Proof. The existence of unique solution \(\{X_{t}\}_{t \in [0,T]}\) to the index-1 SDAE 1 follows from [5]. Multiplying the integral equation ?? by \(R\), we obtain \[R\int_{0}^{t} A_s\,dX_{s} = \int_{0}^{t} R F(s,X_{s})\,ds + \int_{0}^{t} R G(s,X_{s})\,dW_s, \quad t \in [0,T].\] It follows from \(R A_{s} = 0\) and \(R G(s,X_{s}) = 0\) for all \(s \in [0,T]\) that there exists a set \(\Omega_{0} \in \mathcal{F}\) with \(\mathbb{P}(\Omega_0) = 1\) such that for every \(\omega \in \Omega_{0}\), \[\int_{0}^{t} RF(s,X_{s}(\omega))\,ds = 0, \quad t \in [0,T].\] For each fixed \(\omega \in \Omega_{0}\), the map \[t \mapsto \int_{0}^{t} RF(s,X_{s}(\omega))\,ds\] is absolutely continuous and is identically zero on \([0,T]\). Therefore, by the fundamental theorem of calculus for Lebesgue integrals, we obtain \(RF(t,X_t(\omega)) = 0\) for almost every \(t \in [0,T]\). Since \(\{X_{t}\}_{t \in [0,T]}\) has continuous sample paths and the coefficient \(F(t,x)\) is continuous in \((t,x)\), the map \(t \mapsto RF(t,X_t(\omega))\) is continuous on \([0,T]\). From the fact that a continuous function which is equal to zero almost everywhere on \([0,T]\) must be identically zero, we conclude that for every \(\omega \in \Omega_{0}\), \[RF(t,X_t(\omega))=0, \quad t \in [0,T],\] which implies \(\mathbb{P}(X_t \in \mathcal{M}_t\;\text{for all } t\in[0,T]) = 1\) and thus completes the proof. ◻

Proof. Since \(M,N\) are orthogonal, the facts \(A_t=M\Sigma_tN\) and \(A_t^- = N^\top \Sigma_t^- M^\top\) give that \[|A_t|=|\Sigma_t|, \quad |A_t^{-}|=|\Sigma_t^{-}|, \quad |A_t'|=|\Sigma_t'|, \quad t \in [0,T].\] It follows from \(\Sigma_t = \operatorname{diag}\bigl(\sigma_1(t), \cdots, \sigma_r(t), 0, \cdots, 0\bigr)\) and \(\Sigma_t^{-} = \operatorname{diag}(\sigma_1(t)^{-1},\cdots,\sigma_r(t)^{-1},0,\cdots,0)\) with \(r\in\{1,2,\cdots,d-1\}\) that \[|A_{t}|^{2} = \sum_{i=1}^{r} \sigma_{i}(t)^2 \leq r \overline{\sigma}^{2}, \quad |A_t^{-}|^{2} = \sum_{i=1}^{r} (\sigma_{i}(t))^{-2} \leq \frac{r}{\underline{\sigma}^{2}},\] which yields \(|A_t| \leq \sqrt r\,\overline{\sigma}\) and \(|A_t^{-}| \leq \frac{\sqrt r}{\underline{\sigma}}\). Noting that the map \(t \mapsto \Sigma_t'\) is continuous on \([0,T]\), one gets \(|A_t'| \leq \max_{t \in [0,T]}|\Sigma_t'| =: C_{A} < +\infty\).

Owing to \(\sigma_i \in C^1([0,T];\mathbb{R})\) and \(\sigma_i(t) \geq \underline{\sigma}>0\) for \(i = 1,2,\cdots,r, t \in [0,T]\), the map \(t\mapsto \sigma_i(t)^{-1}\) belongs to \(C^1([0,T];\mathbb{R})\) with \[\frac{d}{dt}\sigma_i(t)^{-1} = -\sigma_i'(t)\sigma_i(t)^{-2}, \quad i=1,\cdots,r,\] which implies that \(t\mapsto \Sigma_t^{-}\) is differentiable and satisfies \[\frac{d}{dt}\Sigma_t^{-} = \operatorname{diag}\bigl( -\sigma_1'(t)\sigma_1(t)^{-2}, \cdots, -\sigma_r'(t)\sigma_r(t)^{-2}, 0, \cdots, 0 \bigr).\] Applying \(\frac{d}{dt} A_t^{-} = N^\top(\frac{d}{dt}\Sigma_t^{-})M^\top\) yields \[\begin{align} \bigg|\frac{d}{dt} A_t^{-}\bigg| = \bigg|\frac{d}{dt}\Sigma_t^{-}\bigg| = \left( \sum_{i=1}^r \frac{|\sigma_i'(t)|^2}{\sigma_i(t)^4} \right)^{1/2} \leq \frac{|\Sigma_t'|}{\underline{\sigma}^{2}} \leq \frac{C_A}{\underline{\sigma}^{2}}. \end{align}\] Moreover, the fundamental theorem of calculus shows that for any \(s,t \in [0,T]\) with \(s \leq t\), \[\begin{align} |A_t^{-}-A_s^{-}| = \left|\int_s^t \frac{d}{dr} A_r^{-}\,dr\right| \leq \int_s^t \bigg|\frac{d}{dr} A_r^{-}\bigg|\,dr \leq C|t-s|. \end{align}\] Thus we complete the proof. ◻

Proof. For any \(t \in [0,T]\) and \(u \in \mathbb{R}^{d}\), setting \(\Psi_t(u) := u + \widehat{V}(t,u)\) yields \[\begin{align} f(t,u) = A_t^{-}F(t,\Psi_t(u)), \quad g(t,u) = A_t^{-}G(t,\Psi_t(u)). \end{align}\] By the boundedness of \(A_t^{-}\) in Lemma 1, the global Lipschitz continuity of \(F\), \(G\) in Assumption 3, and the Lipschitz continuity of \(\widehat{V}(t,\cdot)\) in Lemma 2, we have \[\begin{align} |f(t,u_1)-f(t,u_2)| \leq&~ |A_t^{-}| \big|F(t,\Psi_t(u_1))-F(t,\Psi_t(u_2))\big| \\\leq&~ C|\Psi_t(u_1)-\Psi_t(u_2)| \\\leq&~ C\bigl(|u_1-u_2| + |\widehat{V}(t,u_1) - \widehat{V}(t,u_2)|\bigr) \\\leq&~ C|u_1-u_2|, \quad t \in [0,T], u_1, u_2 \in \mathbb{R}^{d}. \end{align}\] The same arguments give \(|g(t,u_1)-g(t,u_2)| \leq C|u_1-u_2|, t \in [0,T], u_1, u_2 \in \mathbb{R}^{d}\). Thus ?? holds. Next, using the linear growth of \(F\), \(G\) in Assumption 3, the boundedness of \(A_t^{-}\) in Lemma 1, and the linear growth of \(\widehat{V}\) in Lemma 2, we obtain that for any \(t \in [0,T]\) and \(u \in \mathbb{R}^{d}\), \[\begin{align} |f(t,u)| + |g(t,u)| \leq |A_t^{-}|\bigl(|F(t,\Psi_t(u))| + |G(t,\Psi_t(u))|\bigr) \leq C(1 + |\Psi_t(u)|) \leq C(1+|u|), \end{align}\] which proves ?? . According to [20], 7 admits a unique strong solution \(\{U_{t}\}_{t \in [0,T]}\). ◻

Proof. Fix \(t \in [0,T]\) and define \(H_t(u,v) := A_{t}v + RF(t,u+v)\) for \(u, v \in \mathbb{R}^{d}\). Since \(F(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d})\), it follows that \(H_t \in C^{2p+2}(\mathbb{R}^{d} \times \mathbb{R}^{d};\mathbb{R}^{d})\). Moreover, the derivative of \(H_t\) with respect to the algebraic variable \(v\) is \[D_vH_t(u,v) = A_t + RD_xF(t,u+v) = J(t,u+v).\] Let \(u_0 \in \mathbb{R}^{d}\) be arbitrary and set \(v_0 := \widehat{V}(t,u_0)\). Then \(H_t(u_0,v_0) = 0\). Since \(D_{v}H_{t}(u_{0},v_{0}) = J(t,u_0 + v_0)\) is invertible by Assumption 2, the implicit function theorem implies that there exist a neighbourhood \(U_0\) of \(u_0\) and a unique function \(\widetilde{V} \in C^{2p+2}(U_0;\mathbb{R}^{d})\) such that \(H_t(u,\widetilde{V}(u)) = 0\) for any \(u \in U_{0}\). By the global uniqueness of the algebraic equation defining \(\widehat{V}(t,u)\), this local solution coincides with \(\widehat{V}(t,u)\) on \(U_0\). Since \(u_0\) is arbitrary, we obtain \(\widehat{V}(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\operatorname{Im}(Q))\).

It remains to prove the uniform boundedness of the derivatives. For fixed \(t\), letting \(V(u) := \widehat{V}(t,u)\) and \(\Psi(u) := u + V(u)\) enables us to rewrite \[\label{eq:AtVuRFtPsiu} A_tV(u) + RF(t,\Psi(u)) = 0.\tag{8}\] First, differentiating 8 with respect to \(u\) gives \(A_tDV(u) + RD_xF(t,\Psi(u))D\Psi(u) = 0\). Since \(D\Psi(u) = I+DV(u)\), we have \[J(t,\Psi(u))DV(u) = \big(A_t+RD_xF(t,\Psi(u))\big)DV(u) = -RD_xF(t,\Psi(u)).\] Using \(RD_xF(t,\Psi(u)) = J(t,\Psi(u))-A_t\) yields \(DV(u) = J(t,\Psi(u))^{-1}A_t-I\) and accordingly \[\begin{align} \label{eq:boundedDVu} |D_u \widehat{V}(t,u)| = |DV(u)| \leq |J(t,\Psi(u))^{-1}|\,|A_t|+|I| \leq C, \end{align}\tag{9}\] where we used the uniform boundedness of \(J^{-1}\) and \(A_t\).

We now prove the higher-order estimates by induction. Let \(2 \leq j \leq 2p+2\), and let \(\xi_1,\cdots,\xi_j \in \mathbb{R}^{d}\). Applying \(D_u^j\) to 8 and using the multivariate Faà di Bruno formula, we get \[\label{eq:AtDjVuequalzero} A_tD^jV(u)[\xi_1,\cdots,\xi_j] + R D^j\big(F(t,\Psi(u))\big)[\xi_1,\cdots,\xi_j] = 0\tag{10}\] with \[\begin{align} D^j\big(F(t,\Psi(u))\big)[\xi_1,\cdots,\xi_j] = \sum_{\pi\in\Pi_j} D_x^{|\pi|}F(t,\Psi(u)) \Big[D^{|B_1|}\Psi(u)[\xi_{B_1}], \cdots, D^{|B_{|\pi|}|}\Psi(u)[\xi_{B_{|\pi|}}]\Big], \end{align}\] where \(\Pi_j\) denotes the set of all partitions \(\pi = \{B_1,\cdots,B_{|\pi|}\}\) of the set \(\{1, \cdots,j\}\). For a block \(B = \{i_1,\cdots,i_\ell\}\), we use the notation \(D^{|B|}\Psi(u)[\xi_B] := D^\ell\Psi(u)[\xi_{i_1},\cdots,\xi_{i_\ell}]\). The term corresponding to the one-block partition \(\pi = \big\{\{1,\cdots,j\}\big\}\) is \[D_xF(t,\Psi(u))D^j\Psi(u)[\xi_1,\cdots,\xi_j].\] Letting \(\pi_0 := \big\{\{1,\cdots,j\}\big\}\) yields \(|\pi_0| = 1\) and \(B_1 = \{1,\cdots,j\}\) as well as \[\begin{align} D_x^{|\pi_0|}F(t,\Psi(u))\Big[D^{|B_1|}\Psi(u)[\xi_{B_1}]\Big] = D_xF(t,\Psi(u))\Big[D^j\Psi(u)[\xi_1,\cdots,\xi_j]\Big]. \end{align}\] It follows that \[\begin{align} D^j\big(F(t,\Psi(u))\big)[\xi_1,\ldots,\xi_j] = D_xF(t,\Psi(u))\Big[D^j\Psi(u)[\xi_1,\ldots,\xi_j]\Big] + \Gamma_j(t,u)[\xi_1,\ldots,\xi_j], \end{align}\] where \[\begin{align} \Gamma_j(t,u)[\xi_1,\ldots,\xi_j] := \sum_{\substack{\pi\in\Pi_j\\ |\pi|\ge2}} D_x^{|\pi|}F(t,\Psi(u))\Big[D^{|B_1|}\Psi(u)[\xi_{B_1}], \cdots, D^{|B_{|\pi|}|}\Psi(u)[\xi_{B_{|\pi|}}]\Big]. \end{align}\] Applying \(D^j\Psi(u) = D^jV(u), j \geq 2\) and 10 leads to \[\begin{align} J(t,\Psi(u))D^jV(u)[\xi_1,\ldots,\xi_j] = \big(A_t+RD_xF(t,\Psi(u))\big)D^jV(u)[\xi_1,\ldots,\xi_j] = -R\Gamma_j(t,u)[\xi_1,\ldots,\xi_j]. \end{align}\] We now observe that \(\Gamma_j(t,u)\) contains only lower-order derivatives of \(\Psi\). Indeed, if \(|\pi| \geq 2\), then every block \(B_\ell\) of the partition satisfies \(|B_\ell| \leq j-1\). Hence \(\Gamma_j(t,u)\) only involves \(D^\ell\Psi(u), 1 \leq \ell \leq j-1\). Moreover, it holds that \[\begin{align} \label{eq:DlpuequalDVu} D\Psi(u) = I + DV(u), \quad D^\ell\Psi(u) = D^\ell V(u),\quad \ell \geq 2. \end{align}\tag{11}\]

As we have already proved the boundedness of \(DV(u)\) in 9 , now we assume, as the induction hypothesis, that for some \(j \geq 2\), \[|D^\ell V(u)| \leq C, \quad \ell=1,\cdots,j-1,\] uniformly in \(t \in [0,T]\) and \(u \in \mathbb{R}^{d}\). It follows from 11 that \(|D^\ell\Psi(u)| \leq C\) for \(\ell = 1,\cdots,j-1\). Since \(j \leq 2p+2\) and the number of partitions of \(\{1,\cdots,j\}\) is finite, we further use Assumption 4 to obtain \(|\Gamma_j(t,u)| \leq C\) and consequently \[\begin{align} |D^jV(u)| \leq |J(t,\Psi(u))^{-1}|\,|R|\,|\Gamma_j(t,u)| \leq C. \end{align}\] This proves the induction step and completes the proof. ◻

Proof. For any \(t \in [0,T]\) and \(u \in \mathbb{R}^{d}\), setting \(\Psi_t(u) := u + \widehat{V}(t,u)\) yields \(f(t,u) = A_t^{-}F(t,\Psi_t(u))\) and \(g(t,u) = A_t^{-}G(t,\Psi_t(u))\). Since \(F(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d})\), \(G(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d \times m})\) and \(\widehat{V}(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\operatorname{Im}(Q))\) by Lemma 3, we have \(u \mapsto \Psi_t(\cdot)\) belonging to \(C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d})\). Utilizing the standard regularity of compositions ensures \(F(t,\Psi_t(\cdot)) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d})\), \(G(t,\Psi_t(\cdot)) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d \times m})\) and accordingly \(f(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d})\), \(g(t,\cdot) \in C^{2p+2}(\mathbb{R}^{d};\mathbb{R}^{d \times m})\). As \(A_t^{-}\) does not depend on \(u\), we use the multivariate Faà di Bruno formula to show that for any \(1 \leq j \leq 2p+2\) and \(\xi_1, \cdots, \xi_j \in \mathbb{R}^{d}\), \[\begin{align} D_u^j f(t,u)[\xi_1,\cdots,\xi_j] =&~ A_t^{-}D_u^j\bigl(F(t,\Psi_t(u))\bigr)[\xi_1,\cdots,\xi_j] \\=&~ A_t^{-} \sum_{\pi\in\Pi_j} D_x^{|\pi|}F(t,\Psi_t(u)) \Big[D_u^{|B_1|}\Psi_t(u)[\xi_{B_1}], \cdots, D_u^{|B_{|\pi|}|}\Psi_t(u)[\xi_{B_{|\pi|}}]\Big], \end{align}\] where \(\Pi_j\) denotes the set of all partitions \(\pi = \{B_1,\cdots,B_{|\pi|}\}\) of \(\{1,\cdots,j\}\). For a block \(B = \{i_1,\cdots,i_\ell\}\), we write \(D_u^{|B|}\Psi_t(u)[\xi_B] := D_u^\ell\Psi_t(u)[\xi_{i_1}, \cdots,\xi_{i_\ell}]\). For \(j = 1\), the boundedness of \(D_xF\) follows from the global Lipschitz continuity of \(F(t,\cdot)\) together with its differentiability. For \(j \geq 2\), the boundedness of \(D_x^kF\), \(2 \leq k \leq j\) follows from Assumption 4. Moreover, using \(D_u\Psi_t(u) = I + D_u\widehat{V}(t,u)\) and \(D_u^\ell\Psi_t(u)=D_u^\ell\widehat{V}(t,u), \ell \geq 2\) as well as Lemma 3 implies \[\begin{align} |D_u^\ell\Psi_t(u)| \leq C, \quad \ell = 1,\cdots,2p+2, \end{align}\] uniformly for \(t \in [0,T]\) and \(u \in \mathbb{R}^{d}\). It follows from the boundedness of \(A_t^{-}\) in Lemma 1 that \[\begin{align} |D_u^j f(t,u)| \leq |A_t^{-}| \sum_{\pi \in \Pi_j} \Big|D_x^{|\pi|}F(t,\Psi_t(u))\Big| \prod_{B \in \pi} \Big| D_u^{|B|}\Psi_t(u)\Big| \leq C, \quad j = 1,\cdots,2p+2. \end{align}\] The same arguments applied to \(G(t,\Psi_t(u))\) yield \(|D_u^j g(t,u)| \leq C\) for \(j = 1,\cdots,2p+2\), i.e., ?? holds.

To prove ?? , we first estimate \(\partial_t \widehat{V}(t,u)\). From \(A_t\widehat{V}(t,u) + RF(t,u + \widehat{V}(t,u)) = 0\) and the parameter-dependent implicit function theorem, the map \(t \mapsto \widehat{V}(t,u)\) is continuously differentiable for each fixed \(u\). Differentiating \(A_t\widehat{V}(t,u) + RF(t,u + \widehat{V}(t,u)) = 0\) with respect to \(t\), we get \[\begin{align} J(t,\Psi_t(u))\partial_t \widehat{V}(t,u) = -A_t' \widehat{V}(t,u) - R\partial_tF(t,\Psi_t(u)). \end{align}\] Using the boundedness of \(J^{-1}\) in Assumption 2, the boundedness of \(A_t'\) in Lemma 1, the linear growth of \(\widehat{V}\) in Lemma 2, and the time regularity assumption on \(F\) in Assumption 5 yields \[\begin{align} |\partial_t \widehat{V}(t,u)| \leq&~ |J(t,\Psi_t(u))^{-1}|\big(| A_t'|\,|\widehat{V}(t,u)| + |R|\,|\partial_t F(t,\Psi_t(u))|\bigr) \\\leq&~ C\big(1 + |u| + 1 + |\Psi_t(u)|^l\big) \\\leq&~ C(1 + |u|^l) \end{align}\] due to \(l \geq 1\) and \(|\Psi_t(u)| \leq C(1+|u|)\). Now differentiating \(f\) and \(g\) with respect to \(t\) gives \[\begin{gather} \partial_t f(t,u) = \bigg(\frac{d}{dt} A_t^{-}\bigg)F(t,\Psi_t(u)) + A_t^{-}\Big(\partial_t F(t,\Psi_t(u)) + D_x F(t,\Psi_t(u))\partial_t \widehat{V}(t,u)\Big), \\ \partial_t g(t,u) = \bigg(\frac{d}{dt} A_t^{-}\bigg)G(t,\Psi_t(u)) + A_t^{-}\Big(\partial_t G(t,\Psi_t(u)) + D_x G(t,\Psi_t(u))\partial_t \widehat{V}(t,u)\Big). \end{gather}\] By the boundedness of \(A_t^{-}\) and \(\frac{d}{dt} A_t^{-}\), the linear growth of \(F\) and \(G\), the time regularity of \(F\) and \(G\), the boundedness of \(D_xF\) and \(D_xG\), and the estimate for \(\partial_t \widehat{V}\), we get \[\begin{align} |\partial_t f(t,u)| + |\partial_t g(t,u)| \leq&~ \bigg|\frac{d}{dt} A_t^{-}\bigg|\big(|F(t,\Psi_t(u))| + |G(t,\Psi_t(u))|\big) \\&~+ |A_t^{-}|\big(|\partial_tF(t,\Psi_t(u))| + |\partial_tG(t,\Psi_t(u))|\big) \\&~+ |A_t^{-}|\big(|D_xF(t,\Psi_t(u))| + |D_xG(t,\Psi_t(u))|\big) |\partial_t \widehat{V}(t,u)| \\\leq&~ C(1 + |\Psi_t(u)|) + C(1 + |\Psi_t(u)|^l) + C(1 + |u|^l) \\\leq&~ C(1 + |u|^l), \end{align}\] which implies ?? .

Finally, we verify the \(C^{1,2}\)-regularity. It suffices to show that \(\partial_t f, \partial_t g, D_u f, D_u g, D_u^2 f, D_u^2 g\) are jointly continuous in \((t,u) \in [0,T] \times \mathbb{R}^{d}\). Indeed, from 6 and \(\Psi_t(u) = u + \widehat{V}(t,u)\), we have \[\begin{gather} \partial_t f(t,u) = \bigg(\frac{d}{dt} A_t^{-}\bigg)F(t,\Psi_t(u)) + A_t^{-}\big(\partial_tF(t,\Psi_t(u)) + D_xF(t,\Psi_t(u))\partial_t\widehat{V}(t,u)\big), \\ \partial_t g(t,u) = \bigg(\frac{d}{dt} A_t^{-}\bigg)G(t,\Psi_t(u)) + A_t^{-}\big(\partial_tG(t,\Psi_t(u)) + D_xG(t,\Psi_t(u))\partial_t\widehat{V}(t,u)\big), \\ D_u f(t,u) = A_t^{-}D_xF(t,\Psi_t(u))D_u\Psi_t(u), \quad D_u g(t,u) = A_t^{-}D_xG(t,\Psi_t(u))D_u\Psi_t(u), \\ D_u^2 f(t,u)[\xi,\eta] = A_t^{-}\big(D_x^2F(t,\Psi_t(u)) [D_u\Psi_t(u)\xi,D_u\Psi_t(u)\eta] + D_xF(t,\Psi_t(u))D_u^2\Psi_t(u)[\xi,\eta]\big), \\ D_u^2 g(t,u)[\xi,\eta] = A_t^{-}\big(D_x^2G(t,\Psi_t(u)) [D_u\Psi_t(u)\xi,D_u\Psi_t(u)\eta] + D_xG(t,\Psi_t(u))D_u^2\Psi_t(u)[\xi,\eta]\big) \end{gather}\] for any \(\xi,\eta \in \mathbb{R}^{d}\). By the assumptions on \(F\) and \(G\), the regularity of \(A_t^{-}\), and the parameter-dependent implicit-function formulas for \(\widehat{V}\), the quantities \(\widehat{V}, \partial_t \widehat{V}, D_u \widehat{V}, D_u^2 \widehat{V}\) are jointly continuous in \((t,u)\). It follows that \(\Psi_t(u), \partial_t \widehat{V}(t,u), D_u \Psi_t(u), D_u^2 \Psi_t(u)\) are also jointly continuous, and that \(\partial_t f, \partial_t g, D_u f, D_u g, D_u^2 f, D_u^2 g\) are jointly continuous on \([0,T] \times \mathbb{R}^{d}\). The proof is complete. ◻

Proof. The standard moment estimate for SDEs with globally Lipschitz and linearly growing coefficients yields ?? ; see, e.g., [20].

By Lemma 2, the classical differentiability theorem for stochastic flows thus applies; see, e.g., [21]. It follows that the solution admits a version such that \(u \mapsto U(t,u;s)\) is \(2p+2\) times continuously differentiable. The derivative estimates follow from the corresponding variational equations. Indeed, the first derivative satisfies a linear SDE with bounded coefficients, while the \(j\)-th derivative satisfies a linear SDE whose nonhomogeneous terms involve only derivatives of the flow of order less than \(j\). Using the Burkholder–Davis–Gundy inequality, the Gronwall inequality, and induction over \(j\), one obtains ?? .

Finally, the flow property follows from the pathwise uniqueness of the strong solution. Indeed, for \(0 \leq t \leq s \leq r \leq T\), both \(U(t,u;r)\) and \(U(s,U(t,u;s);r)\) solve the same SDE on \([s,T]\) with the same initial value \(U(t,u;s)\) at time \(s\). Hence they are indistinguishable. The proof is complete. ◻

Proof. From (i) and the global solvability of the algebraic equation 3 , it follows that \(Y_n = u_n + \widehat{V}(t_n,u_n), n = 0,1,\cdots,N\). Together with \(X_T = U(T) + \widehat{V}(T,U(T))\) and \(\psi(u) := \varphi(u + \widehat{V}(T,u))\), we obtain \[\begin{align} \label{eq:transform95to95w95correct95B} \mathbb{E}\big[\varphi(X_T)\big] - \mathbb{E}\big[\varphi(Y_N)\big] =&~ \mathbb{E}\big[\varphi\big( U(T) + \widehat{V}(T,U(T)) \big)\big] - \mathbb{E}\big[\varphi\big(u_N + \widehat{V}(T,u_N)\big)\big] \notag \\=&~ \mathbb{E}\big[\psi(U(T))\big] - \mathbb{E}\big[\psi(u_N)\big] \notag \\=&~ \mathbb{E}\big[\psi(U(t_0,u_0;T))\big] - \mathbb{E}\big[\psi(U(t_N,u_N;T))\big] \notag \\=&~ \mathbb{E}\big[w(t_0,u_0)\big] - \mathbb{E}\big[w(t_N,u_N)\big]. \end{align}\tag{14}\] By the flow property of the strong solution, it holds that for any \(0 \leq t \leq s \leq T\) and \(u \in \mathbb{R}^{d}\), \[\begin{align} U(t,u;T) = U(s,U(t,u;s);T), \quad \mathbb{P}\text{-a.s.} \end{align}\] Since \(f\) and \(g\) are deterministic and the Brownian increments after time \(s\) are independent of \(\mathcal{F}_s\), the solution \(U(t,u;\cdot)\) is a time-inhomogeneous Markov process [20], we have \[\label{eq:w95semigroup95B} w(t,u) = \mathbb{E}\big[\psi(U(s,U(t,u;s);T))\big] = \mathbb{E}\big[\mathbb{E}\big(\psi(U(s,U(t,u;s);T)) \mid \mathcal{F}_s \big) \big] = \mathbb{E}\big[w(s,U(t,u;s))\big]\tag{15}\] for any \(0 \leq t \leq s \leq T\), which implies \(\mathbb{E}\big[w(t_{n},u_{n})\big] = \mathbb{E}\big[w\big(t_{n+1},U(t_{n},u_{n};t_{n+1})\big)\big], n = 0,1,\cdots,N-1\). As a consequence, we obtain \[\begin{align} \label{eq:A495sde95B} \mathbb{E}\big[w(t_0,u_0)\big] - \mathbb{E}\big[w(t_N,u_N)\big] = \sum_{i=0}^{N-1}\big( \mathbb{E}\big[w(t_{i+1},U(t_i,u_i;t_{i+1}))\big] - \mathbb{E}\big[w(t_{i+1},u_{i+1})\big]\big). \end{align}\tag{16}\] Using \(u_{n+1} = \widetilde{u}(t_{n},u_{n};t_{n+1})\), (iv) and \(|u_{n}| \leq |P|\,|Y_{n}|\) enables us to get \[\begin{align} &~\big|\mathbb{E}\big[w(t_0,u_0)\big] - \mathbb{E}\big[w(t_N,u_N)\big]\big| \\\leq&~ \sum_{n=0}^{N-1}\big| \mathbb{E}\big[w(t_{n+1},U(t_{n},u_{n};t_{n+1}))\big] - \mathbb{E}\big[w(t_{n+1},\widetilde{u}(t_{n},u_{n};t_{n+1}))\big]\big| \\\leq&~ Ch^{p+1} \sum_{n=0}^{N-1} \big(1 + \mathbb{E}\big[|u_{n}|^{q}\big]\big) \leq Ch^{p+1} \sum_{n=0}^{N-1} \big(1 + \mathbb{E}\big[|Y_{n}|^{q}\big]\big), \end{align}\] which together with (ii) implies the desired result and thus completes the proof. ◻

Proof. Since \(Y_{0} = X_{0} \in \mathcal{M}_{0}\), it suffices to argue by mathematical induction. Assume that \(\mathbb{P}(Y_{n} \in \mathcal{M}_{t_n}) = 1\) for some fixed \(n\). Letting \(u_n := PY_n\) gives \(Y_n = u_n + \widehat{V}(t_n,u_n)\). Since \(P = A_t^{-}A_t\), we have \(PA_t^{-} = A_t^{-}\), and hence \(\operatorname{Im}(A_t^{-})\subseteq \operatorname{Im}(P)\). Define \[\begin{align} Z_n := u_n + h(1-\theta)A_{t_n}^{-}F(t_n,Y_n) + A_{t_n}^{-}G(t_n,Y_n)\Delta W_n \in \operatorname{Im}(P). \end{align}\] and \(\Gamma_n(u) := Z_n + h\theta A_{t_n}^{-} F\bigl(t_{n+1},u+\widehat{V}(t_{n+1},u)\bigr), u \in \operatorname{Im}(P)\). Noting that \(\widehat{V}(t_{n+1},u) \in \operatorname{Im}(Q)\) and \(A_{t_n}^{-}F\bigl(t_{n+1},u+\widehat{V}(t_{n+1},u)\bigr) \in \operatorname{Im}(A_{t_n}^{-}) \subseteq\operatorname{Im}(P)\), we have \(\Gamma_{n} \colon \operatorname{Im}(P) \to \operatorname{Im}(P)\). For any \(u_1, u_2 \in \operatorname{Im}(P)\), using the Lipschitz continuity of \(F\) with respect to the space variable in Assumption 3 and the Lipschitz continuity of \(\widehat{V}(t,\cdot)\) in Lemma 2, we obtain \[\begin{align} |\Gamma_n(u_1)-\Gamma_n(u_2)| \leq&~ h\theta |A_{t_n}^{-}|\, \bigl| F(t_{n+1},u_1 + \widehat{V}(t_{n+1},u_1)) - F(t_{n+1},u_2+\widehat{V}(t_{n+1},u_2)) \bigr| \\\leq&~ h\theta |A_{t_n}^{-}|L\big(|u_1-u_2| + |\widehat{V}(t_{n+1},u_1) - \widehat{V}(t_{n+1},u_2)|\big) \\\leq&~ hK_{\theta} |u_1-u_2|, \end{align}\] where \(K_{\theta} := \theta \sup_{t\in[0,T]}|A_t^{-}|\,L(1+L_V)\). Choosing \(h_{\mathrm{wp}} \in (0,1)\) such that \(h_{\mathrm{wp}}K_{\theta} < 1\), \(\Gamma_{n}\) is a contraction on the complete metric space \(\operatorname{Im}(P)\) for any \(h \in (0,h_{\mathrm{wp}})\). It follows that there exists a unique fixed point \(u_{n+1} \in \operatorname{Im}(P)\) satisfying \[\label{eq:un-fixed-point} u_{n+1} = Z_n + h\theta A_{t_n}^{-}F\big(t_{n+1}, u_{n+1}+\widehat{V}(t_{n+1},u_{n+1})\big).\tag{17}\] Defining \[\label{eq:Y-def95final} Y_{n+1} := u_{n+1} + \widehat{V}(t_{n+1},u_{n+1})\tag{18}\] and using the property of \(\widehat{V}\), we have \(RF(t_{n+1},Y_{n+1}) = 0\), which implies \(\mathbb{P}(Y_{n+1} \in \mathcal{M}_{t_{n+1}}) = 1\).

We next show that \(Y_{n+1}\) defined by 18 indeed solves 2 . Letting \[\begin{align} \mathcal{E}_{n} := A_{t_n}Y_{n+1} - A_{t_n}Y_n - h\big( (1-\theta)F(t_n,Y_n) + \theta F(t_{n+1},Y_{n+1})\big) - G(t_n,Y_n)\Delta{W_n} \end{align}\] and using \(A_{t_n}^{-}A_{t_n} = P\), \(PY_n = u_n\), \(PY_{n+1} = u_{n+1}\) yield \[\begin{align} A_{t_n}^{-}\mathcal{E}_{n} = u_{n+1} - u_n - h(1-\theta)A_{t_n}^{-}F(t_n,Y_n) - h\theta A_{t_n}^{-}F(t_{n+1},Y_{n+1}) - A_{t_n}^{-}G(t_n,Y_n)\Delta W_n. \end{align}\] Owing to 17 , we obtain \(A_{t_n}^{-} \mathcal{E}_{n} = 0\), and thus \(\mathcal{E}_{n} \in \ker(A_{t_n}^{-})\). For the Moore–Penrose inverse associated with the decomposition in Assumption 1, one has \(\ker(A_{t_n}^{-}) = \operatorname{Im}(R)\). It follows that \(\mathcal{E}_{n} \in \operatorname{Im}(R)\). Besides, applying \(R\) to \(\mathcal{E}_{n}\), and using \(RA_{t_n} = 0, RG(t_n,\cdot)=0\) lead to \[\begin{align} R\mathcal{E}_{n} = -h\big( (1-\theta)RF(t_n,Y_n) + \theta RF(t_{n+1},Y_{n+1}) \big). \end{align}\] As \(Y_{n} \in \mathcal{M}_{t_n}\) and \(Y_{n+1}\in\mathcal{M}_{t_{n+1}}\), we have \(RF(t_n,Y_n) = 0, RF(t_{n+1},Y_{n+1}) = 0\), which implies \(R\mathcal{E}_{n} = 0\). Together with \(\mathcal{E}_{n} \in \operatorname{Im}(R)\), we deduce that \(\mathcal{E}_{n} = R\mathcal{E}_{n} = 0\), which means that \(Y_{n+1}\) solves 2 . This proves existence.

It remains to prove uniqueness. Let \(\widetilde{Y}_{n+1}\) be any solution of 2 . Applying \(R\) to 2 gives \[\begin{align} 0 = h\big( (1-\theta)RF(t_n,Y_n) + \theta RF(t_{n+1},\widetilde{Y}_{n+1}) \big). \end{align}\] Owing to \(RF(t_n,Y_n) = 0\) and \(\theta > 0\), we get \(RF(t_{n+1},\widetilde{Y}_{n+1}) = 0\), and therefore \(\widetilde{Y}_{n+1} \in \mathcal{M}_{t_{n+1}}\). Setting \(\widetilde{u}_{n+1} := P\widetilde{Y}_{n+1}\) and using the algebraic reconstruction gives \(\widetilde{Y}_{n+1} = \widetilde{u}_{n+1} + \widehat{V}(t_{n+1},\widetilde{u}_{n+1})\). Applying \(A_{t_n}^{-}\) to 2 gives \(\widetilde{u}_{n+1} = Z_n + h\theta A_{t_n}^{-}F(t_{n+1},\widetilde{Y}_{n+1})\). It follows that \[\begin{align} \widetilde{u}_{n+1} = Z_n + h\theta A_{t_n}^{-} F\big(t_{n+1},\widetilde{u}_{n+1} + \widehat{V}(t_{n+1},\widetilde{u}_{n+1})\big) = \Gamma_n(\widetilde{u}_{n+1}), \end{align}\] that is to say, \(\widetilde{u}_{n+1}\) is a fixed point of \(\Gamma_n\). Since \(\Gamma_n\) has a unique fixed point in \(\operatorname{Im}(P)\), we have \(\widetilde{u}_{n+1} = u_{n+1}\), and consequently \(\widetilde{Y}_{n+1} = Y_{n+1}\). This proves uniqueness. ◻

Proof. For any \(n = 0,1,\cdots,N\), using \(u_n := PY_n\) and \(\mathbb{P}(Y_n \in \mathcal{M}_{t_n}) = 1\) yields \(Y_{n} = u_{n} + \widehat{V}(t_n,u_n)\). Together with \(PA_t^{-} = A_t^{-}\) and 6 , applying \(A_{t_n}^{-}\) to 2 gives \[\begin{align} \label{eq:Deltaun} u_{n+1} = u_n + h(1-\theta)f(t_n,u_n) + h\theta A_{t_n}^{-}F(t_{n+1},Y_{n+1}) + g(t_n,u_n)\Delta{W_n}. \end{align}\tag{19}\] We first derive several local estimates for \(\Delta u_n :=u_{n+1}-u_n\). By the linear growth of \(f, F, g\), the boundedness of \(A_t^{-}\), and the linear growth of \(\widehat{V}\) in Lemma 2, there exists a constant \(C_{0} > 0\), independent of \(h,n,N\), such that for any \(n = 0,1,\cdots,N\), \[\begin{align} |\Delta u_n| \leq&~ Ch(1+|u_n|) + Ch(1+|Y_{n+1}|) + C(1+|u_n|)|\Delta W_n| \\\leq&~ C_{0}h(1+|u_n|) + C_{0}h\bigl(1+|u_{n+1}|\bigr) + C_{0}(1+|u_n|)|\Delta W_n| \\\leq&~ C_{0}h(1+2|u_n|+|\Delta u_n|) + C_{0}(1+|u_n|)|\Delta W_n|. \end{align}\] Choosing \(h_{0} := \min\big\{h_{\mathrm{wp}},\frac{1}{2C_{0}},1\big\} > 0\) and employing \(1-C_{0}h \geq \frac{1}{2}\) for any \(h \in (0, h_0]\), we obtain \[\label{eq:Delta-u-pathwise-optimized} |\Delta u_n| \leq C(1+|u_n|)\bigl(h+|\Delta W_n|\bigr).\tag{20}\] It follows from \(\mathbb{E}\big[|\Delta W_n|^{r}\big] \leq C_r h^{\frac{r}{2}}, r \geq 2\) that \[\label{eq:Delta-u-local-moment-optimized} \mathbb{E}\big(|\Delta u_n|^{r} \mid \mathcal{F}_{t_n}\big) \leq C_{r} h^{\frac{r}{2}}(1+|u_n|^r).\tag{21}\] We also need a conditional first-moment estimate for \(\Delta u_n\). By means of \(Y_{n+1} = u_{n+1} + \widehat{V}(t_{n+1},u_{n+1})\), the linear growth of \(\widehat{V}\) in Lemma 2 and \(u_{n+1} = u_{n} + \Delta{u_{n}}\), we have \[\begin{align} |Y_{n+1}|^2 \leq C(1 + |u_{n+1}|^2) \leq C(1 + |u_n|^2 + |\Delta u_n|^2). \end{align}\] Combining with 20 leads to \[\begin{align} \label{eq:EYnplus1Ftn} \mathbb{E}\big(|Y_{n+1}|^{2} \mid \mathcal{F}_{t_n}\big) \leq&~ C(1+|u_n|^2) + C\mathbb{E}\big(|\Delta u_n|^2 \mid \mathcal{F}_{t_n}\big) \notag \\\leq&~ C(1+|u_n|^2) + C(1+|u_n|^2)\mathbb{E}\bigl[(h+|\Delta W_n|)^2\bigr] \notag \\\leq&~ C(1+|u_n|^2)+Ch(1+|u_n|^2) \notag \\\leq&~ C(1+|u_n|^2). \end{align}\tag{22}\] Owing to 19 and \(\mathbb{E}\big(\Delta W_n \mid \mathcal{F}_{t_n}) = 0\), we have \[\begin{align} \mathbb{E}\big(\Delta u_n \mid \mathcal{F}_{t_n}\big) = h(1-\theta)f(t_n,u_n) + h\theta A_{t_n}^{-} \mathbb{E}\big(F(t_{n+1},Y_{n+1}) \mid \mathcal{F}_{t_n} \big). \end{align}\] Using the linear growth of \(f\) and \(F\), the boundedness of \(A_t^{-}\) and 22 , we obtain \[\label{eq:Delta-u-mean-optimized} \left|\mathbb{E}\big(\Delta u_n \mid \mathcal{F}_{t_n}\big)\right| \leq Ch(1+|u_n|).\tag{23}\]

We now prove ?? . Applying the standard inequality \[\begin{align} |x+y|^q \leq |x|^q + q|x|^{q-2} \langle x,y \rangle + C_q |x|^{q-2}|y|^2 + C_q |y|^q, \quad q \geq 2, x,y \in \mathbb{R}^{d} \end{align}\] with \(x = u_n\) and \(y = \Delta u_n\) (see, e.g., [16]) and taking conditional expectation show that \[\begin{align} \mathbb{E}\big(|u_{n+1}|^q \mid \mathcal{F}_{t_n} \big) \leq&~ |u_n|^q + C_q |u_n|^{q-1} \left|\mathbb{E}\big(\Delta u_n \mid \mathcal{F}_{t_n} \big)\right| \\&~ + C_q |u_n|^{q-2}\mathbb{E}\big( |\Delta u_n|^2 \mid \mathcal{F}_{t_n} \big) + C_q\mathbb{E}\big(|\Delta u_n|^{q} \mid \mathcal{F}_{t_n}\big). \end{align}\] Because of 21 and 23 , we get \[\begin{align} \mathbb{E}\big(|u_{n+1}|^q \mid \mathcal{F}_{t_n} \big) \leq&~ |u_n|^q + C_qh|u_n|^{q-1}(1+|u_n|) + C_qh|u_n|^{q-2}(1+|u_n|^2) + C_qh^{q/2}(1+|u_n|^q) \\\leq&~ |u_n|^q+C_qh(1+|u_n|^q), \end{align}\] and thus \(\mathbb{E}\big[|u_{n+1}|^{q}\big] \leq (1+C_qh) \mathbb{E}\big[|u_n|^q\big] + C_{q}h\). Then the discrete Gronwall inequality implies the first result in ?? . Next, the linear growth of \(\widehat{V}\) and \(Y_n = u_n + \widehat{V}(t_n,u_n)\) indicate \(|Y_n|^q \leq C_q(1+|u_n|^q)\) and thus \[\begin{align} \mathbb{E}\big[|Y_n|^{q}\big] \leq C_q\left(1+ \mathbb{E}\big[|u_n|^q\big]\right) \le C_q(1+|u_0|^q), \end{align}\] which together with \(|u_0| \le |P|\,|Y_0|\) proves the second result in ?? .

We now prove the increment estimate ?? . Applying Lemma 2 yields \[\begin{align} |Y_{n+1}-Y_n| \leq&~ |\Delta u_n| + |\widehat{V}(t_{n+1},u_{n+1})-\widehat{V}(t_{n+1},u_n)| + |\widehat{V}(t_{n+1},u_n)-\widehat{V}(t_n,u_n)| \\\leq&~ C|\Delta u_n| + Ch(1+|u_n|). \end{align}\] It follow from 21 that \[\begin{align} \mathbb{E}\big(|Y_{n+1}-Y_n|^q \mid \mathcal{F}_{t_n}\big) \leq C_q\mathbb{E}\big(|\Delta u_n|^q \mid \mathcal{F}_{t_n}\big) + C_qh^q(1+|u_n|^q) \leq C_qh^{\frac{q}{2}}(1+|u_n|^q), \end{align}\] which along with \(|u_n| \leq |P|\,|Y_n|\) ensures ?? . It remains to prove ?? . From the decomposition \[\begin{align} Y_{n+1}-Y_n =&~ \Delta u_n + \widehat{V}(t_{n+1},u_n) - \widehat{V}(t_n,u_n) + \widehat{V}(t_{n+1},u_{n+1}) - \widehat{V}(t_{n+1},u_n) \\=&~ \Delta u_n + \widehat{V}(t_{n+1},u_n) - \widehat{V}(t_n,u_n) + D_u\widehat{V}(t_{n+1},u_n)\Delta u_n \\&~ + \int_0^1(1-\rho) D_u^2 \widehat{V}(t_{n+1},u_n+\rho\Delta u_n) [\Delta u_n,\Delta u_n]\,d\rho, \end{align}\] we utilize the bonunedness of \(D_u\widehat{V}\) and \(D_u^2\widehat{V}\) in Lemma 3, and the \(\mathcal{F}_{t_n}\)-measurability of \(D_u\widehat{V}(t_{n+1},u_n)\) to derive that \[\begin{align} \left|\mathbb{E}\big(Y_{n+1}-Y_n \mid \mathcal{F}_{t_n}\big)\right| \leq C\big|\mathbb{E}\big(\Delta u_n \mid \mathcal{F}_{t_n}\big)\big| + \big|\widehat{V}(t_{n+1},u_n) - \widehat{V}(t_n,u_n)\big| + C\mathbb{E}\big(|\Delta u_n|^2 \mid \mathcal{F}_{t_n}\big). \end{align}\] Making use of 21 , 23 and the time-continuity estimate for \(\widehat{V}\) in Lemma 2 results in \[\begin{align} \left|\mathbb{E}\bigl[Y_{n+1}-Y_n\mid\mathcal{F}_{t_n}\bigr]\right| \leq Ch(1+|u_n|) + Ch(1+|u_n|) + Ch(1+|u_n|^2) \leq Ch(1+|u_n|^2). \end{align}\] which in combination with \(|u_n| \leq |P|\,|Y_n|\) yields ?? . Thus, we complete the proof. ◻

Proof. By Proposition 3, the reduced SDE 7 generates a stochastic flow \(U(t,u;s)\), \(0 \leq t \leq s \leq T\), which is four times continuously differentiable with respect to the initial value \(u\). Moreover, for every \(q \geq 2\), there exists a constant \(C > 0\) such that for any \(t \in [0,T], u \in \mathbb{R}^{d}\), \[\begin{align} \label{eq:estUDuj} \mathbb{E}\big[|U(t,u;T)|^{q}\big] \leq C(1+|u|^{q}), \quad \mathbb{E}\big[|D_u^jU(t,u;T)|^{q}\big] \leq C, \quad j = 1,\cdots,4. \end{align}\tag{26}\]

We first prove the spatial derivative estimate. Since \(\psi \in C_{\mathrm{pol}}^{4}(\mathbb{R}^{d};\mathbb{R})\), there exist constants \(C > 0\) and \(\chi \geq 0\) such that \(|D^j\psi(z)| \leq C(1+|z|^\chi)\) for all \(z \in \mathbb{R}^{d}\) and \(j = 0,1,2,3,4\). It follows from 26 that for the case \(j = 0\), \[\begin{align} |D_u^0 w(t,u)| = |w(t,u)| \leq \mathbb{E}\big[|\psi(U(t,u;T))|\big] \leq C\big(1 + \mathbb{E}\big[|U(t,u;T)|^{\chi}\big]\big) \leq C\big(1 + |u|^{\chi}\big). \end{align}\] To estimate \(D_u^j w(t,u)\) for \(j = 1,2,3,4\), we need to justify the interchange of differentiation and expectation. For every direction \(\xi \in \mathbb{R}^{d}\), the difference quotient of \(w\) satisfies \[\frac{w(t,u+\varepsilon \xi)-w(t,u)}{\varepsilon} = \mathbb{E}\left[\frac{\psi(U(t,u+\varepsilon \xi;T)) - \psi(U(t,u;T))}{\varepsilon}\right].\] Owing to the differentiability of \(U(t,u;T)\) with respect to \(u\) in Proposition 3, we have \[\left.\frac{d}{d\varepsilon}\psi\bigl(U(t,u+\varepsilon\xi;T)\bigr)\right|_{\varepsilon=0} = D\psi\bigl(U(t,u;T)\bigr)\bigl[D_uU(t,u;T)[\xi]\bigr], \quad \text{\mathbb{P}-a.s.}\] Equivalently, \[Q_{\varepsilon} := \frac{\psi\bigl(U(t,u+\varepsilon\xi;T)\bigr) - \psi\bigl(U(t,u;T)\bigr)}{\varepsilon} \longrightarrow D\psi\bigl(U(t,u;T)\bigr)\bigl[D_uU(t,u;T)[\xi]\bigr]\] almost surely as \(\varepsilon \to 0\). We shall show that, for \(\varepsilon_0>0\) sufficiently small, the family \(\{Q_\varepsilon: 0< |\varepsilon| \le \varepsilon_0\}\) is uniformly integrable. Choose a compact neighbourhood \(K\) of \(u\), for instance \(K:=\overline{B(u,1)}\). Taking \(\varepsilon_0 > 0\) sufficiently small (e.g., \(\varepsilon_0 \leq |\xi|^{-1}\)), we may assume that \[u+\rho\varepsilon\xi\in K, \quad 0<|\varepsilon|\le\varepsilon_0, \quad \rho\in[0,1].\] By Proposition 3, the map \(v \mapsto U(t,v;T)\) is differentiable with respect to the initial value. Therefore, for fixed \(\omega\), applying the chain rule along the line segment \(v_\rho := u + \rho\varepsilon\xi\) gives \[\begin{align} |Q_{\varepsilon}| =&~ \left|\int_0^1 D\psi(U(t,u+\rho\varepsilon\xi;T)) \big[D_uU(t,u+\rho\varepsilon\xi;T)[\xi]\big]\,d\rho\right| \\\leq&~ |\xi| \int_0^1 |D\psi(U(t,u+\rho\varepsilon\xi;T))| |D_uU(t,u+\rho\varepsilon\xi;T)|\,d\rho \\\leq&~ C|\xi| \int_0^1 \bigl(1+|U(t,u+\rho\varepsilon\xi;T)|^\chi\bigr) |D_uU(t,u+\rho\varepsilon\xi;T)|\,d\rho. \end{align}\] Now choose some \(\gamma > 1\), for example \(\gamma = 2\). By the Hölder inequality, one has \[\begin{align} \mathbb{E}\big[|Q_\varepsilon|^{\gamma}\big] \leq&~ C|\xi|^\gamma \int_0^1 \mathbb{E}\left[ \bigl(1+|U(t,u+\rho\varepsilon\xi;T)|^\chi\bigr)^\gamma |D_uU(t,u+\rho\varepsilon\xi;T)|^\gamma \right] \,d\rho \\\leq&~ C|\xi|^\gamma \int_0^1 \big(\mathbb{E}\big[ \bigl(1+|U(t,u+\rho\varepsilon\xi;T)|^\chi\bigr)^{2\gamma} \big]\big)^{\frac{1}{2}} \big(\mathbb{E}\left[|D_uU(t,u+\rho\varepsilon\xi;T)|^{2\gamma} \right]\big)^{\frac{1}{2}} \,d\rho. \end{align}\] Since \(K\) is compact, ?? and ?? indicate that \[\begin{align} \sup_{v\in K}\mathbb{E}\big[|U(t,v;T)|^{2\gamma\chi}\big] \leq \sup_{v\in K}C(1+|v|^{2\gamma\chi}) < +\infty, \quad \sup_{v \in K}\mathbb{E}\big[|D_uU(t,v;T)|^{2\gamma}\big] < +\infty. \end{align}\] Together with \(u + \rho\varepsilon\xi \in K\) for all \(0 < |\varepsilon| \leq \varepsilon_{0}\) and \(\rho \in [0,1]\), we obtain \[\begin{align} \sup_{0 < |\varepsilon| \leq \varepsilon_{0}} \mathbb{E}\big[|Q_\varepsilon|^{\gamma}\big] \leq C_{K}|\xi|^{\gamma} < +\infty, \end{align}\] which together with \(\mathbb{E}\big[|Q_\varepsilon|\mathbf{1}_{\{|Q_\varepsilon|>R\}}\big] \leq \mathbb{E}\big[|Q_\varepsilon|^\gamma\big] R^{1-\gamma}\) for any \(R > 0\) implies \[\begin{align} \lim_{R \to \infty} \sup_{0 < |\varepsilon| \leq \varepsilon_{0}} \mathbb{E} \left[|Q_\varepsilon|\mathbf{1}_{\{|Q_\varepsilon|>R\}}\right] = 0. \end{align}\] Thus the family \(\{Q_\varepsilon: 0< |\varepsilon| \le \varepsilon_0\}\) is uniformly integrable. It follows from the Vitali convergence theorem that \[D_uw(t,u)[\xi] = \mathbb{E}\left[D\psi(U(t,u;T))D_uU(t,u;T)[\xi] \right].\] The same arguments apply to the higher-order directional derivatives. Indeed, for \(j = 2,3,4\), the multivariate Faà di Bruno formula expresses \(D_u^j(\psi(U(t,u;T)))\) as a finite sum of terms of the form \[D^k\psi(U(t,u;T))\Big[D_u^{\ell_1}U(t,u;T), \cdots, D_u^{\ell_k}U(t,u;T)\Big], \quad \ell_1 + \cdots + \ell_k = j.\] Each such term is integrable and locally uniformly integrable by the Hölder inequality, the polynomial growth of \(D^k\psi\), and the moment bounds for the flow derivatives. Therefore, \(D_u^j\) and \(\mathbb{E}\) can be interchanged for \(j = 1,\cdots,4\). Differentiating \(w(t,u)=\mathbb{E}[\psi(U(t,u;T))]\) with respect to \(u\) and using the multivariate Faà di Bruno formula, we obtain \[\begin{align} D_u^j w(t,u)[\xi_1,\cdots,\xi_j] = \mathbb{E}\bigg[\sum_{\pi\in\Pi_j}D^{|\pi|}\psi(U) \Big[J^{(|B_1|)}(t,u;T)[\xi_{B_1}], \cdots, J^{(|B_{|\pi|}|)}(t,u;T)[\xi_{B_{|\pi|}}]\Big]\bigg], \end{align}\] where \(\Pi_j\) denotes the set of all partitions \(\pi = \{B_1,\cdots,B_{|\pi|}\}\) of \(\{1,\cdots,j\}\). For a block \(B = \{i_1,\cdots,i_\ell\}\), we use the notation \(J^{(|B|)}(t,u;T)[\xi_B] := D_u^{\ell}U(t,u;T)[\xi_{i_1},\cdots,\xi_{i_\ell}]\). Since the number of partitions of \(\{1,\cdots,j\}\) is finite, using the polynomial growth of \(D^k\psi\), the Hölder inequality, the moment estimate of \(U(t,u;T)\) in 26 , and the derivative moment bounds for \(J^{(\ell)}\), we get \[\begin{align} |D_u^j w(t,u)[\xi_1,\cdots,\xi_j]| \leq C(1+|u|^q). \end{align}\] Taking the supremum over \(|\xi_1|,\cdots,|\xi_j| \leq 1\) yields \(|D_u^j w(t,u)| \leq C(1+|u|^q)\) for \(j = 1,2,3,4\). Together with the case \(j = 0\), this proves ?? .

We next prove the time-derivative estimates. Owing to 24 , 25 , ?? and the linear growth of \(f\) and \(g\) in Lemma 2, we have \[\begin{align} \label{eq:partialtD0wtu} |\partial_t w(t,u)| \leq |f(t,u)|\,|D_u w(t,u)| + \frac{1}{2} |g(t,u)|^2\,|D_u^2 w(t,u)| \leq C(1+|u|^q). \end{align}\tag{27}\] It remains to estimate \(\partial_tD_uw\) and \(\partial_tD_u^2w\). Since \(w\) is a classical solution of the backward Kolmogorov equation 24 , differentiating \(\partial_t w(t,u) = -\mathcal{L}_t w(t,u)\) with respect to \(u\) gives \[\begin{align} \label{eq:partialtDuj} \partial_tD_u^j w(t,u) = -D_u^j(\mathcal{L}_t w)(t,u), \quad j = 1,2. \end{align}\tag{28}\] To compute \(D_u^j(\mathcal{L}_t w)\) explicitly, writting \(g^k(t,u)\) for the \(k\)-th column of \(g(t,u), k = 1,\cdots,m\) yields \[\begin{align} \mathcal{L}_t w(t,u) = D_uw(t,u)[f(t,u)] + \frac{1}{2}\sum_{k=1}^{m}D_u^2w(t,u)[g^k(t,u),g^k(t,u)]. \end{align}\] For any \(\xi,\eta \in \mathbb{R}^{d}\), the first and the second spatial derivatives are given by \[\begin{align} D_u(\mathcal{L}_t w)(t,u)[\xi] =&~ D_u^2w(t,u)[\xi,f(t,u)] + D_uw(t,u)[D_uf(t,u)[\xi]] \\&~+ \frac{1}{2}\sum_{k=1}^m D_u^3w(t,u)[\xi,g^k(t,u),g^k(t,u)] + \sum_{k=1}^m D_u^2w(t,u)[D_ug^k(t,u)[\xi],g^k(t,u)], \end{align}\] and \[\begin{align} &~D_u^2(\mathcal{L}_t w)(t,u)[\xi,\eta] \\=&~ D_u^3w(t,u)[\xi,\eta,f(t,u)] + D_u^2w(t,u)[\xi,D_uf(t,u)[\eta]] + D_u^2w(t,u)[\eta,D_uf(t,u)[\xi]] \\&~+ D_uw(t,u)[D_u^2f(t,u)[\xi,\eta]] + \frac{1}{2}\sum_{k=1}^{m} D_{u}^{4}w(t,u)[\xi,\eta,g^k(t,u),g^k(t,u)] \\&~+ \sum_{k=1}^m D_u^3w(t,u)[\xi,D_ug^k(t,u)[\eta],g^k(t,u)] + \sum_{k=1}^m D_u^3w(t,u)[\eta,D_ug^k(t,u)[\xi],g^k(t,u)] \\&~+ \sum_{k=1}^m D_u^2w(t,u)[D_u^2g^k(t,u)[\xi,\eta],g^k(t,u)] + \sum_{k=1}^m D_u^2w(t,u)[D_ug^k(t,u)[\xi],D_ug^k(t,u)[\eta]]. \end{align}\] Thanks to Lemma 2, there exists a constant \(C > 0\) such that for any \(\xi,\eta \in \mathbb{R}^{d}\), \[\left|D_u(\mathcal{L}_t w)(t,u)[\xi]\right| \leq C(1+|u|^q)|\xi|, \quad \left|D_u^2(\mathcal{L}_t w)(t,u)[\xi,\eta]\right| \leq C(1+|u|^q)|\xi|\,|\eta|.\] Taking the supremum over \(|\xi|,|\eta| \leq 1\) and using 28 yield \[|\partial_t D_u w(t,u)| + |\partial_t D_u^2 w(t,u)| = |D_u(\mathcal{L}_t w)(t,u)| + |D_u^2(\mathcal{L}_t w)(t,u)| \leq C(1+|u|^q),\] which together with 27 proves ?? .

Finally, for \(j=0,1,2\) and \(s,t\in[0,T]\), the fundamental theorem of calculus gives \[D_u^j w(t,u)-D_u^j w(s,u) = \int_s^t \partial_rD_u^j w(r,u)\,dr.\] It follows from ?? that \[|D_u^j w(t,u)-D_u^j w(s,u)| \leq C(1+|u|^q)|t-s|, \quad j=0,1,2,\] which yields ?? . The proof is complete. ◻

Proof. Since \(\Delta{W_{t,h}}\) is independent of \(\mathcal{F}_t\) and satisfies \[\begin{align} \mathbb{E}[\Delta{W_{t,h}}]=0, \quad \mathbb{E}[\Delta{W_{t,h}}^{k}\Delta{W_{t,h}}^{\ell}] = h\delta_{k\ell}, \quad \mathbb{E}\big[|\Delta{W_{t,h}}|^{q}\big] \leq C_{r} h^{\frac{q}{2}}, \quad r \geq 2, \end{align}\] we use the linear growth of \(f\) and \(g\) as well as \(h \leq h_{0} < 1\) to show that for every \(q \geq 2\), \[\begin{align} \label{eq:eta-moment-frozen} \mathbb{E}\big[|hf(t,u) + g(t,u)\Delta{W_{t,h}}|^{q}\big] \leq Ch^{q}|f(t,u)|^{q} + C|g(t,u)|^{q}\mathbb{E}\big[|\Delta{W_{t,h}}|^{q}\big] \leq C(1+|u|^{q})h^{\frac{q}{2}}. \end{align}\tag{29}\] For simplicity, we write \(\eta := hf(t,u) + g(t,u)\Delta{W_{t,h}}\) and \(\phi(x) := w(t+h,x), x \in \mathbb{R}^{d}\). Applying \(\bar{u}(t,u;t+h) = u+\eta\) and the Taylor formula with integral remainder leads to \[\begin{align} \phi(\bar{u}(t,u;t+h)) = \phi(u) + D\phi(u)[\eta] + \frac{1}{2}D^2\phi(u)[\eta,\eta] + \frac{1}{6}D^3\phi(u)[\eta,\eta,\eta] + R_4, \end{align}\] where \(R_{4} := \frac{1}{6}\int_{0}^{1}(1-\rho)^{3} D^{4}\phi(u+\rho\eta)[\eta,\eta,\eta,\eta]\,d\rho\). First, \(\mathbb{E}[D\phi(u)[\eta]] = hD\phi(u)[f(t,u)]\). Second, \[\begin{align} \mathbb{E}\big[D^2\phi(u)[\eta,\eta]\big] = h\sum_{k=1}^mD^2\phi(u)[g^{k}(t,u),g^{k}(t,u)] + h^2D^2\phi(u)[f(t,u),f(t,u)] \end{align}\] with \(\left|h^2D^2\phi(u)[a,a]\right| \leq C(1+|u|^q)h^2\) because of the growth estimate of \(D^2w\) in Lemma 5 and the linear growth of \(f\). For the third-order term, using the vanishing of odd moments of the centered Gaussian increment, we have \[\begin{align} \mathbb{E}\big[D^3\phi(u)[\eta,\eta,\eta]\big] = h^3D^3\phi(u)[f(t,u),f(t,u),f(t,u)] + 3h^2\sum_{k=1}^mD^3\phi(u)[f(t,u),g^{k}(t,u),g^{k}(t,u)] \end{align}\] with \(\big|\mathbb{E}\big[D^{3}\phi(u)[\eta,\eta,\eta]\big]\big| \leq C(1+|u|^q)h^{2}\), which is due to the growth estimate of \(D^3w\) and the linear growth of \(f\) and \(g\). For the remainder, Lemma 5 gives \(|D^4\phi(x)| = |D_u^4w(t+h,x)| \leq C(1+|x|^q)\), which together with 29 yields \[\begin{align} \mathbb{E}\big[|R_4|\big] \leq C\mathbb{E}\left[(1+|u+\rho\eta|^q)|\eta|^4\right] \leq C(1+|u|^q)\mathbb{E}\big[|\eta|^4\big] + C\mathbb{E}\big[|\eta|^{q+4}\big] \leq C(1+|u|^q)h^{2}. \end{align}\] Combining the preceding estimates, we obtain \[\begin{align} \mathbb{E}\big[\phi(\bar{u}(t,u;t+h))\big] &= \phi(u) + hD\phi(u)[f(t,u)] + \frac{h}{2}\sum_{k=1}^mD^2\phi(u)[g^{k}(t,u),g^{k}(t,u)] + \mathcal{R}_h \end{align}\] with \(|\mathcal{R}_{h}| \leq C(1 + |u|^{q})h^{2}\). It follows from \[\begin{align} D\phi(u)[f(t,u)] + \frac{1}{2}\sum_{k=1}^mD^2\phi(u)[g^{k}(t,u),g^{k}(t,u)] = \mathcal{L}_t w(t+h,u) \end{align}\] that \(\mathbb{E}\bigl[w(t+h,\bar u(t,u;t+h))\bigr] = w(t+h,u) + h\mathcal{L}_t w(t+h,u) + \mathcal{R}_h\), which proves ?? . ◻

Proof. By the semigroup property of the Kolmogorov function 15 with \(s = t+h\), we have \[\label{eq:exact-semigroup-local} \mathbb{E}\bigl[w(t+h,U(t,u;t+h))\bigr] = w(t,u).\tag{30}\] Thus it remains to compare \(w(t,u)\) with the first-order frozen expansion at time \(t+h\). From 24 , we get \(w(t+h,u) - w(t,u) = -\int_t^{t+h}\mathcal{L}_s w(s,u)\,ds\), which implies \[\begin{align} \label{eq:kolmogorov-defect} &~w(t+h,u)+h\mathcal{L}_t w(t+h,u)-w(t,u) \notag = \int_t^{t+h} \big(\mathcal{L}_t w(t+h,u)-\mathcal{L}_s w(s,u)\big)\,ds \notag \\=&~ \int_t^{t+h} \big(\mathcal{L}_t w(t+h,u)-\mathcal{L}_s w(t+h,u)\big) + \big(\mathcal{L}_s w(t+h,u)-\mathcal{L}_s w(s,u)\big)\,ds. \end{align}\tag{31}\] By the fundamental theorem of calculus and the time-regularity estimates for \(f\) and \(g\) in ?? , after possibly increasing \(q\), we have \[\label{eq:fg-time-difference-for-frozen} |f(s,u)-f(t,u)| + |g(s,u)-g(t,u)| \leq C(1+|u|^q)|s-t|.\tag{32}\] which together with the linear growth of \(g\) gives \[|g(s,u)g(s,u)^\top-g(t,u)g(t,u)^\top| \leq |g(s,u)-g(t,u)|\bigl(|g(s,u)|+|g(t,u)|\bigr) \leq C(1+|u|^q)|s-t|.\] It follows from the definition of \(\mathcal{L}_t\), ?? and 32 that \[\begin{align} \label{eq:L-coefficient-difference} \big|\mathcal{L}_t w(t+h,u)-\mathcal{L}_s w(t+h,u)\big| \notag =&~ \Big|\big\langle f(t,u)-f(s,u),D_uw(t+h,u)\big\rangle \notag \\&~+ \frac{1}{2}\operatorname{Tr}\!\big(\big(g(t,u)g(t,u)^{\top} - g(s,u)g(s,u)^{\top}\big)D_u^2w(t+h,u)\big)\Big| \notag \\\leq&~ \frac{1}{2} |g(t,u)g(t,u)^\top-g(s,u)g(s,u)^\top|\,|D_u^2w(t+h,u)| \notag \\&+ |f(t,u)-f(s,u)|\,|D_uw(t+h,u)| \notag \\\leq&~ C(1+|u|^q)|s-t|. \end{align}\tag{33}\] Applying the definition of \(\mathcal{L}_s\), the linear growth of \(f\) and \(g\) and ?? leads to \[\begin{align} \label{eq:L-w-time-difference} \big|\mathcal{L}_s w(t+h,u) - \mathcal{L}_s w(s,u) \big| \notag =&~ \Big|\left\langle f(s,u), D_u w(t+h,u)-D_u w(s,u) \right\rangle \notag \\&~ + \frac{1}{2} \operatorname{Tr}\!\left(g(s,u)g(s,u)^\top \bigl(D_u^2w(t+h,u)-D_u^2w(s,u)\bigr)\right)\Big| \notag \\\leq&~ |f(s,u)|\,|D_u w(t+h,u)-D_u w(s,u)| \notag \\&~ + \frac{1}{2} |g(s,u)|^2 \,|D_u^2w(t+h,u)-D_u^2w(s,u)| \notag \\\leq&~ C(1+|u|^q)|t+h-s|. \end{align}\tag{34}\] Combining 31 , 33 , and 34 , one gets \[\label{eq:kolmogorov-defect-bound} \left|w(t+h,u) + h\mathcal{L}_t w(t+h,u)-w(t,u)\right| \leq C(1+|u|^q)h^2.\tag{35}\] Due to 30 , 35 and Lemma 6, we finally obtain \[\begin{align} &~\left|\mathbb{E}\big[w(t+h,U(t,u;t+h))\big] - \mathbb{E}\big[w(t+h,\bar u(t,u;t+h))\big]\right| \\=&~ \left|w(t,u) - \mathbb{E}\bigl[w(t+h,\bar u(t,u;t+h))\bigr]\right| \\\leq&~ \left|w(t,u)-w(t+h,u)-h\mathcal{L}_t w(t+h,u)\right| \\&~ +\left|\mathbb{E}\bigl[w(t+h,\bar u(t,u;t+h))\bigr] - w(t+h,u) - h\mathcal{L}_t w(t+h,u)\right| \\\leq&~ C(1+|u|^q)h^2, \end{align}\] which proves the required and thus complete the proof. ◻

Proof. For simplicity, write \(\bar{u} := \bar{u}(t,u;t+h)\) and \(\delta := \widetilde{u} - \bar{u}\). Using \(f(t,u) = A_t^{-}F(t,Y_t)\) gives \[\label{eq:delta-implicit-correction} \delta = h\theta A_t^{-}\big(F(t+h,\widetilde{Y}_{t+h})-F(t,Y_t)\big).\tag{36}\] Recall that the local stochastic theta step is simply one step of the full method on \([t,t+h]\), with \[\begin{align} t_n=t,\quad t_{n+1}=t+h,\quad Y_n=Y_t:=u+\widehat{V}(t,u), \quad Y_{n+1}=\widetilde{Y}_{t+h}. \end{align}\] Moreover, in the present local estimate, \(u\) is fixed and deterministic. Hence \(Y_t = u + \widehat{V}(t,u)\) is also deterministic, and the only randomness comes from the Brownian increment \(\Delta{W_{t,h}}\). Applying the increment estimates in Proposition 7 to this single step shows that for every \(q \geq 2\), \[\begin{align} \label{eq:local-Y-increment-Lr} \mathbb{E}\big[|\widetilde{Y}_{t+h} - Y_{t}|^{q}\big] \leq Ch^{\frac{q}{2}}(1 + |Y_t|^{q}) \leq Ch^{\frac{q}{2}}(1 + |u|^{q}), \end{align}\tag{37}\] where we have used \(|Y_t| \leq |u| + |\widehat{V}(t,u)| \leq C(1+|u|)\) because of the linear growth of \(\widehat{V}\). Similarly, Proposition 7, applied to the deterministic initial value \(Y_{t}\), further gives \[\label{eq:local-Y-increment-mean} \big|\mathbb{E}\big[\widetilde{Y}_{t+h} - Y_{t}\big]\big| \leq Ch(1+|Y_t|^2) \leq Ch(1+|u|^2).\tag{38}\]

Let us derive two estimates for \(\delta\). Using 36 , 37 , the boundedness of \(A_t^{-}\) and the Lipschitz continuity of \(F\) in the space variable and its time-regularity, we obtain that for every \(q \geq 2\), \[\begin{align} \label{eq:delta-Lr} \mathbb{E}\big[|\delta|^{q}\big] \leq&~ Ch^{q}\mathbb{E}\big[|F(t+h,\widetilde{Y}_{t+h}) - F(t,Y_t)|^{q}\big] \notag \\\leq&~ Ch^{q}\mathbb{E}\big[|F(t+h,\widetilde{Y}_{t+h}) - F(t+h,Y_t)|^{q}\big] \notag \\&~+ Ch^{q}\mathbb{E}\big[|F(t+h,Y_t) - F(t,Y_t)|^{q}\big] \notag \\\leq&~ Ch^{q}\mathbb{E}\big[|\widetilde{Y}_{t+h} - Y_{t}\big|^{q}\big] \notag + Ch^{2q}(1+|Y_{t}|^{q}) \\\leq&~ C(1 + |u|^{q})h^{\frac{3q}{2}}. \end{align}\tag{39}\] Besides, the Taylor formula in the spatial variable implies \[\begin{align} F(t+h,\widetilde{Y}_{t+h}) - F(t+h,Y_t) = D_xF(t+h,Y_t)(\widetilde{Y}_{t+h}-Y_t) + R_F, \end{align}\] where \(R_{F} := \int_0^1(1-\rho) D_x^2F\bigl(t+h,Y_t+\rho(\widetilde{Y}_{t+h}-Y_t)\bigr) [\widetilde{Y}_{t+h} - Y_t,\widetilde{Y}_{t+h}-Y_t]\,d\rho\). By 36 , 37 , 38 , the boundedness of \(D_{x}^{2}F\) in Assumption 4 and Assumption 5, we have \[\begin{align} \label{eq:delta-mean} |\mathbb{E}[\delta]| \leq&~ Ch\big|D_xF(t+h,Y_t)\big|\,\big|\mathbb{E}\big[\widetilde{Y}_{t+h}-{Y}_t)\big]\big| + Ch\mathbb{E}\big[|R_{F}|\big] + Ch\big|F(t+h,Y_t) - F(t,Y_t)\big| \notag \\\leq&~ Ch\big|\mathbb{E}\big[\widetilde{Y}_{t+h} - Y_t\big]\big| + Ch\mathbb{E}\big[|\widetilde{Y}_{t+h}-Y_t|^{2}\big] + Ch^2(1 + |Y_{t}|) \notag \\\leq&~ C(1+|u|^2)h^2. \end{align}\tag{40}\]

We now compare \(w(t+h,\widetilde{u})\) and \(w(t+h,\bar{u})\). The Taylor formula for \(w(t+h,\cdot)\) gives \[\begin{align} w(t+h,\widetilde{u}) - w(t+h,\bar{u}) = D_uw(t+h,\bar{u})[\delta] + R_w \end{align}\] with \(R_{w} = \int_0^1(1-\rho) D_u^2w(t+h,\bar{u} + \rho\delta)[\delta,\delta]\,d\rho\). Utilizing the decomposition \[\begin{align} D_uw(t+h,\bar u)[\delta] = D_uw(t+h,u)[\delta] + \bigl(D_uw(t+h,\bar u)-D_uw(t+h,u)\bigr)[\delta] \end{align}\] leads to \[\begin{align} \label{eq:wtildeuminbaru} \mathbb{E}\big[w(t+h,\widetilde{u})\big] - \mathbb{E}\big[w(t+h,\bar{u})\big] =&~ \mathbb{E}\big[D_uw(t+h,u)[\delta]\big] + \mathbb{E}[R_w] \notag \\&~+ \mathbb{E}\big[\big(D_uw(t+h,\bar{u}) - D_uw(t+h,u)\big)[\delta]\big]. \end{align}\tag{41}\] Let us estimate these three terms separately. For \(\mathbb{E}\big[D_uw(t+h,u)[\delta]\big]\), ?? and 40 show that \[\label{eq:first-weak-correction-term} \big|\mathbb{E}\big[D_uw(t+h,u)[\delta]\big]\big| = \big|D_{u}w(t+h,u)\mathbb{E}\big[\delta\big]\big| \leq C(1 + |u|^{q})\big|\mathbb{E}\big[\delta\big]\big| \leq C(1+|u|^q)h^{2}.\tag{42}\] Concerning \(\mathbb{E}\big[\big(D_uw(t+h,\bar{u}) - D_uw(t+h,u)\big)[\delta]\big]\), applying ?? and the Taylor formula with integral remainder to \(x \mapsto D_{u} w(t+h,x)\) leads to \[\begin{align} |D_u w(t+h,\bar u)-D_u w(t+h,u)| \leq&~ |\bar{u}-u|\int_0^1 |D_u^2w(t+h,u+\rho\Delta\bar u)|\,d\rho \\\leq&~ C\bigl(1+|u|^q+|\bar u-u|^q\bigr)|\bar u-u|. \end{align}\] In combination with \[\begin{align} \label{estbaruminur} \mathbb{E}\big[|\bar{u}-u|^{q}\big] = \mathbb{E}\big[|hf(t,u) + g(t,u)\Delta{W_{t,h}}|^{q}\big] \leq C(1+|u|^{q})h^{\frac{q}{2}}, \quad q \geq 2, \end{align}\tag{43}\] one can use the Hölder inequality and 39 , after increasing \(q\) if necessary, to get \[\begin{align} \label{eq:second-weak-correction-term} &~\left|\mathbb{E}\big[\big(D_uw(t+h,\bar{u})-D_uw(t+h,u)\big)[\delta]\big]\right| \notag \\\leq&~ \mathbb{E}\big[|D_uw(t+h,\bar{u})-D_uw(t+h,u)|\,|\delta|\big] \notag \\\leq&~ C(1+|u|^q)\mathbb{E}\big[|\bar{u}-u|\,|\delta|\big] + C\mathbb{E}\big[|\bar{u}-u|^{q+1}\,|\delta|\big] \notag \\\leq&~ C(1+|u|^q)\big(\mathbb{E}\big[|\bar{u}-u|^{2}\big]\big)^{\frac{1}{2}} \big(\mathbb{E}\big[|\delta|^{2}\big]\big)^{\frac{1}{2}} + C\big(\mathbb{E}\big[|\bar{u}-u|^{2(q+1)}\big]\big)^{\frac{1}{2}} \big(\mathbb{E}\big[|\delta|^{2}\big]\big)^{\frac{1}{2}} \notag \\\leq&~ C(1+|u|^q)h^{\frac{1}{2}}h^{\frac{3}{2}} + C(1+|u|^q)h^{\frac{1+q}{2}}h^{\frac{3}{2}} \notag \\\leq&~ C(1+|u|^{q})h^{2}. \end{align}\tag{44}\] To estimate \(\mathbb{E}[R_w]\), we use the Hölder inequality, ?? and 39 to deduce that \[\begin{align} \label{eq:remainder-weak-correction} \big|\mathbb{E}\big[R_{w}\big]\big| \leq&~ C\mathbb{E}\left[\int_0^1(1-\rho) \big(1 + |\bar{u} + \rho\delta|^{q}\big)|\delta|^{2}\,d\rho\right] \notag \\\leq&~ C\mathbb{E}\left[\big(1 + |\bar{u}|^{q} + |\delta|^{q}\big)|\delta|^{2}\right] \notag \\\leq&~ C\mathbb{E}\left[\big(1 + |u|^{q} + |\bar{u}-u|^{q} + |\delta|^{q}\big)|\delta|^{2}\right] \notag \\\leq&~ C\big(1 + |u|^{q}\big)\mathbb{E}\left[|\delta|^{2}\right] + C\big(\mathbb{E}\left[|\bar{u}-u|^{2q}\right]\big)^{\frac{1}{2}} \big(\mathbb{E}\left[|\delta|^{4}\right]\big)^{\frac{1}{2}} + C\mathbb{E}\left[|\delta|^{q+2}\right] \notag \\\leq&~ C(1+|u|^{q})h^{3}. \end{align}\tag{45}\] Inserting 42 , 44 and 45 into 41 yields the desired result and thus complete the proof. ◻

Proof. From the triangle inequality, Propositions 8 and 9, it follows that \[\begin{align} &~\left|\mathbb{E}\big[w(t+h,U(t,u;t+h))\big] -\mathbb{E}\big[w(t+h,\widetilde{u}(t,u;t+h))\big]\right| \\\leq&~ \left|\mathbb{E}\big[w(t+h,U(t,u;t+h))\big] - \mathbb{E}\bigl[w(t+h,\bar{u}(t,u;t+h))\bigr]\right| \\&~ + \left|\mathbb{E}\big[w(t+h,\bar{u}(t,u;t+h))\bigr] - \mathbb{E}\bigl[w(t+h,\widetilde{u}(t,u;t+h))\bigr]\right| \\\leq&~ C(1+|u|^{q})h^{2}, \end{align}\] which proves ?? and finishes the proof. ◻

Proof. Conditions (i) and (ii) in Theorem 4 are ensured by Propositions 5 and 7, respectlvely. Besides, Proposition 5 implies that \(Y_{n} = u_{n} + \widehat{V}(t_n,u_n)\) for all \(n = 0,1,\cdots,N\), which together with the one-step recursion \(Y_{n+1} = Y(t_{n},Y_{n};t_{n+1})\) yields \[\begin{align} Y_{n+1} = Y(t_{n},u_{n} + \widehat{V}(t_n,u_n);t_{n+1}), \quad n = 0,1,\cdots,N-1. \end{align}\] Applying \(P\) to both sides yields \[\begin{align} u_{n+1} = PY_{n+1} = PY(t_{n},u_{n} + \widehat{V}(t_n,u_n);t_{n+1}) = \widetilde{u}(t_n,u_n;t_{n+1}), \quad n = 0,1,\cdots,N-1, \end{align}\] which verifies Condition (iii) in Theorem 4. Moreover, Proposition 10 validates the one-step weak local error condition, namely Condition (iv), with \(p = 1\). Therefore all assumptions of Theorem 4 are satisfied. Applying that theorem with \(p = 1\) finally completes the proof. ◻