \rightarrow 0\) let us first perform some computations on the non-linear diffusion term \(R_{\mathrm {porMed}}\) and the correction term \(R_{\mathrm {corr}}\). Exploiting that \((\eta _\delta )''\ge 0\) and \(\xi \ge 0\), completing the square \(|\nabla v_{\varepsilon }^m|^2{+}|\nabla v_{{\hat{\varepsilon }}}^m|^2 =|\nabla v_{\varepsilon }^m{-}\nabla v_{{\hat{\varepsilon }}}^m|^2 +2\nabla v_{\varepsilon }^m\cdot \nabla v_{{\hat{\varepsilon }}}^m\) and integrating by parts entails that
$$\begin{aligned}&-\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} |\nabla v_\varepsilon ^m(x,t)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) \xi \,\mathrm {d}x\,\mathrm {d}y\\&\qquad -\int _{\mathbf {R}^d}\int _{\mathbf {R}^d}|\nabla v_{{\hat{\varepsilon }}}^m(y,s)|^2 (\eta _\delta )''\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) \xi \,\mathrm {d}x\,\mathrm {d}y\\&\quad \le \int _{\mathbf {R}^d}\int _{\mathbf {R}^d} 2\xi (\nabla _y\cdot \nabla _x) \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) \,\mathrm {d}x\,\mathrm {d}y\\&\quad = \int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big ) 2(\nabla _x\cdot \nabla _y)\xi \,\mathrm {d}x\,\mathrm {d}y. \end{aligned}$$
Since \((\Delta _x{+}\Delta _y)(\gamma _\theta (x{-}y){\bar{\xi }}_l(\frac{x{+}y}{2})) =\gamma _\theta (x{-}y)\frac{1}{2}\Delta {\bar{\xi }}_l(\frac{x{+}y}{2}) {+}{\bar{\xi }}_l(\frac{x{+}y}{2})2\Delta \gamma _\theta (x{-}y)\) and \((2\nabla _x\cdot \nabla _y)(\gamma _\theta (x{-}y){\bar{\xi }}_l(\frac{x{+}y}{2})) =\gamma _\theta (x{-}y)\frac{1}{2}\Delta {\bar{\xi }}_l(\frac{x{+}y}{2}) {-}{\bar{\xi }}_l(\frac{x{+}y}{2})2\Delta \gamma _\theta (x{-}y)\), we thus obtain the estimate
$$\begin{aligned} R_{\mathrm {porMed}} \le&\int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,s)\big )\\&\qquad \qquad \qquad \qquad \qquad \times \rho _\tau (t{-}s)\varphi \Big (\frac{t{+}s}{2}\Big ) \gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\ =:&R^{(1)}_{\mathrm {porMed}}. \end{aligned}$$
The idea eventually is—after letting \(\tau \rightarrow 0\), \(\delta \rightarrow 0\) and removing the doubling in the spatial variables (the latter by fine-tuning the scales \(\theta >0\) and \(l>0\) as suitably chosen powers of \(\varepsilon \vee {\hat{\varepsilon }}\))—to integrate by parts and to use the sign in condition ii) for the spatial test function \({\bar{\xi }}\). We will make this precise together with all the required error estimates in a later stage of the proof. For the moment, we only wish to mention that because of the convexity of \(\eta _\delta \) and \(\xi \ge 0\) it holds
$$\begin{aligned} R_{\mathrm {corr}}&\le \int _0^{T^*}\int _{\mathbf {R}^d}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,s)\big ) (\Delta _x{+}\Delta _y)\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}t\,\mathrm {d}y\,\mathrm {d}s\\&=: R^{(1)}_{\mathrm {corr}}. \end{aligned}$$
The task therefore reduces to post-process the bound
$$\begin{aligned}&\mathbf {E}R_{\mathrm {dt}} \le \mathbf {E}R_{\mathrm {porMed}}^{(1)} + \mathbf {E}R_{\mathrm {corr}}^{(1)} + \mathbf {E}R_{\mathrm {noise}} + \mathbf {E}R_{\mathrm {error}} \end{aligned}$$
(55)
with the remaining three terms left unchanged from the estimate (54). In a first step we aim to remove the doubling in the time variable by letting \(\tau \rightarrow 0\).
Lemma 9
Let the assumptions and notation of Sect. 4.2 until this point be in place. Define the quantities
$$\begin{aligned} R_{\mathrm {dt}}^{(1)}:=&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) \gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\frac{\mathrm {d}}{\mathrm {d}t}\varphi (t) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
(56)
$$\begin{aligned} R_{\mathrm {porMed}}^{(2)}:=&\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \eta _\delta \big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big ) \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
(57)
$$\begin{aligned} R_{\mathrm {corr}}^{(2)}:=&\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
(58)
$$\begin{aligned} R_{\mathrm {error}}^{(1)}:=&\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) {-}(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big )\big \}\nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \times (\Delta _x v_\varepsilon ^m)(x,t)\gamma _\theta (x{-}y) {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\varphi (t)\,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&+\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big \{(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,t){-}v_{\varepsilon }(x,t)\big ) -(\eta _\delta )'\big (v_{{\hat{\varepsilon }}}^m(y,t){-}v_{\varepsilon }^m(x,t)\big )\big \}\nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \times (\Delta _y v_{{\hat{\varepsilon }}}^m)(y,t)\gamma _\theta (x{-}y) {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\varphi (t)\,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} m\big ((u_\varepsilon ^\leftarrow )^{m-1}{-}v_\varepsilon ^{m-1}\big )(x,t) \nabla _x\Big (\gamma _\theta (x{-}y) {\bar{\xi }}_{l}\Big (\frac{x{+}y}{2}\Big )\Big )\varphi (t)\nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \nabla v_\varepsilon (x,t) (\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&-\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} m\big ((u_{{\hat{\varepsilon }}}^\leftarrow )^{m-1}{-}v_{{\hat{\varepsilon }}}^{m-1}\big )(y,t) \nabla _y\Big (\gamma _\theta (x{-}y) {\bar{\xi }}_{l}\Big (\frac{x{+}y}{2}\Big )\Big )\varphi (t)\nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \nabla v_{{\hat{\varepsilon }}}(y,t) (\eta _\delta )'\big (v_{{\hat{\varepsilon }}}(y,t){-}v_{\varepsilon }(x,t)\big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(59)
Then the estimate
$$\begin{aligned}&\mathbf {E}R_{\mathrm {dt}}^{(1)} \le \mathbf {E}R_{\mathrm {porMed}}^{(2)} + \mathbf {E}R_{\mathrm {corr}}^{(2)} + \mathbf {E}R_{\mathrm {error}}^{(1)} \end{aligned}$$
(60)
holds true for all \((\theta ,l)\) subject to (52), all \(\delta >0\) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\).
Proof
It follows from (51) that \((\partial _t{+}\partial _s)\xi =\varphi '(\frac{t{+}s}{2})\rho _\tau (t{-}s)\gamma _\theta (x{-}y){\bar{\xi }}_l(\frac{x{+}y}{2})\). In particular, the singular terms as \(\tau \rightarrow 0\) cancel. Hence, it follows by Lebesgue’s dominated convergence based on the regularity and the bounds for the Wong–Zakai approximation \(u_\varepsilon \) from Lemma 4 and definition (22) of the shifted densities \(u_\varepsilon ^\leftarrow \) that
$$\begin{aligned} \mathbf {E}R_{\mathrm {dt}} \rightarrow \mathbf {E}R_{\mathrm {dt}}^{(1)} \text { as } \tau \rightarrow 0. \end{aligned}$$
(61)
Relying again on Lebesgue’s dominated convergence due to the regularity and the a priori estimates for the Wong–Zakai approximation \(u_\varepsilon \) from Lemma 4 and the definition (22) of the shifted densities \(u_\varepsilon ^\leftarrow \), we may also easily pass to the limit \(\tau \rightarrow 0\) in all the terms on the right hand side of (55) except for the noise term \(R_{\mathrm {noise}}\). More precisely, we obtain
$$\begin{aligned} \mathbf {E}R_{\mathrm {porMed}}^{(1)}&\rightarrow \mathbf {E}R_{\mathrm {porMed}}^{(2)}&\text { as } \tau \rightarrow 0, \end{aligned}$$
(62)
$$\begin{aligned} \mathbf {E}R_{\mathrm {corr}}^{(1)}&\rightarrow \mathbf {E}{\widetilde{R}}_{\mathrm {corr}}^{(2)},&\text { as } \tau \rightarrow 0, \end{aligned}$$
(63)
$$\begin{aligned} \mathbf {E}R_{\mathrm {error}}&\rightarrow \mathbf {E}R_{\mathrm {error}}^{(1)}&\text { as } \tau \rightarrow 0, \end{aligned}$$
(64)
with the shorthand
$$\begin{aligned} {\widetilde{R}}_{\mathrm {corr}}^{(2)} :=&\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big )\varphi (t)\\&\qquad \qquad \qquad \quad \times (\Delta _x{+}\Delta _y)\Big (\gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
We proceed with the discussion of the noise term \(R_{\mathrm {noise}}\), and claim that as \(\tau \rightarrow 0\)
$$\begin{aligned}&\mathbf {E}R_{\mathrm {noise}} \rightarrow \int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big )\varphi (t)\nonumber \\&\qquad \qquad \qquad \quad \qquad \qquad \quad \quad \times 2(\nabla _x\cdot \nabla _y)\Big (\gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(65)
Recall that \((\Delta _x{+}\Delta _y)(\gamma _\theta (x{-}y){\bar{\xi }}_l(\frac{x{+}y}{2})) =\gamma _\theta (x{-}y)\frac{1}{2}\Delta {\bar{\xi }}_l(\frac{x{+}y}{2}) {+}{\bar{\xi }}_l(\frac{x{+}y}{2})2\Delta \gamma _\theta (x{-}y)\) and \(2(\nabla _x\cdot \nabla _y)(\gamma _\theta (x{-}y){\bar{\xi }}_l(\frac{x{+}y}{2})) =\gamma _\theta (x{-}y)\frac{1}{2}\Delta {\bar{\xi }}_l(\frac{x{+}y}{2}) {-}{\bar{\xi }}_l(\frac{x{+}y}{2})2\Delta \gamma _\theta (x{-}y)\). Hence, the combination of (63) and (65) yields
$$\begin{aligned} \mathbf {E}R_{\mathrm {corr}}^{(1)} + \mathbf {E}R_{\mathrm {noise}} \rightarrow \mathbf {E}R_{\mathrm {corr}}^{(2)} \end{aligned}$$
(66)
as \(\tau \rightarrow 0\), which in view of (61), (62) and (64) entails the desired estimate (60). Hence, it remains to verify (65).
To this end, we again define for \(\lambda >0\) the cut-off \({\bar{\rho }}_\lambda :=\frac{1}{\lambda }{\bar{\rho }}(\frac{\cdot }{\lambda })\) by means of a standard even cut-off function \({\bar{\rho }}\in C^\infty _{\mathrm {cpt}}((-1,1);[0,\infty ))\) such that \(\int _\mathbf {R}{\bar{\rho }}(r)\,\mathrm {d}r{=}1\). Furthermore, let \(X(y,s;{\hat{z}}) := -\int _s^{s{+}\tau }\int _{\mathbf {R}^d} \nu \eta _\delta \big (v_\varepsilon (x,t){-}{\hat{z}}\big ) \nabla _x\xi (x,t,y,s) \,\mathrm {d}x\,\mathrm {d}B_t\) and \(Y(x,t;z) := -\int _{t-\tau }^{t}\int _{\mathbf {R}^d} \nu \eta _\delta \big (z{-}v_{{\hat{\varepsilon }}}(y,s)\big ) \nabla _y\xi (x,t,y,s) \,\mathrm {d}y\,\mathrm {d}B_s\). Hence, using that the mollifier \(\rho \) is supported on the positive real axis, we may rewrite for all \(\tau >0\) subject to (52)
$$\begin{aligned} R_{\mathrm {noise}}&= \int _{0}^{T^*{-}\tau }\int _{\mathbf {R}^d} X(y,s;{\hat{z}})\big |_{{\hat{z}} = v_{{\hat{\varepsilon }}}(y,s)} \,\mathrm {d}y\,\mathrm {d}s+ \int _{\tau }^{T^*}\int _{\mathbf {R}^d} Y(x,t;z)\big |_{z = v_{\varepsilon }(x,t)} \,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
Based on the argument given in Step 4 of the proof of Lemma 8, it holds
$$\begin{aligned} \mathbf {E}R_{\mathrm {noise}} =&\lim _{\lambda \rightarrow 0}\mathbf {E}\int _{0}^{T^*-\tau }\int _{\mathbf {R}^d} \int _{\mathbf {R}} {\bar{\rho }}_{\lambda }\big (v_{{\hat{\varepsilon }}}(y,s){-}{\hat{z}}\big ) X(y,s;{\hat{z}})\,\mathrm {d}{\hat{z}} \,\mathrm {d}y\,\mathrm {d}s\\&+ \lim _{\lambda \rightarrow 0}\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} {\bar{\rho }}_{\lambda }\big (v_{\varepsilon }(x,t){-} z\big ) Y(x,t;z)\,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\\ =:&\lim _{\lambda \rightarrow 0} \mathbf {E}R_{\mathrm {noise}}^{(1),\lambda } + \lim _{\lambda \rightarrow 0} \mathbf {E}R_{\mathrm {noise}}^{(2),\lambda }. \end{aligned}$$
Observe that \(\mathbf {E}R_{\mathrm {noise}}^{(1),\lambda }=0\) since the bounded random variable \(v_{{\hat{\varepsilon }}}(y,s)\) is measurable with respect to \(\mathcal {F}_s\). Similarly, due to the fact that \(v_{\varepsilon }(x,t{-}\tau )\) is bounded and measurable with respect to \(\mathcal {F}_{t-\tau }\) we may actually write for all \(\lambda \in (0,1]\)
$$\begin{aligned} \mathbf {E}R_{\mathrm {noise}}^{(2),\lambda }&= \mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} Y(x,t;z)\Big \{{\bar{\rho }}_{\lambda }\big (v_{\varepsilon }(x,t){-} z\big ){-} {\bar{\rho }}_{\lambda }\big (v_{\varepsilon }(x,t{-}\tau ){-} z\big )\Big \}\,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
(67)
As a preparation for what follows, we present regularity estimates for Y. More precisely, we claim that for all \(m,n\ge 1\) there exists a constant \(C>0\) such that
$$\begin{aligned} \Big (\mathbf {E}\Vert \partial _z^m\nabla _x^nY\Vert ^8_{L^\infty _{x,t,z}(\mathbf {R}^d{\times }[0,T]{\times }\mathbf {R})}\Big )^\frac{1}{8} \le C\tau ^{-\frac{3}{4}}, \end{aligned}$$
(68)
uniformly over all \(\tau \) subject to (52). For a proof of (68), it suffices to show that
$$\begin{aligned} \Big (\mathbf {E}\Vert \partial _z^{{\widehat{m}}}\nabla _x^{\widehat{n}}Y \Vert ^8_{L^8_t([0,T];L^8_{x,z}(\mathbf {R}^d{\times }\mathbf {R}))}\Big )^\frac{1}{8}&\le C_{{\widehat{m}},{\widehat{n}}} \tau ^{-\frac{1}{2}}, \end{aligned}$$
(69)
$$\begin{aligned} \Big (\mathbf {E}\Vert \partial _t\partial _z^{\widehat{m}}\nabla _x^{{\widehat{n}}}Y \Vert ^8_{L^8_t([0,T];L^8_{x,z}(\mathbf {R}^d{\times }\mathbf {R}))}\Big )^\frac{1}{8}&\le C_{{\widehat{m}},{\widehat{n}}} \tau ^{-\frac{3}{2}}, \end{aligned}$$
(70)
uniformly over all \(\tau \) subject to (52). Indeed, by passing through a fractional Sobolev space in the time variable we may interpolate between the previous two estimates to obtain \(\mathbf {E}\Vert \partial _z^{{\widehat{m}}}\nabla _x^{\widehat{n}}Y\Vert ^8_{W^{8,\delta }_t([0,T];L^8_{x,z}(\mathbf {R}^d{\times }\mathbf {R}))} \le C_{\delta ,{\widehat{m}},{\widehat{n}}}\tau ^{8(-\delta -\frac{1}{2})}\), uniformly over all \(\tau \) subject to (52). In particular, fixing \(\delta =\frac{1}{4}\) in combination with Sobolev embedding allows to deduce (68). For a proof of (69), we may appeal to the Burkholder–Davis–Gundy inequality and the maximum principle bound (15), which imply the desired estimate in form of (with the scaling in \(\tau \) resulting from \(\rho _\tau \))
$$\begin{aligned}&\mathbf {E}\Vert \partial _z^{{\widehat{m}}}\nabla _x^{{\widehat{n}}}Y \Vert ^8_{L^8_t([0,T];L^8_{x,z}(\mathbf {R}^d{\times }\mathbf {R}))}\\&\quad \le C_{{\widehat{m}}}\tau ^{-8}\int _{\tau }^{T^*}\int _{\mathbf {R}^d}\mathbf {E}\, \bigg | \int _{t-\tau }^t \int _{B_\theta (x)} \Big | \nabla _x^{\widehat{n}}\nabla _y\Big (\gamma _\theta (x{-}y) {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\Big )\Big |^2 \,\mathrm {d}y\,\mathrm {d}s\,\bigg |^4 \,\mathrm {d}x\,\mathrm {d}t\\&\quad \le C_{{\widehat{m}},{\widehat{n}}} \tau ^{-4}. \end{aligned}$$
Similarly, one derives the estimate (70) with the scaling in \(\tau \) stemming from \(\partial _t\rho _\tau \).
Let now \(f:\mathbf {R}^d\times (0,T^*)\rightarrow [0,\infty )\) be a smooth and compactly supported space-time mollifier. It then follows from (39) that \(f*u_{\varepsilon }^{\leftarrow }\) satisfies pointwise
$$\begin{aligned} \mathrm {d}(f*u_{\varepsilon }^{\leftarrow })_t =\Big \{\Delta _x(f*(u_\varepsilon ^{\leftarrow })^m) + \frac{1}{2}\nu ^2\Delta _x(f*u_{\varepsilon }^{\leftarrow })\Big \}\,\mathrm {d}t + \nu \nabla _x(f*u_\varepsilon ^{\leftarrow })\cdot \mathrm {d}B_t. \end{aligned}$$
Applying Itô’s formula to \(F_z\circ (f*u_\varepsilon ^\leftarrow )\), \(F_z:={\bar{\rho }}_{\lambda }\big (\zeta ^{\varepsilon ^q}_\kappa (\cdot )-z\big )\in C^\infty (\mathbf {R})\), thus yields
$$\begin{aligned}&F_z\big ((f*u_\varepsilon ^\leftarrow )(x,t)\big ) - F_z\big ((f*u_\varepsilon ^\leftarrow )(x,t{-}\tau )\big )\nonumber \\&\quad = \int _{t{-}\tau }^t F'_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big ) \Big \{\Delta _x(f*(u_\varepsilon ^{\leftarrow })^m) + \frac{1}{2}\nu ^2\Delta _x(f*u_{\varepsilon }^{\leftarrow })\Big \} (x,\ell )\,\mathrm {d}\ell \nonumber \\&\qquad + \int _{t{-}\tau }^t \frac{1}{2}\nu ^2 F''_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big ) |\nabla _x(f*u_{\varepsilon }^{\leftarrow })|^2(x,\ell )\,\mathrm {d}\ell \nonumber \\&\qquad +\int _{t{-}\tau }^t \nu F'_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big ) \nabla _x(f*u_{\varepsilon }^{\leftarrow })(x,\ell )\,\mathrm {d}B_\ell \end{aligned}$$
(71)
for all \(x\in \mathbf {R}^d\), all \(t\in (\tau ,T^*)\) and all \(z\in \mathbf {R}\). The left hand side term of (71) represents a proxy for \({\bar{\rho }}_{\lambda }\big (v_{\varepsilon }(x,t){-} z\big ){-} {\bar{\rho }}_{\lambda }\big (v_{\varepsilon }(x,t{-}\tau ){-} z\big )\) in (67). Substituting this proxy into the right hand side term of (67), then inserting the identity (71), integrating by parts once in the terms with second-order spatial derivatives, and exploiting Itô’s isometry for the third right hand side term of (71), we deduce that it holds
$$\begin{aligned}&\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} Y(x,t;z)\Big \{F_z\big ((f*u_\varepsilon ^\leftarrow )(x,t)\big ) - F_z\big ((f*u_\varepsilon ^\leftarrow )(x,t{-}\tau )\big )\Big \}\,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\\&\quad = -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \nabla _x Y(x,t;z) \cdot \int _{t{-}\tau }^t F'_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big )\\&\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \quad \quad \times \big (f*\nabla _x (u_\varepsilon ^{\leftarrow })^m\big )(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} Y(x,t;z) \int _{t{-}\tau }^t F''_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big )\\&\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \quad \times \big (f*m(u_\varepsilon ^{\leftarrow })^{m-1}|\nabla _x u_\varepsilon ^{\leftarrow }|^2\big ) (x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \nabla _x Y(x,t;z) \cdot \int _{t{-}\tau }^t F'_z\big ((f*u_\varepsilon ^\leftarrow )(x,\ell )\big )\\&\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \quad \quad \times \frac{1}{2}\nu ^2(f*\nabla _x u_\varepsilon ^{\leftarrow })(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\\&\qquad - \mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \int _{t{-}\tau }^t\int _{\mathbf {R}^d} \nu ^2 \eta _\delta \big (z{-}v_{{\hat{\varepsilon }}}(y,s)\big )\nabla _y\xi (x,t,y,s)\\&\qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \cdot \nabla _x \big (F_z\circ (f*u_\varepsilon ^{\leftarrow })\big )(x,s) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
Based on the regularity estimates for Y, the estimates for \(u_\varepsilon \) from Lemma 4, and the support properties of the test function \(\xi \), we may let the mollifier f run through a space-time Dirac sequence and pass to the limit. This updates (67) to
$$\begin{aligned}&\mathbf {E}R_{\mathrm {noise}}^{(2),\lambda }\nonumber \\&\quad = -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \partial _z\nabla _x Y(x,t;z) \cdot \int _{t{-}\tau }^t {\bar{\rho }}_\lambda \big (v_\varepsilon (x,\ell ){-}z\big ) (\zeta ^{\varepsilon ^q}_\kappa )'\big (u_\varepsilon ^{\leftarrow }(x,\ell )\big ) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \qquad \times \nabla _x (u_\varepsilon ^{\leftarrow })^m(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \partial _z\nabla _x Y(x,t;z) \cdot \int _{t{-}\tau }^t {\bar{\rho }}_\lambda \big (v_\varepsilon (x,\ell ){-}z\big ) (\zeta ^{\varepsilon ^q}_\kappa )'\big (u_\varepsilon ^{\leftarrow }(x,\ell )\big )\nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \qquad \times \frac{1}{2}\nu ^2\nabla _x u_\varepsilon ^{\leftarrow }(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \partial _z Y(x,t;z) \int _{t{-}\tau }^t {\bar{\rho }}_\lambda \big (v_\varepsilon (x,\ell ){-}z\big ) (\zeta ^{\varepsilon ^q}_\kappa )''\big (u_\varepsilon ^{\leftarrow }(x,\ell )\big )\nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \times m(u_\varepsilon ^{\leftarrow })^{m-1}(x,\ell )|\nabla _x u_\varepsilon ^{\leftarrow }|^2(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad -\mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}} \partial _z^2 Y(x,t;z) \int _{t{-}\tau }^t {\bar{\rho }}_\lambda \big (v_\varepsilon (x,\ell ){-}z\big ) \big |(\zeta ^{\varepsilon ^q}_\kappa )'\big (u_\varepsilon ^{\leftarrow }(x,\ell )\big )\big |^2 \nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \quad \qquad \times m(u_\varepsilon ^{\leftarrow })^{m-1}(x,\ell )|\nabla _x u_\varepsilon ^{\leftarrow }|^2(x,\ell ) \,\mathrm {d}\ell \,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\qquad - \mathbf {E}\int _{\tau }^{T^*}\int _{\mathbf {R}^d} \int _{\mathbf {R}}\int _{t{-}\tau }^t\int _{\mathbf {R}^d} \nu ^2 \eta _\delta \big (z{-}v_{{\hat{\varepsilon }}}(y,s)\big )\nabla _y\xi (x,t,y,s) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \quad \cdot \nabla _x \big ({\bar{\rho }}_\lambda (v_\varepsilon (x,s){-}z)\big ) \,\mathrm {d}y\,\mathrm {d}s\,\mathrm {d}z\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\quad =: \mathbf {E}R_{\mathrm {noise}}^{(3),\lambda } + \mathbf {E}R_{\mathrm {noise}}^{(4),\lambda } + \mathbf {E}R_{\mathrm {noise}}^{(5),\lambda } + \mathbf {E}R_{\mathrm {noise}}^{(6),\lambda } + \mathbf {E}R_{\mathrm {noise}}^{(7),\lambda }, \end{aligned}$$
(72)
where we also integrated by parts in the z-variable to avoid derivatives of \({\bar{\rho }}_{\lambda }\) (which is required for \(\lambda \rightarrow 0\)).
The upshot of the argument is now the following. Based on the regularity estimates for Y, the estimates for \(u_\varepsilon \) from Lemma 4, and the support properties of the test function \(\xi \), we claim that it holds
$$\begin{aligned} \lim _{\tau \rightarrow 0}\lim _{\lambda \rightarrow 0} \sum _{i=3}^6 \mathbf {E}R_{\mathrm {noise}}^{(i),\lambda } = 0. \end{aligned}$$
(73)
We give details for the term \(R_{\mathrm {noise}}^{(5),\lambda }\); the other terms may be dealt with similarly. By Hölder’s inequality with respect to the exponents \((q,\frac{q+1}{q})\), \(q = 8\), the bound \(\int _{\tau }^{T^*}\int _{t-{\tau }}^t |g(\ell )| \,\mathrm {d}\ell \,\mathrm {d}t\le \tau \int _0^{T^*}|g(t)| \,\mathrm {d}t\), and the estimate (15), we obtain
$$\begin{aligned}&\mathbf {E}\big |R_{\mathrm {noise}}^{(5),\lambda }\big |\\&\quad \le \tau \Big (\mathbf {E}\Vert \partial _zY\Vert ^q_{L^\infty _{x,t,z}(\mathbf {R}^d{\times }[0,T]{\times }\mathbf {R})}\Big )^\frac{1}{q} \bigg (\mathbf {E}\,\bigg |\int _0^{T^*}\int _{\mathbf {R}^d}m(u_\varepsilon ^{\leftarrow })^{m-1} |\nabla u^{\leftarrow }_{\varepsilon }|^2\,\mathrm {d}x\,\mathrm {d}t\,\bigg |^\frac{q+1}{q}\bigg )^\frac{q}{q+1} \end{aligned}$$
uniformly over all \(\tau \) subject to (52) and all \(\lambda \in (0,1]\). By a change of variables to switch from \(u_\varepsilon ^{\leftarrow }\) to \(u_\varepsilon \), the energy estimate (16), as well as the the regularity estimate (68) for Y, we may deduce from the previous display that \(\mathbf {E}\big |R_{\mathrm {noise}}^{(5),\lambda }\big | \le C\tau ^{\frac{1}{4}}\) uniformly over all \(\tau \) subject to (52) and all \(\lambda \in (0,1]\). Hence, we obtain as claimed \(\lim _{\tau \rightarrow 0}\lim _{\lambda \rightarrow 0}\mathbf {E}R_{\mathrm {noise}}^{(5),\lambda } = 0\).
It remains to consider the term \(R_{\mathrm {noise}}^{(7),\lambda }\). However, after performing an integration by parts it directly follows (the limits \(\tau \rightarrow 0\) and \(\lambda \rightarrow 0\) are unproblematic by the support properties of \(\xi \), and the regularity properties of \(u_\varepsilon \) due to Lemma 4)
$$\begin{aligned} \begin{aligned} \lim _{\tau \rightarrow 0}\lim _{\lambda \rightarrow 0} \mathbf {E}R_{\mathrm {noise}}^{(7),\lambda } =&\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\eta _\delta \big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big )\varphi (t)\\&\qquad \qquad \qquad \quad \times 2(\nabla _x\cdot \nabla _y)\Big (\gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t. \end{aligned} \end{aligned}$$
(74)
The combination of (72)–(74) finally entails (65), which in turn concludes the proof of Lemma 9 as already argued above. \(\square \)
We continue with the proof of Proposition 5 taking care in the next step of the error terms. More precisely, we may derive the following bound.
Lemma 10
Let the assumptions and notation of Sect. 4.2 until this point be in place. In particular, recall the definition of the error term \(R^{(1)}_{\mathrm {error}}\) from (59). We then have the estimate
$$\begin{aligned} \mathbf {E}R^{(1)}_{\mathrm {error}} \le C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + O_{\delta \rightarrow 0}(1) \end{aligned}$$
(75)
for all \((\theta ,l)\) subject to (52), all \(\delta >0\) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\).
Proof
We start estimating by Hölder’s inequality
$$\begin{aligned}&\mathbf {E}\bigg |\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} (\Delta _x v_\varepsilon ^m)(x,t) \gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\varphi (t)\\&\qquad \qquad \qquad \qquad \times \big \{(\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) {-}(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big )\big \} \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\bigg |\\&\quad \le \bigg (\mathbf {E}\int _{0}^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big |(\eta _\delta )'\big (v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big ) {-}(\eta _\delta )'\big (v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big )\big |^2 \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\bigg )^\frac{1}{2}\\&\qquad \times \frac{\Vert \varphi \Vert _{L^\infty (0,T^*)}}{(\inf {{\,\mathrm{supp}\,}}\varphi )^\frac{1}{2}} \bigg (\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} t|(\Delta _x v_\varepsilon ^m)(x,t)|^2 \,\mathrm {d}x\,\mathrm {d}t\bigg )^\frac{1}{2} \Vert {\bar{\xi }}_l\Vert _{L^\infty (\mathbf {R}^d)}\Vert \gamma _\theta \Vert _{L^\infty (\mathbf {R}^d)}. \end{aligned}$$
However, by the bounds (18) and (30), the convergence (31) and the fact that \(\mathrm {sign}(a{-}b)=\mathrm {sign}(a^m{-}b^m)\) due to the monotonicity of \(r\mapsto r^m\) we infer by an application of Lebesgue’s dominated convergence theorem based on the regularity of the Wong–Zakai approximation \(u_\varepsilon \) from Lemma 4 that the term on the right hand side of the latter bound vanishes as \(\delta \rightarrow 0\).
Recall that \(\kappa \le v_\varepsilon \le u_\varepsilon ^\leftarrow \) on \({{\,\mathrm{supp}\,}}\nabla v_\varepsilon \). Hence, we can estimate by means of (21)
$$\begin{aligned}&\chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon } \big |(u_\varepsilon ^\leftarrow )^{m-1}(x,t){-}v_\varepsilon ^{m-1}(x,t)\big |\\&\quad \le \chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon } \sup _{r\in [v_\varepsilon (x,t),u^\leftarrow _\varepsilon (x,t)]} (m{-}1)r^{m-2} |v_\varepsilon (x,t){-}u_\varepsilon ^\leftarrow (x,t)|\\&\quad \le C(\kappa )\chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon }\varepsilon ^q. \end{aligned}$$
We may then estimate using also (19), the definition (22) of the shifted densities \(u^\leftarrow _\varepsilon \), \(\Vert \nabla {\bar{\xi }}_l\Vert _{L^\infty (\mathbf {R}^d)}\le \Vert \nabla \gamma _l\Vert _{L^1(\mathbf {R}^d)}\Vert {\bar{\xi }}\Vert _{L^\infty (\mathbf {R}^d)}\le Cl^{-1}\) (which follows from Young’s inequality and condition i) for the spatial cut-off \({\bar{\xi }}\)), \(\Vert \varphi \Vert _{L^\infty (0,T^*)}\le 1\), a change of variables \(x\mapsto x{-}\nu (B_t{-}B^\varepsilon _t)\), the fact that \(u_\varepsilon ^\leftarrow \ge \kappa \) on \({{\,\mathrm{supp}\,}}\nabla v_\varepsilon \), and finally the a priori estimate (16)
$$\begin{aligned}&\mathbf {E}\bigg |\int _{0}^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big ((u_\varepsilon ^\leftarrow )^{m-1}(x,t){-}v_\varepsilon ^{m-1}(x,t)\big )\\&\qquad \qquad \qquad \qquad \quad \times \nabla v_\varepsilon (x,t)\cdot \nabla _x\Big (\gamma _\theta (x{-}y) {\bar{\xi }}_{l}\Big (\frac{x{+}y}{2}\Big )\Big )\varphi (t)\,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\bigg |\\&\quad \le Cl^{-1}\theta ^{-1}\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\chi _{{{\,\mathrm{supp}\,}}\nabla v_\varepsilon } \big |(u_\varepsilon ^\leftarrow )^{m-1}(x,t){-}v_\varepsilon ^{m-1}(x,t)\big | |\nabla u_\varepsilon ^\leftarrow (x,t)| \,\mathrm {d}x\,\mathrm {d}t\\&\quad \le C(\kappa )l^{-1}\theta ^{-1}\varepsilon ^q\bigg ( \mathbf {E}\int _0^{T^*}\int _{D} mu_\varepsilon ^{m-1}(x,t)|\nabla u_\varepsilon (x,t)|^2 \,\mathrm {d}x\,\mathrm {d}t\bigg )^\frac{1}{2}\\&\quad \le C(\kappa )l^{-1}\theta ^{-1}\varepsilon ^q. \end{aligned}$$
Since the other two terms of \(R^{(1)}_{\mathrm {error}}\) can be treated along the same lines, we obtain in total the asserted estimate (75). This concludes the proof of Lemma 10. \(\square \)
We continue with the proof of Proposition 5. We have by now everything in place to let \(\delta \rightarrow 0\) in (60). A straightforward application of Lebesgue’s dominated convergence theorem based on the convergence in (31) and the regularity and bounds for the Wong–Zakai approximations \(u_\varepsilon \) from Lemma 4 shows that by letting \(\delta \rightarrow 0\) in (60) and using (75) it holds
$$\begin{aligned}&\mathbf {E}R_{\mathrm {dt}}^{(2)} \le \mathbf {E}R_{\mathrm {porMed}}^{(3)} + \mathbf {E}R_{\mathrm {corr}}^{(3)} + C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q \end{aligned}$$
(76)
for all \((\theta ,l)\) subject to (52), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). The updated terms in this inequality are given by
$$\begin{aligned} R_{\mathrm {dt}}^{(2)}&:= -\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big | \gamma _\theta (x{-}y){\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big )\frac{\mathrm {d}}{\mathrm {d}t}\varphi (t) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {porMed}}^{(3)}&:= \int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big |v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big | \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {corr}}^{(3)}&:= \int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big | \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l\Big (\frac{x{+}y}{2}\Big ) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
respectively. We may proceed with the next step which consists of removing the doubling in the spatial variables. More precisely, the following holds true.
Lemma 11
Let the assumptions and notation of Sect. 4.2 until this point be in place. Define the quantities
$$\begin{aligned} R_{\mathrm {dt}}^{(3)}&:= -\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big | \gamma _\theta (x{-}y){\bar{\xi }}_l(x)\frac{\mathrm {d}}{\mathrm {d}t}\varphi (t) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {porMed}}^{(5)}&:= \int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d} \big |v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(x,t)\big | \Delta {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {corr}}^{(5)}&:= \int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(x,t)\big | \Delta {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
We then have the estimate
$$\begin{aligned} \mathbf {E}R_{\mathrm {dt}}^{(3)}&\le \mathbf {E}R_{\mathrm {porMed}}^{(5)} + \mathbf {E}R_{\mathrm {corr}}^{(5)} + C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + C(\kappa )l^{-3}\theta \end{aligned}$$
(77)
for all \(\varphi \in C^\infty _{\mathrm {cpt}}((0,T^*);[0,1])\) with \(\Vert \varphi \Vert _{W^{1,1}(0,T^*)}\le {\bar{C}}\), all \((\theta ,l)\) subject to (52) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\).
Proof
By a straightforward estimate, we may replace in all three terms \({\bar{\xi }}_l(\frac{\cdot {-}y}{2})\) by \({\bar{\xi }}_l(\cdot )\) and therefore update the estimate (76) to
$$\begin{aligned}&\mathbf {E}R_{\mathrm {dt}}^{(3)} \le \mathbf {E}R_{\mathrm {porMed}}^{(4)} + \mathbf {E}R_{\mathrm {corr}}^{(4)} + C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + Cl^{-3}\theta \Vert \varphi \Vert _{W^{1,1,}(0,T^*)} \end{aligned}$$
(78)
for all \((\theta ,l)\) subject to (52) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\), where
$$\begin{aligned} R_{\mathrm {porMed}}^{(4)}&:= \int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \big |v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(y,t)\big | \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l(x) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t,\\ R_{\mathrm {corr}}^{(4)}&:= \int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)\big | \varphi (t)\gamma _\theta (x{-}y)\Delta {\bar{\xi }}_l(x) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t, \end{aligned}$$
respectively. We estimate using \(\Vert \Delta {\bar{\xi }}_l\Vert _{L^\infty (\mathbf {R}^d)}\le \Vert \nabla ^2\gamma _l\Vert _{L^1(\mathbf {R}^d)}\Vert {\bar{\xi }}\Vert _{L^\infty (\mathbf {R}^d)}\le Cl^{-2}\) (which follows from Young’s inequality and condition i) for the spatial cut-off \({\bar{\xi }}\)), a change of variables \(y\mapsto y{+}x\), \(u_\varepsilon ^\leftarrow \ge \kappa \) on \({{\,\mathrm{supp}\,}}\nabla v_\varepsilon \), and the bounds (19) resp. (16)
$$\begin{aligned}&\bigg |\mathbf {E}R_{\mathrm {corr}}^{(4)} - \mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \frac{1}{2}\nu ^2\big |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(x,t)\big | \varphi (t)\Delta {\bar{\xi }}_l(x)\,\mathrm {d}x\,\mathrm {d}t\bigg |\\&\quad \le C\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d}\big |v_{{\hat{\varepsilon }}}(y,t){-}v_{{\hat{\varepsilon }}}(x,t)\big | \gamma _\theta (x{-}y)|\Delta {\bar{\xi }}_l(x)|\varphi (t)\,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\\&\quad \le Cl^{-2}\theta \mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \gamma _\theta (x{-}y) \int _0^1 |\nabla v_{{\hat{\varepsilon }}}(rx{+}(1{-}r)y,t)| \,\mathrm {d}r \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\\&\quad \!\le \! C(\kappa )l^{-2}\theta \mathbf {E}\int _0^1\int _0^{T^*}\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} \gamma _\theta (y) (u_{{\hat{\varepsilon }}}^\leftarrow )^\frac{m-1}{2}(x{-}ry,t) |\nabla u_{{\hat{\varepsilon }}}^\leftarrow (x{-}ry,t)| \,\mathrm {d}x\,\mathrm {d}y\,\mathrm {d}t\,\mathrm {d}r\\&\quad \le C(\kappa )l^{-2}\theta \bigg (\mathbf {E}\int _0^{T^*}\int _D m u_{{\hat{\varepsilon }}}^{m-1}(x,t)|\nabla u_{{\hat{\varepsilon }}}(x,t)|^2 \,\mathrm {d}x\,\mathrm {d}t\bigg )^\frac{1}{2}\\&\quad \le C(\kappa )l^{-2}\theta \end{aligned}$$
for all \((\theta ,l)\) subject to (52) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). Arguing along the same lines we also get the estimate
$$\begin{aligned} \bigg |\mathbf {E}R_{\mathrm {porMed}}^{(4)} - \mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} \big |v_\varepsilon ^m(x,t){-}v_{{\hat{\varepsilon }}}^m(x,t)\big | \varphi (t)\Delta {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t\bigg | \le C(\kappa )l^{-2}\theta \end{aligned}$$
for all \((\theta ,l)\) subject to (52) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). Summarizing we then obtain the desired inequality (77). This concludes the proof of Lemma 11. \(\square \)
We continue with the proof of Proposition 5. The next step takes care of estimating the non-linear diffusion term and the correction term.
Lemma 12
Let the assumptions and notation of Sect. 4.2 until this point be in place. In particular, recall from the statement of Lemma 11 the definition of the quantities \(R_{\mathrm {porMed}}^{(5)}\) and \(R_{\mathrm {corr}}^{(5)}\). Then there exists some \(l({\bar{\xi }})>0\) small enough and some absolute constant \({\bar{C}}>0\) such that we have the estimate
$$\begin{aligned} \mathbf {E}R_{\mathrm {porMed}}^{(5)} + \mathbf {E}R_{\mathrm {corr}}^{(5)} \le \bar{C}\kappa + C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha \end{aligned}$$
(79)
for all \((\theta ,l)\) subject to (52) resp. \(l<l({\bar{\xi }})\), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\).
Proof
We first aim to replace \(v_\varepsilon \) resp. \(v_{{\hat{\varepsilon }}}\) by \(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon }\) resp. \(\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}}\) in the two terms \(R_{\mathrm {porMed}}^{(5)}\) and \(R_{\mathrm {corr}}^{(5)}\). This can be done by an estimation similar to the proof of Lemma 11, this time using in particular (7), (10) and \(\mathbf {E}C_\alpha ^2<\infty \), which yields the bound
$$\begin{aligned}&\mathbf {E}R_{\mathrm {porMed}}^{(5)} + \mathbf {E}R_{\mathrm {corr}}^{(5)}\\&\quad \le -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d} |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })^m(x,t){-} (\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})^m(x,t)| \Delta {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })(x,t) {-}(\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})(x,t)| \Delta {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad +C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha . \end{aligned}$$
However, it follows from \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\) that \((\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })^m{-} (\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})^m\in H^1_0(D)\) as well as \((\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon }){-} (\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})\in H^1_0(D)\). An integration by parts together with condition ii) of the spatial cut-off function \({\bar{\xi }}\) then entails
$$\begin{aligned}&\mathbf {E}R_{\mathrm {porMed}}^{(5)} + \mathbf {E}R_{\mathrm {corr}}^{(5)} \\&\quad \le -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d} \nabla |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })^m(x,t){-} (\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})^m(x,t)| \cdot \nabla {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \nabla |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })(x,t) {-}(\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})(x,t)| \cdot \nabla {\bar{\xi }}_l(x) \,\mathrm {d}x\,\mathrm {d}t\\ {}&\qquad +C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha \\ {}&\quad \le -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d} \nabla |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })^m(x,t){-} (\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})^m(x,t)| \cdot \nabla ({\bar{\xi }}_l{-}{\bar{\xi }})(x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad -\mathbf {E}\int _0^{T^*}\varphi (t)\int _{\mathbf {R}^d}\frac{1}{2}\nu ^2 \nabla |(\zeta ^{\varepsilon ^q}_\kappa \circ u_{\varepsilon })(x,t) {-}(\zeta ^{{\hat{\varepsilon }}^q}_\kappa \circ u_{{\hat{\varepsilon }}})(x,t)| \cdot \nabla ({\bar{\xi }}_l{-}{\bar{\xi }})(x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad +C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha . \end{aligned}$$
It remains to bound the two terms featuring the difference \(\nabla {\bar{\xi }}_l{-}\nabla {\bar{\xi }}\). By continuity of translations in \(L^2\), however, together with the a priori estimate (16) we find some small enough \(l({\bar{\xi }})>0\) such that
$$\begin{aligned} l<l({\bar{\xi }}) \quad \Rightarrow \quad \Vert {\bar{\xi }}_l{-}{\bar{\xi }}\Vert _{H^1(\mathbf {R}^d)}\le \kappa ^{\frac{m+1}{2}}, \end{aligned}$$
(80)
holds true, and therefore in particular the desired estimate (79) for some absolute constant \({\bar{C}}>0\). \(\square \)
We continue with the proof of Proposition 5. It is straightforward to estimate
$$\begin{aligned}&\bigg |{-}\mathbf {E}R_{\mathrm {dt}}^{(3)}{-}\mathbf {E}\int _0^{T^*}\frac{\mathrm {d}}{\mathrm {d}t}\varphi (t)\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(x,t)|\gamma _\theta (x{-}y){\bar{\xi }}(x) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\bigg | \le {\bar{C}}\kappa \end{aligned}$$
for all l subject to (80) and all \(\varphi \in C^\infty _{\mathrm {cpt}}((0,T^*);[0,1])\) with \(\Vert \varphi \Vert _{W^{1,1}(0,T^*)}\le {\bar{C}}\). In summary, we thus obtain together with (79) the estimate
$$\begin{aligned} -\mathbf {E}R_{\mathrm {dt}}^{(4)}&:=-\mathbf {E}\int _0^{T^*}\frac{\mathrm {d}}{\mathrm {d}t}\varphi (t)\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} |v_\varepsilon (x,t){-}v_{{\hat{\varepsilon }}}(y,t)| \gamma _\theta (x{-}y){\bar{\xi }}(x) \,\mathrm {d}y\,\mathrm {d}x\,\mathrm {d}t\nonumber \\&\le C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha + C(\kappa )l^{-3}\theta + {\bar{C}}\kappa \end{aligned}$$
(81)
for all \(\varphi \in C^\infty _{\mathrm {cpt}}((0,T^*);[0,1])\) with \(\Vert \varphi \Vert _{W^{1,1}(0,T^*)}\le {\bar{C}}\), all \((\theta ,l)\) subject to (52) resp. (80), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\).
In the next step we take care of the term \(\mathbf {E}R_{\mathrm {dt}}^{(4)}\). Employing the same argument leading to [7, (4.20)] (instead of using [7, Lemma 3.2] to treat the initial condition we can also rely on the continuity down to \(t=0\) thanks to Lemma 4) we infer that the estimate (81) entails the bound (recall that the initial condition is deterministic)
$$\begin{aligned}&\mathbf {E}\int _{D}\int _{D} |v_\varepsilon (x,T){-}v_{{\hat{\varepsilon }}}(y,T)| \gamma _\theta (x{-}y){\bar{\xi }}(x) \,\mathrm {d}y\,\mathrm {d}x\nonumber \\&\quad \le \int _{D}\int _{D} |v_\varepsilon (x,0){-}v_{{\hat{\varepsilon }}}(y,0)| \gamma _\theta (x{-}y){\bar{\xi }}(x) \,\mathrm {d}y\,\mathrm {d}x\nonumber \\&\qquad + C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha + C(\kappa )l^{-3}\theta + {\bar{C}}\kappa \end{aligned}$$
(82)
for all \(T\in [0,T^*]\), all \((\theta ,l)\) subject to (52) resp. (80), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). We then estimate for all \(T\in [\kappa ,T^*]\) by means of (17), (19) and \(\Vert {\bar{\xi }}\Vert _{L^\infty (D)}\le 1\)
$$\begin{aligned}&\bigg |\mathbf {E}\int _{D}\int _{D} |v_\varepsilon (x,T){-}v_{{\hat{\varepsilon }}}(y,T)| \gamma _\theta (x{-}y){\bar{\xi }}(x) \,\mathrm {d}y\,\mathrm {d}x{-}\mathbf {E}\int _{D}|v_\varepsilon (x,T){-}v_{{\hat{\varepsilon }}}(x,T)|{\bar{\xi }}(x)\,\mathrm {d}x\bigg | \\ {}&\quad \le \mathbf {E}\int _{D}\int _{D}|v_{{\hat{\varepsilon }}}(y{+}x,T){-}v_{{\hat{\varepsilon }}}(x,T)| \gamma _\theta (y){\bar{\xi }}(x)\,\mathrm {d}x\,\mathrm {d}y\\ {}&\quad \le C\theta \mathbf {E}\int _0^1\int _{\mathbf {R}^d}\int _{\mathbf {R}^d} |\nabla u_{{\hat{\varepsilon }}}^\leftarrow (x{-}ry,T)| \gamma _\theta (y) \,\mathrm {d}x\,\mathrm {d}y\,\mathrm {d}r \\ {}&\quad \le C(\kappa )\theta \mathbf {E}\int _{D} u_{{\hat{\varepsilon }}}^{m-1}|\nabla u_{{\hat{\varepsilon }}}| \,\mathrm {d}x\le C(\kappa ){\hat{\varepsilon }}^{-\beta }\theta . \end{aligned}$$
Since we may assume in this argument without loss of generality that \(\varepsilon \le {\hat{\varepsilon }}\) (otherwise, switch the roles of \(v_{{\hat{\varepsilon }}}\) and \(v_\varepsilon \) in the previous estimate) we obtain together with an analogous estimate based on the regularity of the initial condition and the fact that \({\bar{\xi }}\equiv 1\) on K as well as \({\bar{\xi }}\ge 0\) on D,
$$\begin{aligned}&\mathbf {E}\int _{K}|v_\varepsilon (x,T){-}v_{{\hat{\varepsilon }}}(x,T)| \,\mathrm {d}x\nonumber \\&\quad \le \int _{D}|v_\varepsilon (x,0){-}v_{{\hat{\varepsilon }}}(x,0)|\,\mathrm {d}x+ C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q + C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha \nonumber \\&\qquad + C(\kappa )(\varepsilon \vee {\hat{\varepsilon }})^{-\beta }\theta + C(\kappa )l^{-3}\theta + {\bar{C}}\kappa \end{aligned}$$
(83)
for all \(T\in [\kappa ,T^*]\), all \((\theta ,l)\) subject to (52) resp. (80), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). As a consequence of (21) and the triangle inequality, we may finally switch from \(v_\varepsilon (\cdot )=\zeta ^{\varepsilon ^q}_\kappa \big (u_\varepsilon ^\leftarrow (\cdot )\big ) =\kappa + \zeta _{\varepsilon ^q}\big (u_\varepsilon ^\leftarrow (\cdot ){-}\kappa \big )\) to \(\kappa \vee u_{\varepsilon }^\leftarrow (\cdot )=\kappa + \big (u_\varepsilon ^\leftarrow (\cdot ){-}\kappa \big )_{+}\), which yields
$$\begin{aligned}&\mathbf {E}\int _{K}|\kappa \vee u_{\varepsilon }^\leftarrow (x,T){-} \kappa \vee u_{{\hat{\varepsilon }}}^\leftarrow (x,T)| \,\mathrm {d}x\nonumber \\&\quad \le \mathbf {E}\int _{K}|v_\varepsilon (x,T){-}v_{{\hat{\varepsilon }}}(x,T)| \,\mathrm {d}x+ C\mathcal {L}^d(D)(\varepsilon \vee {\hat{\varepsilon }})^q\nonumber \\&\quad \le \int _{D}|\kappa \vee u_{\varepsilon }^\leftarrow (x,0) {-}\kappa \vee u_{{\hat{\varepsilon }}}^\leftarrow (x,0)|\,\mathrm {d}x+ C(\kappa )l^{-1}\theta ^{-1}(\varepsilon \vee {\hat{\varepsilon }})^q\nonumber \\&\qquad + C(\kappa )l^{-2}(\varepsilon \vee {\hat{\varepsilon }})^\alpha + C(\kappa )(\varepsilon \vee {\hat{\varepsilon }})^{-\beta }\theta + C(\kappa )l^{-3}\theta + {\bar{C}}\kappa \end{aligned}$$
(84)
for all \(T\in [\kappa ,T^*]\), all \((\theta ,l)\) subject to (52) resp. (80), and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). We eventually arrived at the last step of the proof.
In light of the right hand side terms in (84) we first define \(\theta :=(\varepsilon \vee {\hat{\varepsilon }})^{\beta +1}\), then fix \({\bar{\vartheta }}>0\) and \(q>0\) such that \(2{\bar{\vartheta }}<\alpha \), \(3{\bar{\vartheta }}<\beta +1\) as well as \(\beta +1+{\bar{\vartheta }}<q\), and finally define \(l:=(\varepsilon \vee {\hat{\varepsilon }})^{{\bar{\vartheta }}}\). Choosing \(\varepsilon _0>0\) small enough, we can ensure that \((\theta ,l)\) satisfy (52) resp. (80). Hence, these choices guarantee that we may infer from (84) a bound of the type (23) with right hand side terms \(C(\kappa )(\varepsilon \vee {\hat{\varepsilon }})^{2\vartheta }\) for some suitable exponent \(\vartheta >0\). Choosing \(\varepsilon _0\) even smaller, if needed, we can avoid the dependence of the constant on the data by sacrificing a power \(\vartheta \), which entails (23). This concludes the proof of Proposition 5. \(\square \)
Proof of Corollary 6 (\(L^1\) convergence of shifted densities)
Let \(\delta >0\) be fixed but arbitrary. By the triangle inequality we may estimate
$$\begin{aligned}&\sup _{T\in [\kappa ,T^*]}\mathbf {E}\int _D|u^\leftarrow _{\varepsilon }(T){-}u^\leftarrow _{{\hat{\varepsilon }}}(T)|\,\mathrm {d}x\\ {}&\quad \le \sup _{T\in [\kappa ,T^*]}\mathbf {E}\int _D|u^\leftarrow _{\varepsilon }(T)|\chi _{\{u_\varepsilon (T)<\kappa \}} +|u^\leftarrow _{{\hat{\varepsilon }}}(T)|\chi _{\{u_{{\hat{\varepsilon }}}(T)<\kappa \}}\,\mathrm {d}x\\&\qquad +\sup _{T\in [\kappa ,T^*]} \mathbf {E}\int _D |\kappa \vee u^\leftarrow _\varepsilon (x,T) {-}\kappa \vee u^\leftarrow _{{\hat{\varepsilon }}}(x,T)|\,\mathrm {d}x\end{aligned}$$
for all \(\kappa \in (0,T^*\wedge 1)\) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). Hence, it follows from splitting \(D=K\cup (D\setminus K)\) together with the bounds (15) and (23) as well as the definition (22) of the shifted densities that
$$\begin{aligned}&\sup _{T\in [\kappa ,T^*]}\mathbf {E}\int _D|u^\leftarrow _{\varepsilon }(T){-}u^\leftarrow _{{\hat{\varepsilon }}}(T)|\,\mathrm {d}x\\&\quad \le 2\mathcal {L}^d(D)\kappa + 2\mathcal {L}^d(D\setminus K)\big (\kappa \vee (1{+}\Vert u_0\Vert _{L^\infty (D)})\big ) + \bar{C}(\varepsilon \vee {\hat{\varepsilon }})^\vartheta + {\bar{C}}\kappa \\&\qquad +\mathbf {E}\int _D |\kappa \vee (u_0(x){+}\varepsilon ){-}\kappa \vee (u_0(x){+}{\hat{\varepsilon }})|\,\mathrm {d}x. \end{aligned}$$
The term with the initial data is estimated similarly by
$$\begin{aligned}&\mathbf {E}\int _D |\kappa \vee (u_0(x){+}\varepsilon ){-}\kappa \vee (u_0(x){+}{\hat{\varepsilon }})|\,\mathrm {d}x\le 2\mathcal {L}^d(D)\kappa + \mathcal {L}^d(D)|\varepsilon {-}{\hat{\varepsilon }}| \end{aligned}$$
for all \(\kappa >0\) and all \(\varepsilon ,{\hat{\varepsilon }}\le \frac{\kappa }{2}\). Choosing first \(\kappa <\tau \) sufficiently small such that \((4\mathcal {L}^d(D){+}{\bar{C}})\kappa \le \frac{\delta }{2}\), we may then fix a large enough compact set \(K\subset D\) and some small enough \(\varepsilon _0(\kappa ,K)\) such that the bound
$$\begin{aligned} \sup _{T\in [\tau ,T^*]}\mathbf {E}\int _D|u^\leftarrow _{\varepsilon }(T){-}u^\leftarrow _{{\hat{\varepsilon }}}(T)|\,\mathrm {d}x\le \delta \end{aligned}$$
holds true for all \(\varepsilon ,{\hat{\varepsilon }}\le \varepsilon _0\). This proves that the sequence of shifted densities \(u_\varepsilon ^\leftarrow \) is a Cauchy sequence in the space \(C([\tau ,T^*];L^1(\Omega {\times }D,\mathbf {P}\otimes \mathcal {L}^d))\) for all \(\tau >0\). The corresponding assertion in the space \(L^1([0,T^*];L^1(\Omega {\times }D,\mathbf {P}\otimes \mathcal {L}^d))\) is proved similarly based on the additional estimate
$$\begin{aligned}&\mathbf {E}\int _0^{T^*}\int _D|u^\leftarrow _{\varepsilon }{-}u^\leftarrow _{{\hat{\varepsilon }}}|\,\mathrm {d}x\,\mathrm {d}t\\ {}&\quad \le 2\mathcal {L}^d(D)\kappa (1{+}\Vert u_0\Vert _{L^\infty (D)}) +(T^*{-}\kappa )\sup _{T\in [\kappa ,T^*]}\mathbf {E}\int _D|u^\leftarrow _{\varepsilon }(T){-}u^\leftarrow _{{\hat{\varepsilon }}}(T)|\,\mathrm {d}x. \end{aligned}$$
Let us denote the corresponding limit in \(L^1([0,T^*];L^1(\Omega {\times }D,\mathbf {P}\otimes \mathcal {L}^d))\) by u.
On the other side, it follows immediately from the bound (15) that the sequence of densities \(u_\varepsilon \) has a weak limit in the space \(L^{m+1}(\Omega _{T^*},\mathcal {P}_{T^*};L^{m+1}(D))\), which we denote by \({\bar{u}}\). It remains to verify that \(u={\bar{u}}\). To this end, let \(\phi \in C^\infty _{\mathrm {cpt}}(D\times [0,T^*])\) and \(A\in \mathcal {F}_{T^*}\) be fixed. We then have
$$\begin{aligned} \mathbf {E}\chi _A\int _0^{T^*}\int _D (u{-}{\bar{u}})\phi \,\mathrm {d}x\,\mathrm {d}t=\lim _{\varepsilon \rightarrow 0}\,\mathbf {E}\chi _A\int _0^{T^*}\int _D (u_\varepsilon ^\leftarrow {-}u_\varepsilon )\phi \,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
By a simple change of variables and the definition (22) we may write
$$\begin{aligned}&\mathbf {E}\int _0^{T^*}\int _{D} (u_\varepsilon ^\leftarrow {-}u_\varepsilon )\phi \,\mathrm {d}x\,\mathrm {d}t\\ {}&\quad =\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\big (\phi (x{-}\nu (B_t{-}B_t^\varepsilon ),t){-}\phi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t. \end{aligned}$$
Exploiting (7), (10) and (15) we further estimate
$$\begin{aligned}&\bigg |\mathbf {E}\int _0^{T^*}\int _{\mathbf {R}^d} u_\varepsilon (x,t)\big (\phi (x{-}\nu (B_t{-}B_t^\varepsilon ),t){-}\phi (x,t)\big ) \,\mathrm {d}x\,\mathrm {d}t\bigg |\\&\quad \le C\Vert \nabla \phi \Vert _{L^\infty (D\times [0,T^*])}\varepsilon ^\alpha \mathbf {E}C_\alpha \le C\varepsilon ^\alpha . \end{aligned}$$
Hence, we may infer that
$$\begin{aligned} \mathbf {E}\chi _A\int _0^{T^*}\int _D (u{-}{\bar{u}})\phi \,\mathrm {d}x\,\mathrm {d}t= 0 \end{aligned}$$
holds true for all \(\phi \in C^\infty _{\mathrm {cpt}}(D\times [0,T^*])\) and \(A\in \mathcal {F}_{T^*}\). This shows that \(u={\bar{u}}\) and thus concludes the proof of Corollary 6. \(\square \)
Proof of Proposition 7 (Recovering the unique weak solution)
Let a test function \(\phi \in C^\infty _{\mathrm {cpt}}(D)\) be fixed, and define \(K:=\mathrm {supp}\,\phi \). Fix also an integer \(M\ge 1\), and let \(\mathcal {C}_\alpha \) be the square integrable random variable of (10). Let \(\delta \in (0,1)\) be such that \(\{x\in \mathbf {R}^d:\mathrm {dist}(x,K)\le \delta \}\subset D\). Let \(\varepsilon '=\varepsilon '(M,\delta )\) be the constant from (11). By Itô’s formula, the definition (22) and the fact that the \(u_\varepsilon \) solve (12) classically we have for all \(\varepsilon \le \varepsilon '\) and all measurable \(A\in \mathcal {F}_{T^*}\) (cf. the argument in the first step of the proof of Proposition 5)
$$\begin{aligned} \begin{aligned}&\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D u_\varepsilon ^\leftarrow (x,T)\phi (x) \,\mathrm {d}x- \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D (u_0(x){+}\varepsilon )\phi (x)\,\mathrm {d}x\\&\quad =\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \Big (\Delta (u_{\varepsilon }^\leftarrow )^m(x,t) {+} \frac{1}{2}\nu ^2\Delta u_{\varepsilon }^\leftarrow (x,t)\Big )\phi (x) \,\mathrm {d}x\,\mathrm {d}t\\&\qquad + \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \nu \nabla u_{\varepsilon }^\leftarrow (x,t)\phi (x) \,\mathrm {d}x\,\mathrm {d}B_t \end{aligned} \end{aligned}$$
(85)
for all \(T\in (0,T^*)\). Define the shifted test function
$$\begin{aligned} \phi ^\leftarrow _\varepsilon (x,t) := \phi (x{-}\nu (B_t{-}B_t^\varepsilon )), \end{aligned}$$
so that we obtain by a simple change of variables
$$\begin{aligned} \begin{aligned}&\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D u_\varepsilon ^\leftarrow (x,T)\phi (x) \,\mathrm {d}x- \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D (u_0(x){+}\varepsilon )\phi (x)\,\mathrm {d}x\\&\quad =\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \Big (\Delta u_{\varepsilon }^m(x,t) {+} \frac{1}{2}\nu ^2\Delta u_{\varepsilon }(x,t)\Big )\phi ^\leftarrow _\varepsilon (x,t) \,\mathrm {d}x\,\mathrm {d}t\\ {}&\qquad + \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \nu \nabla u_{\varepsilon }(x,t)\phi ^\leftarrow _\varepsilon (x,t) \,\mathrm {d}x\,\mathrm {d}B_t. \end{aligned} \end{aligned}$$
(86)
Note that as a consequence of (11) we have almost surely on \(\{\mathcal {C}_\alpha \le M\}\) for all \(\varepsilon \le \varepsilon '\) and all \(t\in [0,T^*]\) that \(\mathrm {supp}\,\phi ^\leftarrow _\varepsilon (\cdot ,t) \subset \subset D\). Hence, integrating by parts on the right hand side in (86) does not produce any boundary integrals so that after reversing the change of variables we obtain the identity
$$\begin{aligned} \begin{aligned}&\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D u_\varepsilon ^\leftarrow (x,T)\phi (x) \,\mathrm {d}x- \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D (u_0(x){+}\varepsilon )\phi (x)\,\mathrm {d}x\\ {}&\quad =\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \Big ((u_{\varepsilon }^\leftarrow )^m(x,t) {+} \frac{1}{2}\nu ^2 u_{\varepsilon }^\leftarrow (x,t)\Big ) \Delta \phi (x) \,\mathrm {d}x\,\mathrm {d}t\\ {}&\qquad - \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \nu u_{\varepsilon }^\leftarrow (x,t)\nabla \phi (x) \,\mathrm {d}x\,\mathrm {d}B_t. \end{aligned} \end{aligned}$$
(87)
We aim to pass to the limit \(\varepsilon _0\ge \varepsilon \rightarrow 0\) in all four terms. This is possible by the convergence of the shifted densities \(u_\varepsilon ^\leftarrow \) in \(C([\tau ,T^*];L^1(\Omega {\times }D,\mathbf {P}{\otimes }\mathcal {L}^d))\) as well as in \(L^1([0,T^*];L^1(\Omega {\times }D,\mathbf {P}{\otimes }\mathcal {L}^d))\), see Corollary 6. Because of the uniform bound (15) the convergence in \(L^1\) can actually be lifted to convergence in any \(L^q([0,T^*];L^q(\Omega {\times }D,\mathbf {P}{\otimes }\mathcal {L}^d)),\,q\in (1,\infty ),\) which makes the limit passage possible in the non-linear diffusion term as well as the noise term (using for the latter, e.g., the Burkholder–Davis–Gundy inequality). In summary, we obtain from letting \(\varepsilon _0\ge \varepsilon \rightarrow 0\) the identity
$$\begin{aligned}&\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D u(x,T)\phi (x) \,\mathrm {d}x- \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _D u_0(x)\phi (x)\,\mathrm {d}x\\\quad&=\mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \Big (u^m(x,t) {+} \frac{1}{2}\nu ^2u(x,t)\Big )\Delta \phi (x) \,\mathrm {d}x\,\mathrm {d}t\\ {}&\qquad - \mathbf {E}\chi _{\{\mathcal {C}_\alpha \le M\}} \chi _A \int _0^{T}\int _D \nu u(x,t) \nabla \phi (x) \,\mathrm {d}x\,\mathrm {d}B_t \end{aligned}$$
for all \(T\in (0,T^*)\). Since \(M\ge 1\) as well as \(A\in \mathcal {F}_{T^*}\) were arbitrary, and the random variable \(\mathcal {C}_\alpha \) is integrable, we thus recover (4). This concludes the proof of Proposition 7 since the asserted bounds for u follow immediately from (15). \(\square \)