Proof. Postponed to Appendix A. ◻

Proof. Postponed to Appendix A.2. ◻

Proof. Postponed to Appendix A.2. ◻

Proof. Postponed to Appendix A.2. ◻

Proof. Postponed to Appendix C.1. ◻

Proof. Postponed to Appendix C.1. ◻

Proof. Postponed to Appendix C.1. ◻

Proof. Postponed to Appendix C.1. ◻

Proof. Postponed to Appendix C.2. ◻

Proof. Postponed to Appendix C.2. ◻

Proof. Postponed to Appendix C.2. ◻

Proof. Postponed to Appendix C.2. ◻

Proof. Postponed to Appendix C.2. ◻

Proof. Postponed to Appendix C.3. ◻

Proof. Postponed to Appendix C.3. ◻

Proof. Postponed to Appendix C.3. ◻

Proof. Postponed to Appendix C.3. ◻

Proof of Theorem 1. By the triangle inequality for the Wasserstein-2 distance with Propositions 4 and 5, Corollary 1, and Lemma 5, we have, for all \(n\in\mathbb{N}_0\), \(t\in(nT,(n+1)T]\), \(\gamma\geq\gamma_{\min}\) and \(\lambda\leq\lambda_{\max,\gamma}\), that \[\begin{align} W_2(\mathcal{L}(Z_t^\lambda),\pi_\beta)&\leq W_2(\mathcal{L}(Z_t^\lambda),\mathcal{L}(\zeta_t^{\lambda,n}))+W_2(\mathcal{L}(\zeta_t^{\lambda,n}),\mathcal{L}(r_t^\lambda))+ W_2(\mathcal{L}(r_t^\lambda),\pi_\beta^\epsilon)+W_2(\pi_\beta^\epsilon,\pi_\beta)\\ &\leq C_0\left(\lambda\gamma+\sqrt{\frac{S_{\lambda,\gamma}}{\gamma}}\right)+\frac{16C_0e}{m}\left(\lambda\gamma^2+\sqrt{\gamma S_{\lambda,\gamma}}\right)\\ &\quad+4\exp{\left(-\frac{\lambda m}{2\gamma} t\right)}W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta^\epsilon)+W_2(\pi_\beta^\epsilon,\pi_\beta)\\ &\leq 2\max\left\{C_0,\frac{16C_0e}{m}\right\}\left(\lambda\gamma^2+\sqrt{\gamma S_{\lambda\gamma}}\right)\\ &\quad+4\exp{\left(-\frac{\lambda m}{2\gamma} t\right)}W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+5W_2(\pi_\beta^\epsilon,\pi_\beta)\\ &\leq 2\max\left\{C_0,\frac{16C_0e}{m}\right\}\left(\lambda\gamma^2+\sqrt{\gamma S_{\lambda\gamma}}\right)+4\exp{\left(-\frac{\lambda m}{2\gamma} t\right)}W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+5\bar{c}\epsilon\\ &\leq \dot{C}\left(\sqrt{\lambda}\gamma+\lambda\gamma^{5/2}+\gamma^{-1}+\exp\left(-\frac{\lambda m}{2\gamma} t\right)W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+\epsilon\right), \end{align}\] where \(\dot{C}=\mathcal{O}((d/\beta)^{3r+1/2})\), with its explicit expression given in Table [tab46constants]. Since \(\lambda T\geq 1/2\), it follows, for any \(n\in\mathbb{N}_0\), that \[\begin{align} W_2(\mathcal{L}(Z_{(n+1)T}^\lambda),\pi_\beta)&\leq\dot{C}\left(\sqrt{\lambda}\gamma+\lambda\gamma^{5/2}+\gamma^{-1}+\exp\left(-\frac{\lambda m}{2\gamma} (n+1)T\right)W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+\epsilon\right). \end{align}\] Then, we replace \((n+1)T\) on both sides of the above inequality with \(n+1\) to obtain \[W_2(\mathcal{L}(Z_{n+1}^\lambda),\pi_\beta)\leq\dot{C}\left(\sqrt{\lambda}\gamma+\lambda\gamma^{5/2}+\gamma^{-1}+\exp\left(-\frac{\lambda m}{2\gamma} (n+1)\right)W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+\epsilon\right),\] which yields, by using \(\mathcal{L}(V_n^\lambda,Z_n^\lambda)=\mathcal{L}(\nu_n^\lambda,\theta_n^\lambda)\), that \[W_2(\mathcal{L}(\theta_n^\lambda),\pi_\beta)\leq\dot{C}\left(\sqrt{\lambda}\gamma+\lambda\gamma^{5/2}+\gamma^{-1}+\exp\left(-\frac{\lambda m}{2\gamma}n\right)W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+\epsilon\right).\] This completes the proof. ◻

Proof. Postponed to Appendix C.3. ◻

Proof. Postponed to Appendix C.3. ◻

Proof of Theorem 2. Combining Lemmas 11, 12 and 13 yields the result. ◻

Proof of Remark 6. By Assumption 3, for all \(\theta\in\mathbb{R}^d\) and \(x\in\mathbb{R}^k\), it holds that \[\langle\theta,H(\theta,x)-H(0,x)\rangle\geq \langle\theta,A(x)\theta\rangle.\] Consequently, we obtain that \[\begin{align} \langle\theta,H(\theta,x)\rangle&\geq \langle\theta,A(x)\theta\rangle+\langle\theta,H(0,x)\rangle\\ &\geq \langle\theta,A(x)\theta\rangle-|\theta|\,|H(0,x)|\\ &\geq \langle\theta,A(x)\theta\rangle-m|\theta|^2-K_H^2(1+|x|)^{2\rho}/(4m)\\ &\geq \langle\theta,\widetilde{A}(x)\theta\rangle-\widetilde{b}(x) \end{align}\] with \(\widetilde{A}(x)=A(x)-m I_d\) and \(\widetilde{b}(x)=K_H^2(1+|x|)^{2\rho}/(4m)\), where the third inequality follows from ?? together with the elementary inequality \(ab<m a^2+b^2/(4m)\) for all \(a,b>0\). ◻

Proof of Proposition 1. By 12 , \(u\) is nonnegative and twice continuously differentiable. Moreover, its gradient is given by \[h(\theta)=-\frac{1}{\mathsf{N}}\sum_{i=1}^\mathsf{N}\frac{y_ix_i}{1+\exp(y_i\langle x_i,\theta\rangle)}+\eta\theta.\] Since \(I_l\sim \mathrm{Uniform}(\{1,\ldots,\mathsf{N}\})\) for \(l=1,\ldots,\mathsf{K}\), we have \(\mathbb{P}(I_l=i)=1/\mathsf{N}\) for all \(i=1,\ldots,\mathsf{N}\). Consequently, for all \(\theta\in\mathbb{R}^d\), \[\begin{align} \mathbb{E}[H(\theta,U_{\mathbf{z}})]&=\eta\theta-\frac{1}{\mathsf{K}}\sum_{l=1}^\mathsf{K}\mathbb{E}\left[\frac{y_{I_l}x_{I_l}}{1+\exp(y_{I_l}\langle x_{I_l},\theta\rangle)}\right]\\ &=\eta\theta-\frac{1}{\mathsf{K}}\sum_{l=1}^\mathsf{K}\left(\frac{1}{\mathsf{N}}\sum_{i=1}^\mathsf{N}\frac{y_ix_i}{1+\exp(y_i\langle x_i,\theta\rangle)}\right)=h(\theta). \end{align}\]

To show Assumption 2 holds, we have, for all \(\theta,\theta'\in\mathbb{R}^d\) and \(\bar z,\bar z'\in(\mathbb{R}^d\times\{-1,1\})^\mathsf{K}\), that \[|F(\theta,\bar z)-F(\theta',\bar z')|=\eta|\theta-\theta'|,\quad|F(\theta,\bar z)|=\eta|\theta|,\] which implies that Assumption 2(1) is satisfied with \(\rho=1\), \(r=1/2\), and \(L_F=K_F=\eta\). Furthermore, let \(s(t)\coloneq 1/(1+e^t)\) for \(t\in\mathbb{R}\), we have that \[G(\theta,\bar z)=-\frac{1}{\mathsf{K}}\sum_{i=1}^\mathsf{K} \bar y_i\bar x_i s(\bar y_i\langle\bar x_i,\theta\rangle).\] Moreover, \(s'(t)=-e^t/(1+e^t)^2\), so that \(|s'(t)|\leq 1/4\). Let \(M_x\coloneq\max_{1\leq i\leq \mathsf{N}}|x_i|\). Applying Cauchy-Schwarz inequality, together with \(|s'|\leq 1/4\), \(|y_i|=1\), and \(|x_i|\leq M_x\) for all \(i=1,\dots,\mathsf{N}\), yields \[\begin{align} \mathbb{E}[|G(\theta,U_{\mathbf{z}})-G(\theta',U_{\mathbf{z}})|]&\leq\frac{1}{\mathsf{K}}\sum_{l=1}^\mathsf{K}\mathbb{E}\left[|y_{I_l}x_{I_l}|\,|s(y_{I_l}\langle x_{I_l},\theta\rangle)-s(y_{I_l}\langle x_{I_l},\theta'\rangle)|\right]\\ &\leq \frac{1}{4\mathsf{K}}\sum_{l=1}^\mathsf{K}\mathbb{E}\left[|x_{I_l}|\,|y_{I_l}\langle x_{I_l},\theta-\theta'\rangle|\right]\\ &\leq \frac{1}{4\mathsf{K}}\sum_{l=1}^\mathsf{K}\mathbb{E}\left[|x_{I_l}|^2\right]|\theta-\theta'|\\ &= \frac{1}{4\mathsf{N}}\sum_{i=1}^\mathsf{N}|x_i|^2|\theta-\theta'|\\ &\leq \frac{1}{4}M_x^2|\theta-\theta'|. \end{align}\] Furthermore, by using \(|s|\leq 1\), \(|\bar x_i|\leq |\bar z|\) and \(|\bar y_i|=1\) for all \(i=1,\dots,\mathsf{K}\), we obtain that \[|G(\theta,\bar z)|\leq \frac{1}{\mathsf{K}}\sum_{i=1}^\mathsf{K} |\bar y_i \bar x_i|\,|s(\bar y_i\langle\bar x_i,\theta\rangle)|\leq \frac{1}{\mathsf{K}}\sum_{i=1}^\mathsf{K}|\bar x_i|\leq |\bar z|.\] Thus, Assumption 2(2) is satisfied with \(\vartheta=1\), \(\rho=1\), \(L_G=M_x^2/4\) and \(K_G=1\).

It remains to verify Assumption 3. For all \(\theta\in\mathbb{R}^d\) and \(\bar z\in(\mathbb{R}^d\times\{-1,1\})^\mathsf{K}\), the Jacobian matrix of \(H\) with respect to \(\theta\) is given by \[\label{eq46Jacobian32of32H} \nabla_\theta H(\theta,\bar z)=\eta I_d-\frac{1}{\mathsf{K}}\sum_{i=1}^\mathsf{K} (\bar y_i\bar x_i) s'(\bar y_i\langle\bar x_i,\theta\rangle)(\bar y_i\bar x_i)^\top.\tag{37}\] Since \(-s'(t)=e^t/(1+e^t)^2\geq 0\), \(|\bar y_i|=1\), and \(\bar x_i\bar x_i^\top\) is positive semidefinite for all \(i=1,\dots,\mathsf{K}\), 37 can be rewritten as \[\nabla_\theta H(\theta,\bar z)=\eta I_d+\frac{1}{\mathsf{K}}\sum_{i=1}^\mathsf{K} \frac{\exp(\bar y_i\langle\bar x_i,\theta\rangle)}{(1+\exp(\bar y_i\langle\bar x_i,\theta\rangle))^2}\bar x_i\bar x_i^\top\succeq\eta I_d,\] which implies, for all \(\theta,\theta'\in\mathbb{R}^d\) and \(\bar{z}\in(\mathbb{R}^d\times\{-1,1\})^\mathsf{K}\), that \[\begin{align} \langle\theta-\theta',H(\theta,\bar z)-H(\theta',\bar z)\rangle&=\int_0^1\langle\theta-\theta',\nabla_\theta H(\theta'+t(\theta-\theta'),\bar z)(\theta-\theta')\rangle\,\mathrm{d}t\\ &\geq \int_0^1\eta|\theta-\theta'|^2\,\mathrm{d}t=\eta|\theta-\theta'|^2. \end{align}\] Thus, Assumption 3 is satisfied by taking \(A(\bar z)=\eta I_d\), with \(m=\eta\). ◻

Proof of Proposition 2. Recall \(H\) defined in 18 , \((X_n)_{n\in\mathbb{N}_0}\) consists of i.i.d.standard Gaussian vectors, \(\widetilde{A}\) is positive definite and \(\widetilde{B}\) is positive semidefinite. Therefore, we obtain, for any \(\theta\in\mathbb{R}^d\), that \[u(\theta)=\frac{1}{2}\langle\theta,\widetilde{A}\theta\rangle+\frac{\eta}{4}\langle\theta,\widetilde{B}\theta\rangle^2\geq 0,\] and \[\begin{align} \mathbb{E}[H(\theta,X_0)]&=\mathbb{E}[\widetilde{A}\theta+X_0+\eta\langle\theta,\widetilde{B}\theta\rangle\widetilde{B}\theta]=\widetilde{A}\theta+\eta\langle\theta,\widetilde{B}\theta\rangle\widetilde{B}\theta=h(\theta). \end{align}\] Moreover, \(u\) is twice continuously differentiable, and \(\mathbb{E}[|X_0|^{12\rho r}]<\infty\).

We next verify Assumption 2. From the definition of \(\widetilde{A}\) and \(0<\bar{\varepsilon}<1\), we have, for all \(\theta,\theta'\in\mathbb{R}^d\) and \(x,x'\in\mathbb{R}^d\), that \[|F(\theta,x)-F(\theta',x')|=|\widetilde{A}(\theta-\theta')+(x-x')|\leq |\theta-\theta'|+|x-x'|,\] and \[|F(\theta,x)|=|\widetilde{A}\theta+x|\leq |\theta|+|x|\leq(1+|x|)(1+|\theta|).\] Thus, Assumption 2(1) is satisfied with \(\rho=1\), \(r\geq 1\) and \(L_F=K_F=1\). Moreover, recall that \(\widetilde{B}=vv^\top\) for some fixed \(v\in\mathbb{R}^d\), we have, for any \(\theta\in\mathbb{R}^d\), that \[\langle\theta,\widetilde{B}\theta\rangle=\langle\theta, vv^\top\theta\rangle=\langle v,\theta\rangle^2\quad\text{and}\quad \widetilde{B}\theta=vv^\top\theta=\langle v,\theta\rangle v,\] which implies that \[\label{eq46bar32G} G(\theta,x)=\eta\langle v,\theta\rangle^3 v.\tag{38}\] By using Cauchy-Schwarz inequality and 38 , we obtain, for all \(\theta,\theta'\in\mathbb{R}^d\), that \[\begin{align} \mathbb{E}[|G(\theta,X_0)-G(\theta',X_0)|]&\leq\eta|\langle v,\theta\rangle^3-\langle v,\theta'\rangle^3|\,|v|\\ &\leq\eta|\langle v,\theta-\theta'\rangle|\,\left(|\langle v,\theta\rangle|^2+|\langle v,\theta\rangle|\,|\langle v,\theta'\rangle|+|\langle v,\theta'\rangle|^2\right)|v|\\ &\leq \eta|v|^2|\theta-\theta'|(|v|^2|\theta|^2+|v|^2|\theta|\,|\theta'|+|v|^2|\theta'|^2)\\ &\leq \eta|v|^4(1+|\theta|+|\theta'|)^2|\theta-\theta'|, \end{align}\] and \[|G(\theta,x)|=|\eta\langle v,\theta\rangle^3 v|\leq \eta|v|^4|\theta|^3\leq \eta|v|^4(1+|\theta|)^3.\] Hence, Assumption 2(2) holds with \(L_G=K_G=\eta|v|^4\), \(\rho=1\) and \(\vartheta=3\). Since \(2r\in[\vartheta,\infty)\cap\mathbb{N}\), we take \(r=3/2\).

It remains to verify Assumption 3. For any \(\theta,\theta'\in\mathbb{R}^d\) and \(x\in\mathbb{R}^d\), \[\label{eq46Artificial-SG-difference} H(\theta,x)-H(\theta',x)=\widetilde{A}(\theta-\theta')+\eta[\langle v,\theta\rangle^3-\langle v,\theta'\rangle^3]\,v.\tag{39}\] The definition of \(\widetilde{A}\) and \(0<\bar{\varepsilon}<1\) imply \[\label{eq46Artificial-convex-1} \langle\theta-\theta',\widetilde{A}(\theta-\theta')\rangle\geq\bar{\varepsilon}|\theta-\theta'|^2.\tag{40}\] In addition, \[\begin{align} \left\langle\theta-\theta',\eta(\langle v,\theta\rangle^3-\langle v,\theta'\rangle^3)v\right\rangle&=\eta(\langle v,\theta\rangle^3-\langle v,\theta'\rangle^3)\langle\theta-\theta',v\rangle\notag\\ &=\eta(\langle v,\theta\rangle^3-\langle v,\theta'\rangle^3)(\langle v,\theta\rangle-\langle v,\theta'\rangle)\notag\\ &=\eta(\langle v,\theta\rangle-\langle v,\theta'\rangle)^2[\langle v,\theta\rangle^2+\langle v,\theta\rangle\langle v,\theta'\rangle+\langle v,\theta'\rangle^2]\notag\\ &\geq 0,\label{eq46Artificial-convex-2} \end{align}\tag{41}\] where the inequality follows from \(a^2+ab+b^2=(a+b/2)^2+3b^2/4\geq 0\) for any \(a,b\in\mathbb{R}\). Substituting 40 and 41 into 39 yields \[\langle\theta-\theta',H(\theta,x)-H(\theta',x)\rangle\geq \bar{\varepsilon}|\theta-\theta'|^2.\] Therefore, Assumption 3 holds with \(A(x)=\bar{\varepsilon} I_d\) and \(m=\bar{\varepsilon}\). ◻

Proof of Proposition 3. We recall that \(\bar{A}\) is positive definite, \(\kappa>0\), \(h_i,s_i>0\) for all \(i=1,\ldots,d\), and \(X\coloneq(X^1,\dots,X^d)\sim\mathcal{N}(\mu,\Sigma_X)\). Then, we have, for any \(\theta\in\mathbb{R}^d\), that \[\begin{align} \label{eq46Newsvendor-u} \begin{aligned} u(\theta)&=\frac{1}{2}\langle\theta,\bar A\theta\rangle+\sum_{i=1}^d\left(h_i\mathbb{E}[\max\{\theta_i-X^i,0\}]+s_i\mathbb{E}[\max\{X^i-\theta_i,0\}]\right)+\frac{\kappa}{4}|\theta|^4\\ &=\frac{1}{2}\langle\theta,\bar A\theta\rangle+\sum_{i=1}^d\left(h_i\int_{-\infty}^{\theta_i}(\theta_i-t)f_i(t)\,\mathrm{d}t+s_i\int_{\theta_i}^\infty(t-\theta_i)f_i(t)\,\mathrm{d}t\right)+\frac{\kappa}{4}|\theta|^4\geq 0, \end{aligned} \end{align}\tag{42}\] where we denote by \(f_i\) the marginal density of \(X_i\) for \(i=1,\dots,d\). By differentiating 42 and using \(X_0\sim\mathcal{N}(\mu,\Sigma_X)\), we obtain \[\begin{align} h(\theta)&=\bar A\theta+\sum_{i=1}^d\left(h_i\int_{-\infty}^{\theta_i}f_i(t)\,\mathrm{d}t-s_i\int_{\theta_i}^\infty f_i(t)\,\mathrm{d}t\right)e_i+\kappa|\theta|^2\theta\\ &=\bar A\theta+\sum_{i=1}^d\left(h_i\mathbb{P}(X^i<\theta_i)-s_i\mathbb{P}(X^i>\theta_i)\right)e_i+\kappa|\theta|^2\theta. \end{align}\] Furthermore, recall \(H\) in 22 , it holds that, for all \(\theta\in\mathbb{R}^d\), \[\begin{align} \mathbb{E}[H(\theta,X_0)]&=\bar{A}\theta+\sum_{i=1}^d\left(h_i\mathbb{E}[\mathbb{1}_{\{\theta_i>X_0^i\}}]-s_i\mathbb{E}[\mathbb{1}_{\{X_0^i>\theta_i\}}]\right)e_i+\kappa|\theta|^2\theta\\ &=\bar{A}\theta+\sum_{i=1}^d\left(h_i\mathbb{P}(X_0^i<\theta_i)-s_i\mathbb{P}(X_0^i>\theta_i)\right)e_i+\kappa|\theta|^2\theta=h(\theta). \end{align}\] Moreover, \(u\) is twice continuously differentiable, and \(\mathbb{E}[|X_0|^{12\rho r}]<\infty\).

We next verify Assumption 2. Let \(\bar c_d\) be a uniform upper bound for \(f_i\), \(i=1,\ldots,d\). For any \(\theta,\theta'\in\mathbb{R}^d\), and \(x,x'\in\mathbb{R}^d\), we have \[\begin{align} |F(\theta,x)-F(\theta',x')|&\leq\|\bar A\|_{\mathrm{op}}|\theta-\theta'|+\kappa||\theta|^2\theta-|\theta'|^2\theta'|\\ &\leq \|\bar A\|_{\mathrm{op}}|\theta-\theta'|+\kappa\left(|\theta|^2|\theta-\theta'|+||\theta|^2-|\theta'|^2|\,|\theta'|\right)\\ &\leq \|\bar A\|_{\mathrm{op}}|\theta-\theta'|+\kappa\left(|\theta|^2|\theta-\theta'|+|\theta-\theta'|(|\theta|+|\theta'|)|\theta'|\right)\\ &\leq (\|\bar A\|_{\mathrm{op}}+\kappa)(1+|\theta|+|\theta'|)^2|\theta-\theta'|, \end{align}\] where \(\|\bar A\|_{\mathrm{op}}\coloneq \sup_{|\theta|=1}|\bar A\theta|\) is the operator norm of \(\bar A\). Moreover, using \((a+b)^3\leq 4(a^3+b^3)\) for all \(a,b\geq 0\), we obtain that \[\begin{align} |F(\theta,x)|&\leq \|\bar A\|_{\mathrm{op}}|\theta|+\kappa|\theta|^3\leq 4(\|\bar A\|_{\mathrm{op}}+\kappa)(1+|\theta|^3). \end{align}\] Thus, Assumption 2(1) holds with \(\rho=1\), \(r=3/2\) and \(L_F=\|\bar A\|_{\mathrm{op}}+\kappa\) and \(K_F=4(\|\bar A\|_{\mathrm{op}}+\kappa)\). Moreover, for any \(\theta,\theta'\in\mathbb{R}^d\) and \(x\in\mathbb{R}^d\), it holds that \[\begin{align} |G(\theta,x)-G(\theta',x)|&\leq\sum_{i=1}^d h_i|\mathbb{1}_{\{\theta_i>x_i\}}-\mathbb{1}_{\{\theta_i'>x_i\}}|+\sum_{i=1}^d s_i|\mathbb{1}_{\{x_i>\theta_i\}}-\mathbb{1}_{\{x_i>\theta_i'\}}|\\ &\leq \sum_{i=1}^d (h_i+s_i)\mathbb{1}_{\{\min\{\theta_i,\theta_i'\}\leq x_i\leq\max\{\theta_i,\theta_i'\}\}}. \end{align}\] Taking expectations, using Cauchy-Schwarz inequality, and \(\sup_{1\leq i\leq d}|f_i|\leq \bar{c}_d\) for some \(\bar{c}_d>0\), we obtain \[\begin{align} \mathbb{E}[|G(\theta,X_0)-G(\theta',X_0)|]&\leq\sum_{i=1}^d (h_i+s_i)\int_{\min\{\theta_i,\theta_i'\}}^{\max\{\theta_i,\theta_i'\}}f_i(t)\,\mathrm{d}t\\ &\leq \bar{c}_d\sum_{i=1}^d (h_i+s_i)|\theta_i-\theta_i'|\\ &\leq \bar{c_d}(|h|+|s|)\,|\theta-\theta'|, \end{align}\] where \(h\coloneq (h_1,\dots,h_d)\) and \(s\coloneq(s_1,\dots,s_d)\). Furthermore, we have that \[\begin{align} |G(\theta,x)|&\leq \left|\sum_{i=1}^d \left(h_i\mathbb{1}_{\{\theta_i>x_i\}}-s_i\mathbb{1}_{\{x_i>\theta_i\}}\right)e_i\right|\\ &\leq \left|\sum_{i=1}^d h_i\mathbb{1}_{\{\theta_i>x_i\}}e_i\right|+\left|\sum_{i=1}^ds_i\mathbb{1}_{\{x_i>\theta_i\}}e_i\right|\\ &=\left(\sum_{i=1}^d h_i^2\mathbb{1}_{\{\theta_i>x_i\}}^2\right)^{1/2}+\left(\sum_{i=1}^d s_i^2\mathbb{1}_{\{x_i>\theta_i\}}^2\right)^{1/2}\\ &\leq |h|+|s|, \end{align}\] which implies that Assumption 2(2) is satisfied with \(\rho=\vartheta=1\), \(L_G=\bar{c}_d(|h|+|s|)\) and \(K_G=|h|+|s|\).

Finally, we verify Assumption 3. The mapping \(\theta\mapsto|\theta|^4\) is convex and differentiable, which implies that \(\langle\theta-\theta',|\theta|^2\theta-|\theta'|^2\theta'\rangle\geq 0\). In addition, for each \(i=1,\dots,d\), \[(\theta_i-\theta_i')(\mathbb{1}_{\{\theta_i>x_i\}}-\mathbb{1}_{\{\theta_i'>x_i\}})\geq 0,\] and \[-(\theta_i-\theta_i')(\mathbb{1}_{\{x_i>\theta_i\}}-\mathbb{1}_{\{x_i>\theta_i'\}})\geq 0.\] Consequently, we have \[\begin{align} \langle\theta-\theta',H(\theta,x)-H(\theta',x)\rangle&=\langle\theta-\theta',\bar{A}(\theta-\theta')\rangle+\kappa\langle\theta-\theta',|\theta|^2\theta-|\theta'|^2\theta'\rangle\\ &\quad+\sum_{i=1}^d h_i (\theta_i-\theta_i')(\mathbb{1}_{\{\theta_i>x_i\}}-\mathbb{1}_{\{\theta_i'>x_i\}})\\ &\quad-\sum_{i=1}^d s_i(\theta_i-\theta_i')(\mathbb{1}_{\{x_i>\theta_i\}}-\mathbb{1}_{\{x_i>\theta_i'\}})\\ &\geq \langle\theta-\theta',\bar{A}(\theta-\theta')\rangle. \end{align}\] Thus, Assumption 3 holds with \(A(x)=\bar{A}\). ◻

Proof of Lemma 2. By Lemma 1, we have, for all \(\theta\in\mathbb{R}^d\), that \[\label{eq46hprox} h(\operatorname{prox}_u^\epsilon(\theta))=h_{MY,\epsilon}(\theta)=\frac{1}{\epsilon}\left(\theta-\operatorname{prox}_u^\epsilon(\theta)\right).\tag{43}\] If \(\theta=\operatorname{prox}_u^\epsilon(\theta)\) for some \(\theta\in\mathbb{R}^d\), then the desired result holds. Otherwise, assuming \(\theta\neq\operatorname{prox}_u^\epsilon(\theta)\), we apply Remark 5, 43 and Cauchy-Schwarz inequality to obtain \[\begin{align} |h(\theta)|\,|\theta-\operatorname{prox}_u^\epsilon(\theta)|&\geq\langle h(\theta),\theta-\operatorname{prox}_u^\epsilon(\theta)\rangle\\ &=\langle h(\theta)-h(\operatorname{prox}_u^\epsilon(\theta)),\theta-\operatorname{prox}_u^\epsilon(\theta)\rangle+\langle h(\operatorname{prox}_u^\epsilon(\theta)),\theta-\operatorname{prox}_u^\epsilon(\theta)\rangle\\ &\geq m|\theta-\operatorname{prox}_u^\epsilon(\theta)|^2+\frac{1}{\epsilon}|\theta-\operatorname{prox}_u^\epsilon(\theta)|^2\\ &\geq \frac{1}{\epsilon}|\theta-\operatorname{prox}_u^\epsilon(\theta)|^2. \end{align}\] Combining this lower bound with 43 yields \[|h_{MY,\epsilon}(\theta)|= \frac{1}{\epsilon}|\theta-\operatorname{prox}_u^\epsilon(\theta)|\leq |h(\theta)|,\] which completes the proof. ◻

Proof of Lemma 3. By Remark 3, Lemma 2, 43 , and \(h(\theta^*)=0\), we have, for all \(\theta\in\mathbb{R}^d\), that \[\begin{align} |h(\theta)-h_{MY,\epsilon}(\theta)|&= |h(\theta)-h(\operatorname{prox}_u^\epsilon(\theta))|\notag\\ &\leq L_h(1+|\theta|+|\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1}|\theta-\operatorname{prox}_u^\epsilon(\theta)|\notag\\ &\leq L_h(1+|\theta|+|\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1}|h_{MY,\epsilon}(\theta)|\epsilon\notag\\ &\leq L_h(1+|\theta|+|\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1}|h(\theta)|\epsilon\notag\\ &= L_h(1+|\theta|+ |\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1} |h(\theta)-h(\theta^\ast)|\,\epsilon \notag\\ &\leq L_h^2(1+|\theta|+|\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1}(1+|\theta|+|\theta^*|)^{2r-1}|\theta-\theta^*|\epsilon\notag\\ &\leq L_h^2(1+|\theta|+|\operatorname{prox}_u^\epsilon(\theta)|)^{2r-1}(1+|\theta|+|\theta^*|)^{2r}\epsilon.\label{eq46h-MY} \end{align}\tag{44}\] According to [62], the proximal operator \(\operatorname{prox}_u^\epsilon\) is \(1\)-Lipschitz continuous. Additionally, Lemma 1 and definition of \(\operatorname{prox}_u^\epsilon\) imply that \(\operatorname{prox}_u^\epsilon(\theta^*)=\theta^*\). Combining these results with Remark 5, we obtain \[\begin{align} |\operatorname{prox}_u^\epsilon(\theta)|&\leq|\operatorname{prox}_u^\epsilon(\theta^*)|+|\operatorname{prox}_u^\epsilon(\theta)-\operatorname{prox}_u^\epsilon(\theta^*)|\notag\\ &\leq|\theta^*|+|\theta-\theta^*|\leq |\theta|+2|\theta^*|\leq |\theta|+2R_0.\label{eq46bound-prox} \end{align}\tag{45}\] Therefore, by substituting 45 back into 44 , we obtain \[\begin{align} |h(\theta)-h_{MY,\epsilon}(\theta)|&\leq L_h^2(1+2|\theta|+2R_0)^{2r-1}(1+|\theta|+R_0)^{2r}\epsilon\\ &\leq L_h^2\big(2(1+|\theta|+R_0)\big)^{2r-1}(1+|\theta|+R_0)^{2r}\epsilon\\ &\leq 2^{2r-1}L_h^2(1+|\theta|+R_0)^{4r-1}\epsilon, \end{align}\] which completes the proof. ◻

Proof of Lemma 4. Let \((x_t)_{t\geq 0}\) be the overdamped Langevin SDE defined by \[\mathrm{d}x_t=-h(x_t)\,\mathrm{d}t+\sqrt{2\beta^{-1}}\,\mathrm{d}B_t,\] where \(x_0=\theta_0\) and \((B_t)_{t\geq 0}\) is a \(d\)-dimensional Brownian motion. By Assumption 1, we note that \(\theta_0\) has \(2q\)-th finite moment. Moreover, by Remark 5, \(u\) satisfies a dissipativity condition. Hence, we can apply [42] with \(Z_t^\lambda\curvearrowleft x_t\), \(\lambda=1\), \(p\curvearrowleft q\), \(a_h\curvearrowleft m/2\), \(b_h\curvearrowleft u(0)\) to obtain, for all \(t\geq 0\), that \[\label{eq462q-bound32overdamped} \mathbb{E}[|x_t|^{2q}]\leq \exp\left(-\frac{mq}{2}t\right)\mathbb{E}[|\theta_0|^{2q}]+\left(\frac{4u(0)}{m}+\frac{4(d+2(q-1))}{\beta m}\right)^q.\tag{46}\] Since \(\mathcal{L}(x_t)\) converges to \(\pi_\beta\) in Wasserstein-2 distance [14], it also converges weakly to \(\pi_\beta=\mathcal{L}(Y)\) [66]. For any \(R>0\), define \(f_R(\theta)\coloneq \min\{|\theta|^{2q},R\}\), \(\theta\in\mathbb{R}^d\). We have, for any \(R>0\), that \(f_R\) is nonnegative, bounded and continuous. Thus, by the definition of weak convergence, \[\mathbb{E}[f_R(Y)]=\lim_{t\to\infty}\mathbb{E}[f_R(x_t)].\] Moreover, since \(f_R(\theta)\leq |\theta|^{2q}\) for all \(\theta\in\mathbb{R}^d\), we have, for any \(R>0\), that \[\label{eq46f95R} \mathbb{E}[f_R(Y)]=\lim_{t\to\infty}\mathbb{E}[f_R(x_t)]\leq\liminf_{t\to\infty}\mathbb{E}[|x_t|^{2q}].\tag{47}\] Since \(f_R(\theta)\uparrow |\theta|^{2q}\) as \(R\to\infty\), by monotone convergence theorem, 46 and 47 , we obtain \[\mathbb{E}[|Y|^{2q}]=\lim_{R\to\infty}\mathbb{E}[f_R(Y)]\leq\lim_{R\to\infty}\liminf_{t\to\infty}\mathbb{E}[|x_t|^{2q}]\leq\left(\frac{4u(0)}{m}+\frac{4(d+2(q-1))}{\beta m}\right)^q \eqcolon C_{\pi_\beta,2q},\] where \(C_{\pi_\beta,2q}=\mathcal{O}((d/\beta)^q)\). Similarly, let \((v_t)_{t\geq 0}\) be the overdamped Langevin SDE defined by \[\mathrm{d}v_t=-h_{MY,\epsilon}(v_t)\,\mathrm{d}t+\sqrt{2\beta^{-1}}\,\mathrm{d}B_t,\] where \(v_0=\theta_0\). By Assumption 1 and Remark 9, we apply [42] with \(Z_t^\lambda\curvearrowleft v_t\), \(h\curvearrowleft h_{MY,\epsilon}\), \(\lambda=1\), \(p\curvearrowleft q\), \(a_h\curvearrowleft m/4\), and \(b_h\curvearrowleft u(0)\) to obtain, for all \(t\geq 0\), that \[\mathbb{E}[|v_t|^{2q}]\leq\exp\left(-\frac{mq}{4}t\right)\mathbb{E}[|\theta_0|^{2q}]+\left(\frac{8u(0)}{m}+\frac{8(d+2(q-1))}{\beta m}\right)^q.\] By [14], \(\mathcal{L}(v_t)\) converges to \(\pi_\beta^\epsilon\) in Wasserstein-2 distance. Thus, using the same argument gives \[\mathbb{E}[|\bar Y|^{2q}]\leq \left(\frac{8u(0)}{m}+\frac{8(d+2(q-1))}{\beta m}\right)^q.\] This completes the proof. ◻

Proof of Lemma 5. Let \((\widetilde{x}_t)_{t\geq 0}\) be the overdamped Langevin SDE defined by \[\label{eq46overdamped32x} \mathrm{d}\widetilde{x}_t=-h(\widetilde{x}_t)\,\mathrm{d}t+\sqrt{2\beta^{-1}}\,\mathrm{d}B_t\tag{48}\] and let \((\widetilde{v}_t)_{t\geq 0}\) be defined by \[\mathrm{d}\widetilde{v}_t=-h_{MY,\epsilon}(\widetilde{v}_t)\,\mathrm{d}t+\sqrt{2\beta^{-1}}\,\mathrm{d}B_t,\] where both processes are driven by the same Brownian motion \((B_t)_{t\geq 0}\) and have the same initial condition \(\widetilde{x}_0=\widetilde{v}_0=\widetilde{\theta}_0\) satisfying \(\mathcal{L}(\widetilde{\theta}_0)=\pi_\beta\). By Itô’s formula, we obtain, for all \(t\geq 0\), that \[\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]=-2\int_0^t\mathbb{E}[\langle\widetilde{x}_s-\widetilde{v}_s,h(\widetilde{x}_s)-h_{MY,\epsilon}(\widetilde{v}_s)\rangle]\,\mathrm{d}s.\] Differentiating with respect to \(t\), and then using Young’s inequality together with Remark 9, yields \[\begin{align} \label{eq46time-derivative32x-y} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]&=-2\mathbb{E}[\langle\widetilde{x}_t-\widetilde{v}_t,h(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{v}_t)\rangle]\\ &=-2\mathbb{E}[\langle\widetilde{x}_t-\widetilde{v}_t,h(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{x}_t)\rangle]-2\mathbb{E}[\langle\widetilde{x}_t-\widetilde{v}_t,h_{MY,\epsilon}(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{v}_t)\rangle]\\ &\leq \frac{m}{2}\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]+\frac{2}{m}\mathbb{E}[|h(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{x}_t)|^2]-m\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]\\ &\leq -\frac{m}{2}\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]+\frac{2}{m}\mathbb{E}[|h(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{x}_t)|^2]. \end{aligned} \end{align}\tag{49}\] Then, by using Lemma 3 and 46 , we obtain that \[\begin{align} \frac{2}{m}\mathbb{E}[|h(\widetilde{x}_t)-h_{MY,\epsilon}(\widetilde{x}_t)|^2]&\leq \frac{2}{m} 2^{4r-2}L_h^4\mathbb{E}\left[(1+|\widetilde{x}_t|+R_0)^{8r-2}\right]\epsilon^2\\ &\leq \frac{2^{4r-1}}{m}L_h^4 2^{8r-3}\left((1+R_0)^{8r-2}+\mathbb{E}\left[|\widetilde{x}_t|^{8r-2}\right]\right)\epsilon^2\\ &\leq\frac{2^{12r-4}}{m} L_h^4\left((1+R_0)^{8r-2}+C_{\pi_\beta,8r-2}\right)\epsilon^2\leq c_\pi\epsilon^2, \end{align}\] where \(c_\pi\coloneq 2^{12r-4}L_h^4\left((1+R_0)^{8r-2}+C_{\pi_\beta,8r-2}\right)/m=\mathcal{O}((d/\beta)^{4r-1})\), and where \(C_{\pi_\beta,8r-2}\) is given in Lemma 4. Consequently, it holds that \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]&\leq -\frac{m}{2}\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]+c_\pi\epsilon^2, \end{align}\] which implies \[\begin{align} \mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]&\leq c_\pi\epsilon^2\frac{2}{m}(1-e^{-mt/2})\leq \frac{2c_\pi}{m}\epsilon^2. \end{align}\] By using the definitions of \((\widetilde{x}_t)_{t\geq 0}\) and \((\widetilde{v}_t)_{t\geq 0}\), we have, for any \(t\geq 0\), that \[W_2(\pi_\beta,\mathcal{L}(\widetilde{v}_t))=W_2(\mathcal{L}(\widetilde{x}_t),\mathcal{L}(\widetilde{v}_t))\leq\left(\mathbb{E}[|\widetilde{x}_t-\widetilde{v}_t|^2]\right)^{1/2}\leq\sqrt{\frac{2c_\pi}{m}}\epsilon.\] Therefore, we obtain that \[W_2(\pi_\beta,\pi_\beta^\epsilon)\leq W_2(\pi_\beta,\mathcal{L}(\widetilde{v}_t))+W_2(\mathcal{L}(\widetilde{v}_t),\pi_\beta^\epsilon)\leq \bar{c}\epsilon+W_2(\mathcal{L}(\widetilde{v}_t),\pi_\beta^\epsilon),\] where \(\bar{c}\coloneq \sqrt{2c_\pi/m}=\mathcal{O}((d/\beta)^{2r-1/2})\). By [14], \(\widetilde{v}_t\) converges to its invariant measure \(\pi_\beta^\epsilon\) in Wasserstein-2 distance. Hence, taking the limit as \(t\to\infty\) and noting the inequality \(W_2(\Pi_\beta,\Pi_\beta^\epsilon)\leq W_2(\pi_\beta,\pi_\beta^\epsilon)\) yields the desired result. ◻

Proof. By the definition of \(H_\gamma\) in 9 , together with Remark 6 and ?? , we have that \[\begin{align} \langle\theta,H_\gamma(\theta,x)\rangle&=m|\theta|^2+\frac{\langle\theta,H(\theta,x)\rangle-m|\theta|^2}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}\\ &\geq\left(1-\frac{1}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}\right)m|\theta|^2+\frac{\langle\theta,A(x)\theta\rangle}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}-\frac{m|\theta|^2+\widetilde{b}(x)}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}\\ &\geq -m|\theta|^2-\widetilde{b}(x). \end{align}\] Moreover, By using the definition of \(H_\gamma\) in 9 again, we have, for all \(\theta\in\mathbb{R}^d\), that \[\langle\mathbb{E}[H_\gamma(\theta,X_0)],\theta\rangle=\left\langle\frac{\mathbb{E}[H(\theta,X_0)]-m\theta}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}},\theta\right\rangle+m|\theta|^2.\] First, we consider the case where \(\langle\mathbb{E}[H(\theta,X_0)]-m\theta,\theta\rangle\geq 0\). Then, it follows that \[\left\langle\frac{\mathbb{E}[H(\theta,X_0)]-m\theta}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}},\theta\right\rangle\geq 0,\] which implies that \[\langle\mathbb{E}[H_\gamma(\theta,X_0)],\theta\rangle\geq m|\theta|^2\geq\frac{m}{2}|\theta|^2-u(0).\] Next, we consider \(\langle\mathbb{E}[H(\theta,X_0)]-m\theta,\theta\rangle< 0\), which implies that \[\left\langle\frac{\mathbb{E}[H(\theta,X_0)]-m\theta}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}},\theta\right\rangle\geq \langle\mathbb{E}[H(\theta,X_0)]-m\theta,\theta\rangle.\] Thus we obtain \[\langle\mathbb{E}[H_\gamma(\theta,X_0)],\theta\rangle\geq\langle\mathbb{E}[H(\theta,X_0)],\theta\rangle=\langle h(\theta),\theta\rangle\geq\frac{m}{2}|\theta|^2-u(0),\] where the last inequality follows from Remark 5. ◻

Proof. Recall \(H_\gamma\) defined in 9 and \(\gamma\geq\gamma_{\min}\geq 1\). For all \(\theta\in\mathbb{R}^d\) and \(x\in\mathbb{R}^k\), by using Remark 3, we obtain that \[\begin{align} |H_\gamma(\theta,x)|&\leq m|\theta|+\frac{|H(\theta,x)|+m|\theta|}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}\\ &\leq m|\theta|+\frac{\sqrt{2}(|H(\theta,x)|+m|\theta|)}{1+\gamma^{-1/2}|\theta|^{2r}}\\ &\leq m|\theta|+\frac{\sqrt{2}\,m|\theta|}{\gamma^{-1/2}(1+|\theta|^{2r})}+\frac{\sqrt{2}\,|H(\theta,x)|}{\gamma^{-1/2}(1+|\theta|^{2r})}\\ &\leq m|\theta|+\frac{\sqrt{2}m}{\gamma^{-1/2}}+\frac{\sqrt{2}K_H(1+|x|)^{\rho}}{\gamma^{-1/2}}, \end{align}\] where the second inequality follows from the inequality \((\sqrt{a}+\sqrt{b})/\sqrt{2}\leq\sqrt{a+b}\) for all \(a,b\geq 0\). Taking squares on both sides yields \[|H_\gamma(\theta,x)|^2\leq 2m^2|\theta|^2+8\gamma\left(m^2+K_H^2(1+|x|)^{2\rho}\right),\] which establishes the desired result. ◻

Proof of Lemma 6. Recall that \(\gamma_{\min}\) and \(\lambda_{\max,\gamma}\) are defined in 10 and 11 , respectively. For any \(\gamma\geq\gamma_{\min}\), \(\lambda\leq\lambda_{\max,\gamma}\), and \(n\in\mathbb{N}_0\), define \[\label{eq46Delta32n32and32En} \Delta_n\coloneq \theta_n^\lambda + \gamma^{-1}\nu_n^\lambda - \lambda\gamma^{-1} H_\gamma(\theta_n^\lambda,X_{n+1}),\quad E_n \coloneq \nu_n^\lambda - \lambda[\gamma \nu_n^\lambda+H_\gamma(\theta_n^\lambda,X_{n+1})].\tag{51}\] By using 8 9 , we obtain that \[\label{eq46expansion32of32theta32gamma32nu} |\theta_{n+1}^\lambda+\gamma^{-1}\nu_{n+1}^\lambda|^2=|\Delta_n|^2 +2\sqrt{\frac{2\lambda}{\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle +\frac{2\lambda}{\gamma\beta}|\xi_{n+1}|^2.\tag{52}\] Furthermore, by using 51 , Young’s inequality and Cauchy-Schwarz inequality, we have \[\begin{align} \label{eq46upper32bound32of32Delta95n} \begin{aligned} |\Delta_n|^2 &=|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2-2\lambda\gamma^{-1}\langle\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda, H_\gamma(\theta_n^\lambda,X_{n+1})\rangle+\lambda^2\gamma^{-2}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\leq |\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2 -2\lambda\gamma^{-1}{\langle\theta_n^\lambda, H_\gamma(\theta_n^\lambda,X_{n+1})\rangle}\\ &\quad+2\lambda\gamma^{-2}|\nu_n^\lambda|\,|H_\gamma(\theta_n^\lambda,X_{n+1})|+\lambda^2\gamma^{-2}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\leq|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2 +\lambda(2\gamma)^{-1}|\nu_n^\lambda|^2+2\lambda\gamma^{-3}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\quad-2\lambda\gamma^{-1}\langle\theta_n^\lambda, H_\gamma(\theta_n^\lambda,X_{n+1})\rangle+\lambda^2\gamma^{-2}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2. \end{aligned} \end{align}\tag{53}\] Then, by using 52 and the last term on the right hand side of 53 , we obtain that \[\begin{align} \label{eq46M95n-1} \begin{aligned} &\frac{\gamma^2}{4}\big(|\theta_{n+1}^\lambda+\gamma^{-1}\nu_{n+1}^\lambda|^2-|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2\big)\\ &\leq\frac{\lambda \gamma}{8}|\nu_n^\lambda|^2+\frac{\lambda}{2\gamma}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\quad-\frac{\lambda\gamma}{2}\langle\theta_n^\lambda, H_\gamma(\theta_n^\lambda,X_{n+1})\rangle+\frac{\lambda^2}{4}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\quad+\gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle +\frac{\lambda\gamma}{2\beta}|\xi_{n+1}|^2. \end{aligned} \end{align}\tag{54}\] Similarly, by using 8 and 51 , we have that \[\begin{align} \nu_{n+1}^\lambda&=(1-\lambda\gamma)\nu_n^\lambda-\lambda H_\gamma(\theta_n^\lambda,X_{n+1})+\sqrt{2\lambda\gamma\beta^{-1}}\xi_{n+1}\\ &=E_n+\sqrt{2\lambda\gamma\beta^{-1}}\xi_{n+1}. \end{align}\] Further calculations yield \[\begin{align} \label{eq46v32n431322} \begin{aligned} |\nu_{n+1}^\lambda|^2&=|E_n|^2+2\lambda\gamma\beta^{-1}|\xi_{n+1}|^2+2\sqrt{2\lambda\gamma\beta^{-1}}\langle E_n,\xi_{n+1}\rangle\\ &=(1-\lambda\gamma)^2|\nu_n^\lambda|^2+\lambda^2|H_\gamma(\theta_n^\lambda,X_{n+1})|^2+2\lambda\gamma\beta^{-1}|\xi_{n+1}|^2\\ &\quad -2\lambda(1-\lambda\gamma)\langle\nu_n^\lambda,H_\gamma(\theta_n^\lambda,X_{n+1})\rangle+2\sqrt{2\lambda\gamma\beta^{-1}}\langle E_n,\xi_{n+1}\rangle. \end{aligned} \end{align}\tag{55}\] By using 51 , 55 , and Young’s inequality, we have \[\begin{align} \label{eq46M95n-2} \begin{aligned} \frac{1}{4}(|\nu_{n+1}^\lambda|^2-|\nu_n^\lambda|^2) &\leq -\frac{\lambda\gamma}{2}|\nu_n^\lambda|^2+\frac{\lambda^2\gamma^2}{4}|\nu_n^\lambda|^2+\frac{\lambda}{2}|\langle\nu_n^\lambda,H_\gamma(\theta_n^\lambda,X_{n+1})\rangle|\\ &\quad+\frac{\lambda^2}{4}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2+\frac{\lambda\gamma}{2\beta}|\xi_{n+1}|^2+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle\\ &\leq -\frac{\lambda\gamma}{4}|\nu_n^\lambda|^2+\frac{\lambda^2\gamma^2}{4}|\nu_n^\lambda|^2+\frac{\lambda}{4\gamma}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2\\ &\quad+\frac{\lambda^2}{4}|H_\gamma(\theta_n^\lambda,X_{n+1})|^2+\frac{\lambda\gamma}{2\beta}|\xi_{n+1}|^2+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle. \end{aligned} \end{align}\tag{56}\] Furthermore, by using the definition of \(M_n\) in 50 , we have that \[\begin{align} M_n&=\frac{\gamma^2}{4}\left(|\theta_n^\lambda|^2+2\gamma^{-1}\langle\theta_n^\lambda,\nu_n^\lambda\rangle+\gamma^{-2}|\nu_n^\lambda|^2\right)+\frac{1}{4}|\nu_n^\lambda|^2-\frac{\sigma\gamma^2}{4}|\theta_n^\lambda|^2\\ &\leq \frac{\gamma^2}{4}|\theta_n^\lambda|^2+\frac{\gamma}{2}\langle\theta_n^\lambda,\nu_n^\lambda\rangle+\frac{1}{2}|\nu_n^\lambda|^2. \end{align}\] This, together with 8 9 , yields \[\begin{align} -\frac{\sigma\gamma^2}{4}(|\theta_{n+1}^\lambda|^2-|\theta_n^\lambda|^2)&=-\frac{\sigma\gamma^2}{4}(\lambda^2|\nu_n^\lambda|^2+2\lambda\langle\theta_n^\lambda,\nu_n^\lambda\rangle)\notag\\ &\leq -\frac{\sigma\lambda^2\gamma^2}{4}|\nu_n^\lambda|^2-\lambda\sigma\gamma M_n+\frac{\lambda\sigma\gamma}{2}|\nu_n^\lambda|^2+\frac{\lambda\sigma\gamma^3}{4}|\theta_n^\lambda|^2\notag\\ &\leq -\lambda\sigma\gamma M_n+\lambda\sigma\gamma^3|\theta_n^\lambda|^2+\frac{\lambda\sigma\gamma}{2}|\nu_n^\lambda|^2.\label{eq46M95n-3} \end{align}\tag{57}\] Using 50 , 54 , 56 , and 57 yields \[\begin{align} \label{eq46M95n431} \begin{aligned} M_{n+1}&\leq (1-\lambda\sigma\gamma)M_n+\lambda\left(\frac{\gamma}{8}-\frac{\gamma}{4}+\frac{\lambda\gamma^2}{4}+\frac{\sigma\gamma}{2}\right)|\nu_n^\lambda|^2+\lambda\sigma\gamma^3|\theta_n^\lambda|^2 \\ &\quad+\lambda\left(\frac{1}{2\gamma}+\frac{\lambda}{2}+\frac{1}{4\gamma}\right){|H_\gamma(\theta_n^\lambda,X_{n+1})|^2}-\frac{\lambda \gamma}{2}{\langle\theta_n^\lambda,H_\gamma(\theta_n^\lambda,X_{n+1})\rangle}\\ &\quad+\gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle+\frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle. \end{aligned} \end{align}\tag{58}\] Taking conditional expectations on both sides of 58 , denoted as \(\mathbb{E}_n[\cdot]=\mathbb{E}[\cdot\mid\theta_n^\lambda,\nu_n^\lambda]\) for notational simplicity, and using Lemmas 14 and 15 along with the fact that \(\Delta_n\) and \(E_n\) are independent of \(\xi_{n+1}\), yields \[\begin{align} \label{eq46coeffs32of32theta32nu} \begin{aligned} \mathbb{E}_n[M_{n+1}]&\leq (1-\lambda\sigma\gamma)M_n+\lambda\left(-\frac{\gamma}{8}+\frac{\lambda\gamma^2}{4}+\frac{\sigma\gamma}{2}\right)|\nu_n^\lambda|^2+\lambda\sigma\gamma^3|\theta_n^\lambda|^2\\ &\quad+\lambda\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)\left(2m^2|\theta_n^\lambda|^2+C_H\gamma\right)-\frac{\lambda\gamma}{2}\left(\frac{m}{2}|\theta_n^\lambda|^2-u(0)\right)+\frac{\lambda\gamma}{\beta}d\\ &= (1-\lambda\sigma\gamma)M_n+\lambda\left(-\frac{\gamma}{8}+\frac{\lambda\gamma^2}{4}+\frac{\sigma\gamma}{2}\right)|\nu_n^\lambda|^2\\ &\quad+\lambda\left[\sigma\gamma^3+2m^2\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)-\frac{\gamma m}{4}\right]|\theta_n^\lambda|^2\\ &\quad +\lambda\left[C_H\gamma\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)+\frac{\gamma}{2}u(0)+\frac{\gamma}{\beta}d\right]. \end{aligned} \end{align}\tag{59}\] Recall that \(\sigma=m/(8\gamma^2)\). Since \(\gamma\geq \gamma_{\min}\geq \max\{1,14m\}\) and \(\lambda\leq \lambda_{\max,\gamma}\leq 1/(4\gamma)\leq 1/4\), the coefficients of \(|\nu_n^\lambda|^2\) and \(|\theta_n^\lambda|^2\) in 59 are non-positive, i.e., \[\label{eq46non-positive32coefficients32of32theta32and32nu} \lambda\left(-\frac{\gamma}{8}+\frac{\lambda\gamma^2}{4}+\frac{\sigma\gamma}{2}\right)\leq 0,\quad \lambda\left[\sigma\gamma^3+2m^2\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)-\frac{\gamma m}{4}\right]\leq 0.\tag{60}\] Therefore, by using 60 , \(\gamma^{-1}\leq 1\), and \(\lambda\leq 1/(4\gamma)\leq 1/4\), we obtain that \[\begin{align} \mathbb{E}_n[M_{n+1}]&\leq\left(1-\frac{\lambda m}{8\gamma}\right)M_n+\lambda\gamma\left[C_H\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)+\frac{u(0)}{2}+\frac{d}{\beta}\right]\notag\\ &\leq\left(1-\frac{\lambda m}{8\gamma}\right)M_n+\lambda\gamma\left[\frac{7C_H}{8}+\frac{u(0)}{2}+\frac{d}{\beta}\right].\label{eq46expectation32Lyapunov} \end{align}\tag{61}\] Consequently, by induction and by taking expectations, we obtain \[\label{expectation32of32M95123n431125} \mathbb{E}[M_{n+1}]\leq \left(1-\frac{\lambda m}{8\gamma}\right)^{n+1} \mathbb{E}[M_0]+\frac{8\gamma^2}{m}\left[\frac{7C_H}{8}+\frac{u(0)}{2}+\frac{d}{\beta}\right].\tag{62}\] Moreover, since for \(\gamma\geq\gamma_{\min}\geq \sqrt{m}\), we have that \[\label{eq461-2sigma} 1-2\sigma\geq\frac{3}{4}>0,\tag{63}\] and therefore, we have \[\begin{align} M_n&=\frac{\gamma^2}{4}\left(|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2+|\gamma^{-1}\nu_n^\lambda|^2-\sigma|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda-\gamma^{-1}\nu_n^\lambda|^2\right)\notag\\ &\geq \frac{\gamma^2}{4}\left(|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2+|\gamma^{-1}\nu_n^\lambda|^2-2\sigma(|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2+|\gamma^{-1}\nu_n^\lambda|^2)\right)\notag\\ &=\frac{\gamma^2(1-2\sigma)}{4}|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda|^2+\frac{1-2\sigma}{4}|\nu_n^\lambda|^2\notag\\ &\geq \max\left\{\frac{(1-2\sigma)\gamma^2}{8}|\theta_n^\lambda|^2,\frac{1-2\sigma}{4}|\nu_n^\lambda|^2\right\},\label{eq46lower32bound32of32M95n} \end{align}\tag{64}\] where the last inequality follows from the inequality \(|a+b|^2\geq |a|^2/2-|b|^2\) for \(a,b>0\). This further implies \(M_n\geq 0\) for all \(n\in\mathbb{N}_0\). In addition, \[\label{eq46M950} M_0=\frac{\gamma^2}{4}|\theta_0+\gamma^{-1}\nu_0|^2+\frac{1}{4}|\nu_0|^2-\frac{\sigma\gamma^2}{4}|\theta_0|^2.\tag{65}\] Then, for \(\gamma\geq\gamma_{\min}\geq 1\), combining 62 65 , we have that \[\begin{align} \mathbb{E}[|\theta_n^\lambda|^2]&\leq \frac{8}{(1-2\sigma)\gamma^2}\mathbb{E}[M_n]\\ &\leq \frac{8}{(1-2\sigma)\gamma^2}\mathbb{E}[M_0]+\frac{64}{(1-2\sigma)m}\left[\frac{7C_H}{8}+\frac{u(0)}{2}+\frac{d}{\beta}\right]\\ &\leq \frac{8}{3}\mathbb{E}[|\theta_0+\gamma^{-1}\nu_0|^2]+\frac{8}{3}\mathbb{E}[|\nu_0|^2]+\frac{256}{3m}\left[\frac{7C_H}{8}+\frac{u(0)}{2}+\frac{d}{\beta}\right]\\ &\leq 6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2]+\frac{86}{m}\left[\frac{7C_H}{8}+\frac{u(0)}{2}+\frac{d}{\beta}\right]. \end{align}\] Thus, we obtain that \[\label{eq46sup-position} \sup_{n\in\mathbb{N}_0}\mathbb{E}[|\theta_n^\lambda|^2]\leq \bar{C}_2,\tag{66}\] where \(\bar{C}_2\coloneq 6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2]+86\big(7C_H/8+u(0)/2+d/\beta\big)/m=\mathcal{O}(d/\beta)\).

Finally, we establish a bound for the second moment of \(\nu_n^\lambda\). By using 55 , we obtain that \[\begin{align} |\nu_{n+1}^\lambda|^2&\leq (1-\lambda\gamma)|\nu_n^\lambda|^2+(\lambda^2+2\lambda\gamma^{-1})|H_\gamma(\theta_n^\lambda,X_{n+1})|^2+\frac{\lambda\gamma}{2}|\nu_n^\lambda|^2\\ &\quad+2\lambda\gamma\beta^{-1}|\xi_{n+1}|^2+2\sqrt{2\lambda\gamma\beta^{-1}}\langle E_n,\xi_{n+1}\rangle. \end{align}\] By taking conditional expectations, using Lemma 15, 66 , \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\leq 1/4\), we obtain that \[\begin{align} \mathbb{E}_n[|\nu_{n+1}^\lambda|^2]&\leq\left(1-\frac{\lambda\gamma}{2}\right)|\nu_n^\lambda|^2+(\lambda^2+2\lambda\gamma^{-1})(2m^2|\theta_n^\lambda|^2+C_H\gamma)+\frac{2\lambda\gamma}{\beta}d\\ &\leq \left(1-\frac{\lambda\gamma}{2}\right)|\nu_n^\lambda|^2+\lambda\gamma\left[2m^2(\lambda\gamma^{-1}+2\gamma^{-2})\bar{C}_2+(\lambda+2\gamma^{-1})C_H+\frac{2d}{\beta}\right]\\ &\leq \left(1-\frac{\lambda\gamma}{2}\right)|\nu_n^\lambda|^2+\lambda\gamma\left[\frac{9m^2}{2}\bar{C}_2+\frac{9}{4}C_H+\frac{2d}{\beta}\right]. \end{align}\] Iterating the above inequality and taking expectations yield, \[\mathbb{E}[|\nu_{n+1}^\lambda|^2]\leq \left(1-\frac{\lambda\gamma}{2}\right)^{n+1}\mathbb{E}[|\nu_0|^2]+9m^2\bar{C}_2+\frac{9}{2}C_H+\frac{4d}{\beta}.\] So we conclude that \[\sup_{n\in\mathbb{N}_0}\mathbb{E}[|\nu_n^\lambda|^2]\leq\bar{B}_2,\] where \(\bar{B}_2\coloneq \mathbb{E}[|\nu_0|^2]+9m^2\bar{C}_2+9C_H/2+4d/\beta=\mathcal{O}(d/\beta)\). This completes the proof. ◻

Proof of Lemma 7. Recall that \(\gamma_{\min}\) and \(\lambda_{\max,\gamma}\) are defined in 10 and 11 , respectively. For any \(q\in[2,\infty)\cap\mathbb{N}\), \(\gamma\geq\gamma_{\min}\), \(\lambda\leq\lambda_{\max,\gamma}\), and \(n\in\mathbb{N}_0\), recall the definition of \(\Delta_n\) and \(E_n\) defined in 51 . Since \(\sigma=m/(8\gamma^2)\), \(\gamma\geq\gamma_{\min}\geq \max\{1,14m,\sqrt{m}\}\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\), the coefficients of \(|\nu_n^\lambda|^2\) in 58 are non-positive, i.e., \[\lambda\left(\frac{\gamma}{8}-\frac{\gamma}{4}+\frac{\lambda\gamma^2}{4}+\frac{\sigma\gamma}{2}\right)\leq 0,\] and hence \[\begin{align} \label{upper32bound32of32M95123n431125} M_{n+1}\leq (1-\lambda\sigma\gamma)M_n+K_n, \end{align}\tag{67}\] where \[\begin{align} \label{eq46K95n} \begin{aligned} K_n&\coloneq \lambda\sigma\gamma^3|\theta_n^\lambda|^2+\lambda\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right){|H_\gamma(\theta_n^\lambda,X_{n+1})|^2}-\frac{\lambda \gamma}{2}{\langle\theta_n^\lambda,H_\gamma(\theta_n^\lambda,X_{n+1})\rangle}\\ &\quad+\gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle+\frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle. \end{aligned} \end{align}\tag{68}\] Let \(s\coloneq 1-\lambda\sigma\gamma\). By 63 and 64 , we note that \(M_n\geq 0\) for all \(n\in\mathbb{N}_0\). Recall the notation \(\mathbb{E}_n[\cdot]=\mathbb{E}[\cdot\mid\theta_n^\lambda,\nu_n^\lambda]\). Then, by taking conditional expectations on both sides of 67 , using binomial theorem and the fact that \(\Delta_n\) and \(E_n\) are independent of \(\xi_{n+1}\), we obtain that \[\begin{align} \label{eq46even-bounds32of32M95n} \begin{aligned} \mathbb{E}_n[M_{n+1}^q]&\leq(sM_n)^{q}+q(sM_n)^{q-1}\mathbb{E}_n[K_n]+\sum_{k=2}^{q}\binom{q}{k}\mathbb{E}_n\left[(sM_n)^{q-k}K_n^k\right]\\ &= (sM_n)^q+q(sM_n)^{q-1}\mathbb{E}_n[K_n]+\sum_{k=0}^{q-2}\binom{q}{k+2}\mathbb{E}_n\left[(sM_n)^{q-k-2}K_n^{k+2}\right]\\ &\leq (sM_n)^q+q(sM_n)^{q-1}\mathbb{E}_n[K_n]+\binom{q}{2}\sum_{k=0}^{q-2}\binom{q-2}{k}\mathbb{E}_n\left[|sM_n|^{q-k-2}|K_n|^{k+2}\right] \\ &\leq (sM_n)^q+q(sM_n)^{q-1}\mathbb{E}_n[K_n] \\ &\quad+q(q-1)\mathbb{E}_n\left[|K_n|^2\sum_{k=0}^{q-2}\binom{q-2}{k}|sM_n|^{q-k-2}|K_n|^k\right] \\ &=(sM_n)^q+q(sM_n)^{q-1}\mathbb{E}_n[K_n]+q(q-1)\mathbb{E}_n\left[|K_n|^2 (|sM_n|+|K_n|)^{q-2}\right] \\ &\leq (sM_n)^q+q(sM_n)^{q-1}\mathbb{E}_n[K_n]\\ &\quad+q(q-1)2^{q-3}|sM_n|^{q-2}\mathbb{E}_n[|K_n|^2]+q(q-1)2^{q-3}\mathbb{E}_n[|K_n|^{q}]. \end{aligned} \end{align}\tag{69}\] To establish an upper bound for the second term on the right-hand side of 69 , we use similar argument to 58 61 , \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\leq 1/4\), to obtain \[\label{eq46condition32expectation32of32K95n} \mathbb{E}_n[K_n]\leq\lambda\gamma \bar{C}_K,\tag{70}\] where \(\bar{C}_K\coloneq 7C_H/8+u(0)/2+d/\beta=\mathcal{O}(d/\beta)\). Recall that \(\sigma=m/(8\gamma^2)\). By 68 , Lemmas 14 and 15, \(\gamma\geq 1\) and \(\lambda\leq 1/4\), we have \[\begin{align} K_n&\leq \frac{\lambda m\gamma}{8}|\theta_n^\lambda|^2+\lambda\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)\left(2m^2|\theta_n^\lambda|^2+8\gamma\big[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}\big]\right)\notag\\ &\quad+\frac{\lambda\gamma}{2}\left(m|\theta_n^\lambda|^2+\widetilde{b}(X_{n+1})\right)+\gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle+\frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle\notag\\ &\leq\left(\frac{\lambda m\gamma}{8}+2m^2\lambda\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)+\frac{\lambda m\gamma}{2}\right)|\theta_n^\lambda|^2+\gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle\notag\\ &\quad+\frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2+8\lambda\gamma\left(\frac{\lambda}{2}+\frac{3}{4\gamma}\right)\left[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}\right]+\frac{\lambda\gamma}{2}\widetilde{b}(X_{n+1})\notag\\ &\leq \lambda\gamma C_m|\theta_n^\lambda|^2+f_n+g_n,\label{eq46232upper32bound32of32K95n} \end{align}\tag{71}\] where \(C_m\coloneq 5m/8+7m^2/4\) and \[\begin{align} f_n &\coloneq \gamma^2\sqrt{\frac{\lambda}{2\gamma\beta}}\langle\Delta_n,\xi_{n+1}\rangle+\sqrt{\frac{\lambda\gamma}{2\beta}}\langle E_n,\xi_{n+1}\rangle,\\ g_n &\coloneq \frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2+\lambda\gamma\left(\frac{1}{8m}+7\right)\left[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}\right]. \end{align}\] To establish an upper estimate for the third term on the right-hand side of 69 , we use 63 to obtain that \[\label{eq46position4432momentum32and32M95n} |\theta_n^\lambda|^2 \leq \frac{32}{3\gamma^2}M_n,\quad|\nu_n^\lambda|^2\leq \frac{16}{3}M_n.\tag{72}\] By using 51 , 72 , Lemma 15, Young’s inequality, \(\gamma\geq 1\), and \(\lambda\leq 1/(4\gamma)\), we obtain \[\begin{align} &|f_n|^2\leq \frac{\lambda\gamma^3}{\beta}|\Delta_n|^2|\xi_{n+1}|^2+\frac{\lambda\gamma}{\beta}|E_n|^2|\xi_{n+1}|^2\notag\\ &\leq \frac{\lambda\gamma^3}{\beta}|\xi_{n+1}|^2\left|\theta_n^\lambda+\gamma^{-1}\nu_n^\lambda-\lambda\gamma^{-1} H_\gamma(\theta_n^\lambda,X_{n+1})\right|^2\notag\\ &\quad+\frac{\lambda\gamma}{\beta}|\xi_{n+1}|^2\left|(1-\lambda\gamma)\nu_n^\lambda-\lambda H_\gamma(\theta_n^\lambda,X_{n+1})\right|^2\notag\\ &\leq \frac{3\lambda\gamma^3}{\beta}|\xi_{n+1}|^2\left(|\theta_n^\lambda|^2+\gamma^{-2}|\nu_n^\lambda|^2+\lambda^2\gamma^{-2}\big(2m^2|\theta_n^\lambda|^2+8\gamma[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}]\big)\right)\notag\\ &\quad+\frac{2\lambda\gamma}{\beta}|\xi_{n+1}|^2\left((1-\lambda\gamma)^2|\nu_n^\lambda|^2+\lambda^2\big(2m^2|\theta_n^\lambda|^2+8\gamma[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}]\big)\right)\notag\\ &\leq \frac{48\lambda\gamma}{\beta}|\xi_{n+1}|^2 M_n+\frac{64m^2\lambda^3}{\gamma\beta}|\xi_{n+1}|^2M_n+\frac{24\lambda^3\gamma^2}{\beta}|\xi_{n+1}|^2[m^2+K_H^2(1+|X_{n+1}|)^{2\rho}]\notag\\ &\quad+\frac{32\lambda\gamma}{3\beta}|\xi_{n+1}|^2M_n+\frac{128m^2\lambda^3}{3\gamma\beta}|\xi_{n+1}|^2M_n+\frac{16\lambda^3\gamma^2}{\beta}|\xi_{n+1}|^2 [m^2+K_H^2(1+|X_{n+1}|)^{2\rho}] \notag\\ &\leq C_f\lambda\gamma\beta^{-1}M_n|\xi_{n+1}|^2+10\lambda\gamma\beta^{-1}|\xi_{n+1}|^2\left(m^2+K_H^2(1+|X_{n+1}|)^{2\rho}\right),\label{eq46upper32bound32of32124f95n124942} \end{align}\tag{73}\] where \(C_f\coloneq 59(1+m^2)\). Similarly, we have \[\begin{align} |g_n|^2&\leq \frac{2\lambda^2\gamma^2}{\beta^2}|\xi_{n+1}|^4 + 4\lambda^2\gamma^2\left(\frac{1}{8m}+7\right)^2[m^4+K_H^4(1+|X_{n+1}|)^{4\rho}]\notag\\ &\leq \frac{2\lambda\gamma}{\beta^2}|\xi_{n+1}|^4 + 4\lambda\gamma\left(\frac{1}{8m}+7\right)^2[m^4+K_H^4(1+|X_{n+1}|)^{4\rho}].\label{eq46upper32bound32of32124g95n124942} \end{align}\tag{74}\] Substituting 73 74 back into 71 and using 72 yields \[\begin{align} \mathbb{E}_n[|K_n|^2]&\leq \mathbb{E}_n\left[3\lambda^2\gamma^2 C_m^2|\theta_n^\lambda|^4+3|f_n|^2+3|g_n|^2\right]\notag\\ &\leq \frac{32^2 C_m^2}{3\gamma^4}\lambda^2\gamma^2|M_n|^2+3C_f\lambda\gamma\beta^{-1} d M_n+c_f\lambda\gamma, \label{eq462nd32bound32of32K95n} \end{align}\tag{75}\] where \(c_f\coloneq 6d(d+2)/\beta^2+4 C_H d/\beta+12(1/(8m)+7)^2(m^4+K_H^4\mathbb{E}[(1+|X_0|)^{4\rho}])=\mathcal{O}((d/\beta)^2)\). To establish an upper bound for the last term on the right-hand side of 69 , we first note, for any \(q\in[2,\infty)\cap\mathbb{N}\), that 73 implies \[\begin{align} \label{eq46f95n32q} \begin{aligned} |f_n|^q=(|f_n|^2)^{q/2}&\leq 2^{q/2-1}C_f^{q/2}\lambda^{q/2}\gamma^{q/2}\beta^{-q/2}|M_n|^{q/2}|\xi_{n+1}|^{q}\\ &\quad+10^{q/2} 2^{q-2}\lambda^{q/2}\gamma^{q/2}\beta^{-q/2}|\xi_{n+1}|^{q}\big(m^{q}+K_H^{q}(1+|X_{n+1}|)^{\rho q}\big). \end{aligned} \end{align}\tag{76}\] By taking conditional expectations on both sides of 76 , by noticing \(q/2\geq 1\), and by using the fact that \[\mathbb{E}[|\xi_{n+1}|^q]=2^{q/2}\frac{\Gamma\left((d+q)/2\right)}{\Gamma\left(d/2\right)}\leq (d+q)^{q/2},\] we obtain \[\begin{align} \mathbb{E}_n[|f_n|^{q}]&\leq 2^{q/2-1}C_f^{q/2}\lambda\gamma\beta^{-q/2}(d+q)^{q/2}\,|M_n|^{q/2}\notag\\ &\quad+10^{q/2} 2^{q-2}\lambda\gamma\beta^{-q/2}(d+q)^{q/2}\big(m^{q}+K_H^{q}\mathbb{E}[(1+|X_0|)^{\rho q}]\big)\notag\\ &\leq K_f\lambda\gamma(|M_n|^{q/2}+1),\label{eq46estimates32of32f95n} \end{align}\tag{77}\] where \(K_f\coloneq 2^{3q-2}((d+q)/\beta)^{q/2}(C_f^{q/2}+m^{q}+K_H^{q}\mathbb{E}[(1+|X_0|)^{\rho q}])=\mathcal{O}((d/\beta)^{q/2})\). Similarly, 74 implies \[\begin{align} \label{eq46g95n32q} \begin{aligned} |g_n|^q&\leq 2^{q-1}\lambda^{q/2}\gamma^{q/2}\beta^{-q}|\xi_{n+1}|^{2q}\\ &\quad+2^{2q-2}\lambda^{q/2}\gamma^{q/2}\left(\frac{1}{8m}+7\right)^q\big(m^{2q}+K_H^{2q}(1+|X_{n+1}|)^{2\rho q}\big). \end{aligned} \end{align}\tag{78}\] Taking conditional expectations on both sides of 78 yields \[\begin{align} \label{eq46estimates32of32g95n} \mathbb{E}_n[|g_n|^q]\leq K_g\lambda\gamma, \end{align}\tag{79}\] where \(K_g\coloneq 2^{q-1}((d+2q)/\beta)^q+2^{2q-2}(1/(8m)+7)^q(m^{2q}+K_H^{2q}\mathbb{E}[(1+|X_0|)^{2\rho q}])=\mathcal{O}((d/\beta)^q)\). Combining 71 , 77 , and 79 , and using 72 , yields \[\begin{align} \mathbb{E}_n[|K_n|^q]&\leq 3^{q-1}\lambda^q\gamma^q C_m^q|\theta_n^\lambda|^{2q}+3^{q-1}\mathbb{E}_n[|f_n|^q]+3^{q-1}\mathbb{E}_n[|g_n|^q]\notag\\ &\leq 3^{q-1}C_m^q\frac{32^q\lambda^q\gamma^q}{3^q\gamma^{2q}}|M_n|^q+3^{q-1}K_f\lambda\gamma(|M_n|^{q/2}+1)+3^{q-1} K_g \lambda\gamma \notag\\ &\leq\frac{32^q C_m^q}{3\gamma^{2q}}\lambda^q\gamma^q|M_n|^q+3^{q-1}K_f\lambda\gamma|M_n|^{q/2}+C_K\lambda\gamma,\label{eq46even-bounds32of32K95n} \end{align}\tag{80}\] where \(C_K\coloneq 3^{q-1}(K_f+K_g)=\mathcal{O}((d/\beta)^q)\). By noticing \(0<s=1-\lambda\sigma\gamma=1-\lambda m/(8\gamma)<1\), and by substituting 70 , 75 and 80 into 69 , we obtain \[\begin{align} \label{eq46even-bound32of32M95n431} \begin{aligned} \mathbb{E}_n[M_{n+1}^q]&\leq (sM_n)^q+q(sM_n)^{q-1}\lambda\gamma \bar{C}_K\\ &\quad+q(q-1)2^{q-3}|sM_n|^{q-2}\left(\frac{32^2 \lambda^2 C_m^2}{3\gamma^2}|M_n|^2+3C_f\lambda\gamma dM_n/\beta+c_f\lambda\gamma\right)\\ &\quad+q(q-1)2^{q-3}\left(\frac{32^{q}\lambda^{q}C_m^{q}}{3\gamma^{q}}|M_n|^q+3^{q-1}K_f\lambda\gamma|M_n|^{q/2}+C_K\lambda\gamma\right)\\ &\leq \left(1-\frac{\lambda m}{8\gamma}+q(q-1)2^{q-3}\frac{32^2\lambda^2C_m^2}{3\gamma^2}+q(q-1)2^{q-3}\frac{32^{q}\lambda^{q}C_m^{q}}{3\gamma^{q}}\right)|M_n|^q\\ &\quad+\left(q \bar{C}_K+3q(q-1)2^{q-3}C_f \beta^{-1}d\right)\lambda\gamma|M_n|^{q-1}\\ &\quad+q(q-1)2^{q-3}c_f\lambda\gamma|M_n|^{q-2}+q(q-1)2^{q-3}3^{q-1}K_f\lambda\gamma|M_n|^{q/2}\\ &\quad+q(q-1)2^{q-3}C_K\lambda\gamma. \end{aligned} \end{align}\tag{81}\] By using \(\gamma\geq\gamma_{\min}\geq 32 C_m\) and \(\lambda\leq\lambda_{\max,\gamma}\leq \min\{1, 3m/(16q(q-1)2^{q-2}\gamma)\}\), we have that \[\begin{align} \label{eq46Mq32lambda} \begin{aligned} &1-\frac{\lambda m}{8\gamma}+q(q-1)2^{q-3}\frac{32^2\lambda^2C_m^2}{3\gamma^2}+q(q-1)2^{q-3}\frac{32^{q}\lambda^{q}C_m^{q}}{3\gamma^{q}}\\ &\leq 1-\frac{\lambda m}{8\gamma}+\frac{q(q-1)2^{q-2}\lambda^2}{3}\leq 1-\frac{\lambda m}{16\gamma} \end{aligned} \end{align}\tag{82}\] Substituting 82 into 81 yields \[\begin{align} \mathbb{E}_n[M_{n+1}^q]&\leq\left(1-\frac{\lambda m}{16\gamma}\right)|M_n|^q+\left(q\bar{C}_K+3q(q-1)2^{q-3}C_f \beta^{-1}d\right)\lambda\gamma|M_n|^{q-1}\notag\\ &\quad+q(q-1)2^{q-3}c_f\lambda\gamma|M_n|^{q-2}+q(q-1)2^{q-3}3^{q-1}K_f\lambda\gamma|M_n|^{q/2}\notag\\ &\quad+q(q-1)2^{q-3}C_K\lambda\gamma\notag\\ &\leq \left(1-\frac{\lambda m}{64\gamma}\right)|M_n|^q+q(q-1)2^{q-3}C_K\lambda\gamma+\sum_{i=1}^3 T_i(|M_n|),\label{eq46upper32bound32of32Lyapunov-2q} \end{align}\tag{83}\] where for any \(s\in\mathbb{R}^d\), \[\begin{align} T_1(s)&\coloneq \left(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d\right)\lambda\gamma|s|^{q-1}-\frac{\lambda m}{64\gamma}|s|^{q},\\ T_2(s)&\coloneq q(q-1)2^{q-3}c_f\lambda\gamma|s|^{q-2}-\frac{\lambda m}{64\gamma}|s|^q,\\ T_3(s)&\coloneq q(q-1)2^{q-3}3^{q-1}K_f\lambda\gamma|s|^{q/2}-\frac{\lambda m}{64\gamma}|s|^q. \end{align}\] Define \[\begin{align} \mathsf{M}_{q,1}&\coloneq \frac{64\gamma^2\left(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d\right)}{m},\\ \mathsf{M}_{q,2}&\coloneq 8\gamma\left(\frac{q(q-1)2^{q-3}c_f}{m}\right)^{1/2},\\ \mathsf{M}_{q,3}&\coloneq \gamma^{4/q}\left(\frac{64q(q-1)2^{q-3}3^{q-1}K_f}{m}\right)^{2/q}. \end{align}\] Then, \(|s|\geq \mathsf{M}_{q,i}\) implies \(T_i(s)\leq 0\), for all \(i=1,2,3\). Thus, we have \[T_i(s)=T_i(s)\mathbb{1}_{\{s\leq \mathsf{M}_{q,i}\}}+T_i(s)\mathbb{1}_{\{s\geq \mathsf{M}_{q,i}\}}\leq T_i(s)\mathbb{1}_{\{s\leq \mathsf{M}_{q,i}\}}.\] By using the above inequality, we obtain, for all \(s\in\mathbb{R}\), that \[\begin{align} T_1(s)&\leq \left(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d\right)\lambda\gamma|s|^{q-1}\mathbb{1}_{\{s\leq \mathsf{M}_{q,1}\}}\notag\\ &\leq \left(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d\right)\lambda\gamma \mathsf{M}_{q,1}^{q-1}\notag\\ &=\left(\frac{64}{m}\right)^{q-1}\left(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d\right)^{q}\lambda\gamma^{2q-1}\notag\\ &=N_1\lambda\gamma^{2q-1},\label{eq46T1} \end{align}\tag{84}\] where \(N_1\coloneq(64/m)^{q-1}(q \bar{C}_K+3q(q-1)2^{q-3}C_f\beta^{-1}d)^{q}=\mathcal{O}((d/\beta)^{q})\). Moreover, by \(\gamma\geq\gamma_{\min}\geq 1\), it holds that \[\begin{align} T_2(s)&\leq q(q-1)2^{q-3}c_f\lambda\gamma|s|^{q-2}\mathbb{1}_{\{s\leq \mathsf{M}_{q,2}\}}\notag\\ &\leq q(q-1)2^{q-3}c_f\lambda\gamma \mathsf{M}_{q,2}^{q-2}\notag\\ &= \left(\frac{8}{\sqrt{m}}\right)^{q-2}\left(q(q-1)2^{q-3}c_f\right)^{q/2}\lambda\gamma^{q-1}\notag\\ &\leq N_2\lambda\gamma^{2q-1},\label{eq46T2} \end{align}\tag{85}\] where \(N_2\coloneq (8/\sqrt{m})^{q-2}(q(q-1)2^{q-3}c_f)^{q/2}=\mathcal{O}((d/\beta)^q)\). Similarly, by \(\gamma\geq\gamma_{\min}\geq 1\) and \(q\geq 2\), we have that \[\begin{align} T_3(s)&\leq q(q-1)2^{q-3}3^{q-1}K_f\lambda\gamma|s|^{q/2}\mathbb{1}_{\{s\leq \mathsf{M}_{q,3}\}}\notag\\ &\leq q(q-1)2^{q-3}3^{q-1}K_f\lambda\gamma \mathsf{M}_{q,3}^{q/2}\notag\\ &=\frac{64}{m}\left(q(q-1)2^{q-3}3^{q-1}K_f\right)^2\lambda\gamma^3\notag\\ &\leq N_3\lambda\gamma^{2q-1},\label{eq46T3} \end{align}\tag{86}\] where \(N_3\coloneq 64(q(q-1)2^{q-3}3^{q-1}K_f)^2/m=\mathcal{O}((d/\beta)^q)\). Substituting 84 86 into 83 yields \[\begin{align} \label{eq46upper32bound32Mn1} \mathbb{E}_n[|M_{n+1}|^q]\leq\left(1-\frac{\lambda m}{64\gamma}\right)|M_n|^q+N_0\lambda\gamma^{2q-1}, \end{align}\tag{87}\] where \(N_0\coloneq N_1+N_2+N_3+q(q-1)2^{q-3}C_K=\mathcal{O}((d/\beta)^q)\). Iterating 87 yields \[\sup_{n\in\mathbb{N}_0}\mathbb{E}[|M_{n+1}|^q]\leq \mathbb{E}[|M_0|^q]+\frac{64N_0}{m}\gamma^{2q}.\] By 72 , 87 , and \(\gamma\geq 1\), we obtain \[\begin{align} \sup_{n\in\mathbb{N}_0}\mathbb{E}[|\theta_n^\lambda|^{2q}]&\leq \frac{32^q}{3^q\gamma^{2q}}\left(\mathbb{E}[|M_0|^q]+\frac{64N_0}{m}\gamma^{2q}\right)\\ &\leq \frac{8^q 4^q}{3^q\gamma^{2q}}\mathbb{E}[|M_0|^q]+\frac{8^{q+2}4^q N_0}{3^qm}\\ &\leq \frac{8^q 2^q}{\gamma^{2q}}\left(2^{q-2}\gamma^{2q}\mathbb{E}[|\theta_0|^{2q}]+2^{q-1}\mathbb{E}[|\nu_0|^{2q}]\right)+\frac{8^{q+2}2^qN_0}{m}\\ &\leq 2^{5q-2}\mathbb{E}[|\theta_0|^{2q}]+2^{5q-1}\mathbb{E}[|\nu_0|^{2q}]+2^{4q+6}N_0/m\eqcolon\bar{C}_{2q}, \end{align}\] where \(\bar{C}_{2q}=\mathcal{O}((d/\beta)^q)\). ◻

Proof of Lemma 8. For any \(t\geq 0\), by 34 , we have that \[V_t^\lambda=V_{\lfloor t \rfloor}^\lambda-\lambda\gamma\int_{\lfloor t \rfloor}^{t}V_{\lfloor s \rfloor}^\lambda\,\mathrm{d}s-\lambda\int_{\lfloor t \rfloor}^t H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})\,\mathrm{d}s+\sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda).\] By using the inequality that \(|a+b+c|^2\leq 3(|a|^2+|b|^2+|c|^2)\) for \(a,b,c>0\) and Lemma 15, we have that \[\begin{align} \mathbb{E}[|V_t^\lambda-V_{\lfloor t \rfloor}^\lambda|^2]&\leq 3\lambda^2\gamma^2\mathbb{E}\left[\bigg|\int_{\lfloor t \rfloor}^t V_{\lfloor s \rfloor}^\lambda\,\mathrm{d}s\bigg|^2\right]+3\lambda^2\mathbb{E}\left[\bigg|\int_{\lfloor t \rfloor}^t H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})\,\mathrm{d}s\bigg|^2\right]\notag\\ &\quad+\frac{6\lambda\gamma}{\beta}\mathbb{E}[|W_t^\lambda-W_{\lfloor t \rfloor}^\lambda|^2]\notag\\ &\leq 3\lambda^2\gamma^2\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda|^2]+3\lambda^2\left(2m^2\mathbb{E}[|Z_{\lfloor s \rfloor}^\lambda|^2]+C_H\gamma\right)+\frac{6\lambda\gamma d}{\beta}.\label{eq46lemma46mse32of32interpolation} \end{align}\tag{88}\] Since the law of the continuous-time interpolation process 34 agrees with tSGHMC 8 9 at grid points, applying Lemma 6 yields \[\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda|^2]\leq \bar{B}_2,\quad \mathbb{E}[|Z_{\lfloor s \rfloor}^\lambda|^2]\leq \bar{C}_2,\] where \(\bar{B}_2=\mathcal{O}(d/\beta)\) and \(\bar{C}_2=\mathcal{O}(d/\beta)\) are given explicitly in Table [tab46constants]. Hence, we obtain, by noticing \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\leq 1/4\), that \[\begin{align} \mathbb{E}[|V_t^\lambda-V_{\lfloor t \rfloor}^\lambda|^2]\leq \lambda\gamma C_{1,v}, \end{align}\] where \(C_{1,v}\coloneq 3\bar{B}_2/4+3m^2\bar{C}_2/2+3C_H/4+6d/\beta=\mathcal{O}(d/\beta)\). Moreover, by similar argument, we obtain that \[\mathbb{E}[|Z_t^\lambda-Z_{\lfloor t \rfloor}^\lambda|^2]\leq \lambda\bar{B}_2,\] which completes the proof. ◻

Proof of Lemma 9. For all \((\theta,\nu)\in\mathbb{R}^{2d}\), consider the following Lyapunov function \[\label{eq46bar32V} \bar{\mathcal{V}}(\theta,\nu)\coloneq\beta u_{MY,\epsilon}(\theta)+\frac{\beta}{4}\gamma^2\left(|\theta+\gamma^{-1}\nu|^2+|\gamma^{-1}\nu|^2-\sigma|\theta|^2\right),\tag{89}\] where \(\sigma=m/(8\gamma^2)\) as before. Moreover, by Lemma 2 and Remark 9, we have \[\label{eq46Lip32MY32h} |h_{MY,\epsilon}(\theta)|\leq K|\theta|+|h_{MY,\epsilon}(0)|\leq K|\theta|+|h(0)|.\tag{90}\] By the fundamental theorem of calculus, Cauchy-Schwarz inequality, and 90 , we obtain \[\begin{align} u_{MY,\epsilon}(\theta)-u_{MY,\epsilon}(0)&=\int_0^1 \langle\theta,h_{MY,\epsilon}(t\theta)\rangle\,\mathrm{d}t\leq \int_0^1|h_{MY,\epsilon}(t\theta)|\,|\theta|\,\mathrm{d}t\notag\\ &\leq \int_0^1\left(Kt|\theta|+|h(0)|\right)|\theta|\,\mathrm{d}t=\frac{K}{2}|\theta|^2+|h(0)|\,|\theta|.\label{eq46upper32bound32uMY} \end{align}\tag{91}\] Since \(\gamma\geq\gamma_{\min}\geq \sqrt{2(K+2|h(0)|)/3}\), we have \[\frac{m}{4}\geq 2\sigma\left(\frac{K}{2}+|h(0)|+\frac{\gamma^2}{4}\right).\] Therefore, by Remark 9, 91 , the inequality \(|\theta|\leq 1+|\theta|^2\) for all \(\theta\in\mathbb{R}^d\), and \(\gamma\geq\gamma_{\min}\geq \sqrt{m}\), we obtain that \[\begin{align} \langle\theta,h_{MY,\epsilon}(\theta)\rangle&\geq \frac{m}{4}|\theta|^2-u(0)\notag\\ &\geq 2\sigma\left(\frac{K}{2}+|h(0)|+\frac{\gamma^2}{4}\right)|\theta|^2-u(0)\notag\\ &\geq 2\sigma\left(\frac{K}{2}|\theta|^2+|h(0)|\,|\theta|-|h(0)|\,|\theta|+|h(0)|\,|\theta|^2+\frac{\gamma^2}{4}|\theta|^2\right)-u(0)\notag\\ &\geq 2\sigma\left(u_{MY,\epsilon}(\theta)-u_{MY,\epsilon}(0)-|h(0)|(1+|\theta|^2)+|h(0)|\,|\theta|^2+\frac{\gamma^2}{4}|\theta|^2\right)-u(0)\notag\\ &\geq 2\sigma\left(u_{MY,\epsilon}(\theta)+\frac{\gamma^2}{4}|\theta|^2\right)-2\sigma \left(u_{MY,\epsilon}(0)+|h(0)|\right)-u(0)\notag\\ &\geq 2\sigma\left(u_{MY,\epsilon}(\theta)+\frac{\gamma^2}{4}|\theta|^2\right)-\frac{1}{4}\left(u(0)+|h(0)|\right)-u(0)\notag\\ &\geq 2\sigma\left(u_{MY,\epsilon}(\theta)+\frac{\gamma^2}{4}|\theta|^2\right)-\frac{2A_c}{\beta},\label{eq46u32h32dissipativity} \end{align}\tag{92}\] where \(A_c\coloneq (5u(0)+|h(0)|)\beta/8=\mathcal{O}(\beta)\).

Next, we follow the argument in [30] to establish an upper bound for \(\mathbb{E}[|r_t|^2]\). By Remark 9 and 92 , we note that \(u_{MY,\epsilon}\) is gradient Lipschitz and satisfies a dissipativity condition. Moreover, since \(\gamma\geq\gamma_{\min}\geq\sqrt{m}\), \(\sigma=m/(8\gamma^2)\leq 1/4\). These properties fulfill the assumptions of [30], see also Remarks 2.2-2.4 therein. Therefore, we can apply [30] with \(\lambda\curvearrowleft\sigma\), \(\mathcal{V}\curvearrowleft\bar{\mathcal{V}}\), \(\theta_t\curvearrowleft r_t\), \(V_0\curvearrowleft \nu_0\) to obtain \[\frac{\beta}{8}(1-2\sigma)\gamma^2\mathbb{E}[|r_t|^2]\leq\mathbb{E}[\bar{\mathcal{V}}(\theta_0,\nu_0)]+\frac{d+A_c}{\sigma}.\] Using \(\gamma\geq\gamma_{\min}\geq \max\{1,\sqrt{m}\}\), 65 , 89 and Remark 9, we have \[\begin{align} \mathbb{E}[|r_t|^2]&\leq \frac{\mathbb{E}[\bar{\mathcal{V}}(\theta_0,\nu_0)]+(d+A_c)/\sigma}{\beta(1-2\sigma)\gamma^2/8}\notag\\ &\leq \frac{\beta\mathbb{E}[u_{MY,\epsilon}(\theta_0)]+\beta\mathbb{E}[M_0]+(d+A_c)/\sigma}{\beta(1-2\sigma)\gamma^2/8}\notag\\ &\leq \frac{8\mathbb{E}[u(\theta_0)]}{(1-2\sigma)\gamma^2}+\frac{8\mathbb{E}[M_0]}{(1-2\sigma)\gamma^2}+\frac{8(d+A_c)}{\beta\sigma(1-2\sigma)\gamma^2}.\label{eq462nd32moment32of32Zt} \end{align}\tag{93}\] Moreover, by 63 and \(\gamma\geq 1\), we have \[\label{eq46Zt1} \frac{8\mathbb{E}[u(\theta_0)]}{(1-2\sigma)\gamma^2}\leq 11\mathbb{E}[u(\theta_0)].\tag{94}\] Besides, by 65 , we obtain that \[M_0\leq\frac{\gamma^2}{4}\left(2|\theta_0|^2+2\gamma^{-2}|\nu_0|^2\right)+\frac{1}{4}|\nu_0|^2\leq \frac{\gamma^2}{2}|\theta_0|^2+\frac{3}{4}|\nu_0|^2,\] which, by 63 and \(\gamma\geq 1\), implies \[\label{eq46Zt2} \frac{8\mathbb{E}[M_0]}{(1-2\sigma)\gamma^2}\leq 6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2].\tag{95}\] Finally, by \(\sigma=m/(8\gamma^2)\) and 63 , we have \[\label{eq46Zt3} \frac{8(d+A_c)}{\beta\sigma(1-2\sigma)\gamma^2}=\frac{64(d+A_c)}{\beta m(1-2\sigma)}\leq \frac{86(d+A_c)}{\beta m}.\tag{96}\] Substituting 94 96 into 93 yields \[\label{eq462nd32Zt} \sup_{t\geq 0}\mathbb{E}[|r_t|^2]\leq C_r,\tag{97}\] where \(C_r\coloneq 11\mathbb{E}[u(\theta_0)]+6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2]+86(d+A_c)/(\beta m)=\mathcal{O}(d/\beta)\).

To establish an upper bound for \(\mathbb{E}[|R_t|^2]\), we first set \(a\land b\coloneq\min\{a,b\}\) for \(a,b\in\mathbb{R}\) and define the stopping time \[\tau_m\coloneq\inf\{t\geq 0: |R_t|\geq m\}.\] Applying Itô’s formula to the stopped process \(e^{\gamma (t\land\tau_m)}|R_{t\land\tau_m}|^2\) yields \[\begin{align} \label{eq46stopped32process} \begin{aligned} e^{\gamma(t\land\tau_m)}|R_{t\land\tau_m}|^2-|R_0|^2&=\int_0^{t\land\tau_m}e^{\gamma s}\left(-\gamma |R_s|^2-2\langle R_s,h_{MY,\epsilon}(r_s)\rangle+2\gamma\beta^{-1}d\right)\,\mathrm{d}s\\ &\quad+2\sqrt{2\gamma\beta^{-1}}\int_0^{t\land\tau_m}e^{\gamma s}\langle R_s,\mathrm dW_s\rangle. \end{aligned} \end{align}\tag{98}\] By taking expectations and by using Young’s inequality together with 90 and 97 , we obtain \[\begin{align} &\mathbb{E}[e^{\gamma(t\land\tau_m)}|R_{t\land\tau_m}|^2]-\mathbb{E}[|R_0|^2]\notag\\ &\leq\mathbb{E}\left[\int_0^{t\land\tau_m}e^{\gamma s}\left(-\gamma|R_s|^2+\gamma|R_s|^2+\gamma^{-1}|h_{MY,\epsilon}(r_s)|^2+\frac{2\gamma d}{\beta}\right)\,\mathrm{d}s\right]\notag\\ &\leq \mathbb{E}\left[\int_0^{t\land\tau_m}e^{\gamma s}\left(\frac{2|h(0)|^2+2K^2|r_s|^2}{\gamma}+\frac{2\gamma d}{\beta}\right)\,\mathrm{d}s\right]\notag\\ &\leq \int_0^t e^{\gamma s}\left(\frac{2|h(0)|^2+2K^2\mathbb{E}[|r_s|^2]}{\gamma}+\frac{2\gamma d}{\beta}\right)\,\mathrm{d}s\notag\\ &\leq \int_0^t e^{\gamma s}\left(\frac{2|h(0)|^2+2K^2 C_r}{\gamma}+\frac{2\gamma d}{\beta}\right)\,\mathrm{d}s\notag\\ &=\left(\frac{2|h(0)|^2+2K^2 C_r}{\gamma^2}+\frac{2 d}{\beta}\right)(e^{\gamma t}-1),\label{eq46stop-inequality} \end{align}\tag{99}\] where the third inequality is satisfied since \(t\land\tau_m\leq t\). By [67] and Remark 9, \(R_t\) is a strong solution of 32 with continuous sample paths. Therefore, we have \(\tau_m\uparrow\infty\) almost surely. As a result, \(t\land\tau_m\uparrow t\) and \(e^{\gamma(t\land\tau_m)}R_{t\land \tau_m}\to e^{\gamma t}R_t\) almost surely as \(m\to\infty\). By Fatou’s lemma, we have \[\begin{align} \label{eq46Fatou} \mathbb{E}[e^{\gamma t}|R_t|^2]\leq \liminf_{n\to\infty}\mathbb{E}[e^{\gamma(t\land\tau_m)}|R_{t\land\tau_m}|^2]. \end{align}\tag{100}\] Combining 99 and 100 , and letting \(m\to\infty\), we obtain \[\begin{align} e^{\gamma t}\mathbb{E}[|R_t|^2]\leq\mathbb{E}[|R_0|^2]+\left(\frac{2|h(0)|^2+2K^2 C_r}{\gamma^2}+\frac{2 d}{\beta}\right)(e^{\gamma t}-1), \end{align}\] which implies that \[\begin{align} \mathbb{E}[|R_t|^2]\leq e^{-\gamma t}\mathbb{E}[|R_0|^2]+\left(\frac{2|h(0)|^2+2K^2 C_r}{\gamma^2}+\frac{2 d}{\beta}\right)(1-e^{-\gamma t}). \end{align}\] Since \(\gamma\geq\gamma_{\min}\geq \max\{1,K\}\) and \(R_0=\nu_0\), we have that \[\begin{align} \sup_{t\geq 0}\mathbb{E}[|R_t|^2]\leq C_R, \end{align}\] where \(C_R\coloneq \mathbb{E}[|\nu_0|^2]+2|h(0)|^2+2C_r+2d/\beta=\mathcal{O}(d/\beta)\). ◻

Proof of Lemma 10. Recall the process \((\zeta_t^{\lambda,n},V_t^{\lambda,n})_{t\geq nT}\) defined in ?? . According to Lemma 9, we have, for all \(n\in\mathbb{N}_0\), that \[\begin{align} \sup_{t\geq nT}\mathbb{E}[|\zeta_t^{\lambda,n}|^2]\leq \frac{\mathbb{E}[\bar{\mathcal{V}}(Z_{nT}^\lambda,V_{nT}^\lambda)]+(d+A_c)/\sigma}{\beta(1-2\sigma)\gamma^2/8}. \end{align}\] By Remark 9 and 92 , we note that \(u_{MY,\epsilon}\) is gradient Lipschitz and satisfies a dissipativity condition. Moreover, since \(\gamma\geq\gamma_{\min}\geq\sqrt{m}\), \(\sigma=m/(8\gamma^2)\leq 1/4\). These properties fulfill the assumptions of [30], see also Remarks 2.2-2.4 therein. Thus, we apply [30] with \(\bar\zeta_t^{\eta,n}\curvearrowleft\zeta_t^{\lambda,n}\), \(\bar{Z}_t^{\eta,n}\curvearrowleft V_t^{\lambda,n}\) \(\bar\theta_{nT}^\eta\curvearrowleft Z_{nT}^\lambda\), \(\bar V_{nT}^\eta\curvearrowleft V_{nT}^\lambda\), \(\eta\curvearrowleft\lambda\), \(\lambda\curvearrowleft\sigma\) and \(\mathcal{V}\curvearrowleft\bar{\mathcal{V}}\) to obtain \[\begin{align} \sup_{t\geq nT}\mathbb{E}[|\zeta_t^{\lambda,n}|^2]&\leq \frac{\mathbb{E}[\bar{\mathcal{V}}(\theta_0,\nu_0)]+5(d+A_c)/\beta}{\beta(1-2\sigma)\gamma^2/8}\\ &\leq 11\mathbb{E}[u(\theta_0)]+6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2]+\frac{430(d+A_c)}{\beta m}, \end{align}\] where the last inequality follows using the same argument from 93 to 97 in the proof of Lemma 9. Thus we have \[\label{eq46bound32zeta} \sup_{n\in\mathbb{N}_0}\sup_{t\geq nT}\mathbb{E}[|\zeta_t^{\lambda,n}|^2]\leq C_\zeta^{\#},\tag{101}\] where \(C_\zeta^\#\coloneq 11\mathbb{E}[u(\theta_0)]+6\mathbb{E}[|\theta_0|^2]+8\mathbb{E}[|\nu_0|^2]+430(d+A_c)/(\beta m)=\mathcal{O}(d/\beta)\).

To establish an upper bound for \(\mathbb{E}[|V_t^{\lambda,n}|^2]\), fix \(n\in\mathbb{N}_0\) and \(t\geq nT\). For each \(m\in\mathbb{N}\), define the stopping time \[\rho_m\coloneq\inf\{t\geq nT:|V_t^{\lambda,n}|\geq m\}.\] Applying Itô’s formula to the stopped process \(e^{\lambda\gamma(t\land\rho_m-nT)}|V_{t\land\rho_m}^{\lambda,n}|^2\), we obtain \[\begin{align} \label{eq46ito32V} \begin{aligned} &e^{\lambda\gamma((t\land\rho_m)-nT)}|V_{t\land\rho_m}^{\lambda,n}|^2-|V_{nT}^{\lambda,n}|^2\\ &=\int_{nT}^{t\land\rho_m}e^{\lambda\gamma(s-nT)} \left(-\lambda\gamma |V_s^{\lambda,n}|^2-2\lambda\langle V_s^{\lambda,n},h_{MY,\epsilon}(\zeta_s^{\lambda,n})\rangle+\frac{2\lambda\gamma d}{\beta}\right)\,\mathrm{d}s\\ &\quad+2\sqrt{2\lambda\gamma\beta^{-1}}\int_{nT}^{t\land\rho_m}e^{\lambda\gamma(s-nT)}\langle V_s^{\lambda,n},\mathrm{d}W_s^\lambda\rangle. \end{aligned} \end{align}\tag{102}\] Then, it holds that \[\label{eq46martingale2} 2\sqrt{2\lambda\gamma\beta^{-1}}\mathbb{E}\left[\int_{nT}^{t\land\rho_m}e^{\lambda\gamma(t-nT)}\langle V_s^{\lambda,n},\mathrm{d}W_s^\lambda\rangle\right]=0.\tag{103}\] Taking expectation in 102 , using 90 , 101 and 103 , and applying Young’s inequality, we obtain \[\begin{align} &\mathbb{E}[e^{\lambda\gamma((t\land\rho_m)-nT)}|V_{t\land\rho_m}^{\lambda,n}|^2]-\mathbb{E}[|V_{nT}^{\lambda,n}|^2]\notag\\ &\leq \mathbb{E}\left[\int_{nT}^{t\land\rho_m}e^{\lambda\gamma(s-nT)} \left(-\lambda\gamma |V_s^{\lambda,n}|^2+\lambda\gamma |V_s^{\lambda,n}|^2+\lambda\gamma^{-1}|h_{MY,\epsilon}(\zeta_s^{\lambda,n})|^2+\frac{2\lambda\gamma d}{\beta}\right)\,\mathrm{d}s\right]\notag\\ &\leq \mathbb{E}\left[\int_{nT}^{t\land\rho_m}e^{\lambda\gamma(s-nT)} \left(\frac{2\lambda(|h(0)|^2+K^2|\zeta_s^{\lambda,n}|^2)}{\gamma}+\frac{2\lambda\gamma d}{\beta}\right)\,\mathrm{d}s\right]\notag\\ &\leq \int_{nT}^{t}e^{\lambda\gamma(s-nT)} \left(\frac{2\lambda(|h(0)|^2+K^2\mathbb{E}[|\zeta_s^{\lambda,n}|^2])}{\gamma}+\frac{2\lambda\gamma d}{\beta}\right)\,\mathrm{d}s\notag\\ &=\left(\frac{2|h(0)|^2+2K^2C_\zeta^\#}{\gamma^2}+\frac{2d}{\beta}\right)(e^{\lambda\gamma(t-nT)}-1).\label{eq46stop-inequality1} \end{align}\tag{104}\] where the third inequality is satisfied since \(t\land\rho_m\leq t\). By [67] and Remark 9, \(V_t^{\lambda,n}\) is a strong solution of ?? with continuous sample paths. Thus, we have \(\rho_m\uparrow\infty\) almost surely. Hence, \(t\land\rho_m\uparrow t\) and \(e^{\lambda\gamma((t\land\rho_m)-nT)}|V_{t\land\rho_m}^{\lambda,n}|^2 \to e^{\lambda\gamma(t-nT)} |V_t^{\lambda,n}|^2\) almost surely as \(m\to\infty\). Therefore, by Fatou’s lemma, \[\mathbb{E}\left[e^{\lambda\gamma(t-nT)}|V_t^{\lambda,n}|^2\right]\leq\liminf_{m\to\infty}\mathbb{E}\left[e^{\lambda\gamma((t\land\rho_m)-nT)}|V_{t\land\rho_m}^{\lambda,n}|^2\right].\] Letting \(m\to\infty\) in 104 , we obtain \[e^{\lambda\gamma(t-nT)}\mathbb{E}[|V_t^{\lambda,n}|^2]\le\mathbb{E}[|V_{nT}^{\lambda,n}|^2]+\left(\frac{2|h(0)|^2+2K^2C_\zeta^\#}{\gamma^2}+\frac{2d}{\beta}\right)(e^{\lambda\gamma(t-nT)}-1).\] By 34 and ?? , we have \(V_{nT}^{\lambda,n}=V_{nT}^\lambda=\nu_{nT}^\lambda\). Combining this and the above inequality yields \[\mathbb{E}[|V_t^{\lambda,n}|^2]\leq e^{-\lambda\gamma(t-nT)}\mathbb{E}[|\nu_{nT}^\lambda|^2]+\left(\frac{2|h(0)|^2+2K^2C_\zeta^\#}{\gamma^2}+\frac{2d}{\beta}\right)(1-e^{-\lambda\gamma(t-nT)}).\] Using Lemma 6 and \(\gamma\ge\gamma_{\min}\geq\max\{1,K\}\), we conclude that \[\mathbb{E}[|V_t^{\lambda,n}|^2]\leq \bar B_2+2|h(0)|^2+2C_\zeta^\#+\frac{2d}{\beta}.\] Thus we have \[\sup_{n\in\mathbb{N}_0}\sup_{t\geq nT}\mathbb{E}[|V_t^{\lambda,n}|^2]\leq C_V^\#,\] where \(C_V^\#\coloneq \bar{B}_2+2|h(0)|^2+2C_\zeta^\#+2d/\beta=\mathcal{O}(d/\beta)\). ◻

Proof. Recall the expression of \(H_\gamma\) defined in 9 . For all \(\theta\in\mathbb{R}^d\) and \(x\in\mathbb{R}^k\), we have \[\begin{align} H_\gamma(\theta,x)-H(\theta,x)&=\frac{1-\sqrt{1+\gamma^{-1}|\theta|^{4r}}}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}}\left(H(\theta,x)-m\theta\right)\\ &=-\frac{\gamma^{-1}|\theta|^{4r}}{\sqrt{1+\gamma^{-1}|\theta|^{4r}}(1+\sqrt{1+\gamma^{-1}|\theta|^{4r}})}(H(\theta,x)-m\theta). \end{align}\] By Remark 3, and the inequalities \(|\theta|^{8r}\leq 1+|\theta|^{12r}\) and \(|\theta|^{8r+2}\leq 1+|\theta|^{12r}\) for \(r\geq 1/2\), we have \[\begin{align} |H_\gamma(\theta,x)-H(\theta,x)|^2&\leq 2\gamma^{-2}|\theta|^{8r}\left(|H(\theta,x)|^2+m^2|\theta|^2\right)\\ &\leq 2\gamma^{-2}|\theta|^{8r}\left(2K_H^2(1+|x|)^{2\rho}(1+|\theta|^{4r})+m^2|\theta|^2\right)\\ &\leq 2\gamma^{-2}\left(2K_H^2(1+|x|)^{2\rho}(|\theta|^{8r}+|\theta|^{12r})+m^2|\theta|^{8r+2}\right)\\ &\leq 2\gamma^{-2}\left(4K_H^2(1+|x|)^{2\rho}(1+|\theta|^{12r})+m^2(1+|\theta|^{12r})\right)\\ &\leq 2\gamma^{-2}\big[4K_H^2(1+|x|)^{2\rho}+m^2\big](1+|\theta|^{12r}). \end{align}\] This completes the proof. ◻

Proof of Proposition 4. Throughout the proof, we use symbols, e.g., \(\mathsf{V},\mathsf{Z}\), to denote the expressions on the right-hand side of the corresponding (in)equalities. By the definition of Wasserstein-2 distance, we have \[\label{eq46auxiliary-decomposition} W_2(\mathcal{L}(Z_t^\lambda,V_t^\lambda),\mathcal{L}(\zeta_t^{\lambda,n},V_t^{\lambda,n}))\leq (\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2])^{1/2}+(\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2])^{1/2}.\tag{105}\] We first establish an upper bound for the first term on the right-hand side of 105 . By Itô’s formula together with the definitions in 34 and ?? , we obtain, for all \(t\in(nT,(n+1)T]\), that \[\begin{align} \mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]&=\mathbb{E}[|V_{nT}^\lambda-V_{nT}^{\lambda,n}|^2]-2\lambda\gamma\int_{nT}^t\mathbb{E}[\langle V_s^\lambda-V_s^{\lambda,n},V_{\lfloor s \rfloor}^\lambda-V_s^{\lambda,n}\rangle]\,\mathrm{d}s\\ &\quad-2\lambda\int_{nT}^t\mathbb{E}[\langle V_s^\lambda-V_s^{\lambda,n},H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})-h_{MY,\epsilon}(\zeta_s^{\lambda,n})\rangle]\,\mathrm{d}s, \end{align}\] which implies that \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]&=-2\lambda\gamma\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},V_{\lfloor t \rfloor}^{\lambda}-V_t^{\lambda,n}\rangle]_1}\tag{106}\\ &\quad-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h_{MY,\epsilon}(\zeta_t^{\lambda,n})\rangle]._2}\tag{107} \end{align}\] By using 34 , we can further decompose \(\mathsf{V}_1\) as \[\begin{align} \mathsf{V}_1&=2\lambda\gamma\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},V_t^\lambda-V_{\lfloor t \rfloor}^\lambda\rangle]-2\lambda\gamma\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2] \notag\\ &=-2\lambda^2\gamma\mathbb{E}\left[\left\langle V_t^\lambda-V_t^{\lambda,n},\int_{\lfloor t \rfloor}^t\big(\gamma V_{\lfloor s \rfloor}^\lambda+H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})\big)\,\mathrm{d}s\right\rangle\right] _{1,1}}\\ &\quad+2\lambda\gamma\mathbb{E}\left[\left\langle V_t^\lambda-V_t^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{1,2}}\\ &\quad-2\lambda\gamma\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2].\notag \end{align}\] By Lemmas 6 and 15, together with Young’s inequality, we have \[\begin{align} \label{eq46V951441} \mathsf{V}_{1,1}&\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+18\lambda^3\gamma \left(\gamma^2\mathbb{E}[|V_{\lfloor t \rfloor}^\lambda|^2]+\mathbb{E}[|H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]\right)\notag\\ &\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+18\lambda^3\gamma^3\bar{B}_2+36m^2\lambda^3\gamma \bar{C}_2+18\lambda^3\gamma^2C_H. \end{align}\tag{108}\] Furthermore, by using 34 and ?? , we have \[\begin{align} \mathsf{V}_{1,2}&=2\lambda\gamma\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda)\rangle]_{1,2,1}}\\ &\quad-2\lambda\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t\lambda\gamma(V_{\lfloor s \rfloor}^\lambda-V_s^{\lambda,n})\,\mathrm{d}s,\sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{1,2,2}}\\ &\quad-2\lambda\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t\lambda(H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})-h_{MY,\epsilon}(\zeta_s^{\lambda,n}))\,\mathrm{d}s, \sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{1,2,3}}. \end{align}\] Define the filtration \(\mathcal{H}_t^\lambda\coloneq \mathcal{F}_t^\lambda\vee\mathcal{G}_{\lfloor t \rfloor}\vee\sigma(\theta_0,\nu_0)\) for all \(t\geq 0\). We notice that \[\begin{align} \mathsf{V}_{1,2,1}&=2\lambda\gamma\mathbb{E}[\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_t^\lambda-W_{\lfloor t \rfloor}^\lambda)\rangle\mid\mathcal{H}_{\lfloor t \rfloor}^\lambda]]\notag\\ &=2\lambda\gamma\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}\mathbb{E}[W_t^\lambda-W_{\lfloor t \rfloor}^\lambda\mid\mathcal{H}_{\lfloor t \rfloor}^\lambda]\rangle]=0.\label{eq46V951442441} \end{align}\tag{109}\] Next, by applying Young’s inequality, Cauchy-Schwarz inequality, and Lemma 8, we derive \[\begin{align} \mathsf{V}_{1,2,2}&\leq\frac{2\lambda^2\gamma^2}{9}\int_{\lfloor t \rfloor}^t\mathbb{E}[|V_s^\lambda-V_s^{\lambda,n}|^2]\,\mathrm{d}s+\frac{2\lambda^2\gamma^2}{9}\int_{\lfloor t \rfloor}^t\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda-V_s^\lambda|^2]\,\mathrm{d}s\notag\\ &\quad+18\lambda^3\gamma^3\beta^{-1}\mathbb{E}[|W_t^\lambda-W_{\lfloor t \rfloor}^\lambda|^2]\notag\\ &\leq \frac{2\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\frac{2}{9}\lambda^3\gamma^3 C_{1,v}+18\lambda^3\gamma^3 d/\beta.\label{eq46V951442442} \end{align}\tag{110}\] By Lemmas 6 and 15, together with 90 , the condition \(\gamma\geq\gamma_{\min}\geq \max\{1,K\}\), and Young’s inequality, we obtain \[\begin{align} \mathsf{V}_{1,2,3}&\leq 2\lambda^2\gamma^{-1}\left(\int_{\lfloor t \rfloor}^t\mathbb{E}[|H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})|^2]\,\mathrm{d}s+\int_{\lfloor t \rfloor}^t\mathbb{E}[|h_{MY,\epsilon}(\zeta_s^{\lambda,n})|^2]\,\mathrm{d}s\right)+2\lambda^3\gamma^4 d/\beta\notag\\ &\leq 2\lambda^2\gamma^{-1}(2m^2\bar{C}_2+C_H\gamma)+4\lambda^2\gamma^{-1}(K^2 C_\zeta^\#+|h(0)|^2)+2\lambda^3\gamma^4 d/\beta\notag\\ &\leq 4m^2\lambda^2\bar{C}_2+2\lambda^2 C_H+4\lambda^2\gamma C_\zeta^\#+4\lambda^2|h(0)|^2+2\lambda^3\gamma^4 d/\beta.\label{eq46V951442443} \end{align}\tag{111}\] Combining 109 111 and using \(\gamma\geq\gamma_{\min}\geq \max\{1,K\}\) yields \[\begin{align} \label{eq46V951442} \begin{aligned} \mathsf{V}_{1,2}&\leq\frac{2\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+4|h(0)|^2\right)\\ &\quad+\lambda^3\gamma^4\left(\frac{2}{9}C_{1,v}+\frac{20d}{\beta}\right). \end{aligned} \end{align}\tag{112}\] Hence, substituting 108 and 112 into \(\mathsf{V}_1\), together with \(\gamma\geq\gamma_{\min}\geq 1\), yields \[\begin{align} \label{eq46V951} \begin{aligned} \mathsf{V}_1&\leq-2\lambda\gamma\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+ \frac{3\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\\ &\quad+\lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+4|h(0)|^2\right)\\ &\quad+\lambda^3\gamma^4\left(\frac{2}{9}C_{1,v}+\frac{20d}{\beta}+18\bar{B}_2+36m^2\bar{C}_2+18C_H\right). \end{aligned} \end{align}\tag{113}\] Now, we establish an upper bound for \(\mathsf{V}_2\), which can be decomposed as follows: \[\begin{align} \mathsf{V}_2&=-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})\rangle]_{2,1}}\tag{114}\\ &\quad-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\rangle]_{2,2}}\tag{115}\\ &\quad-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},h(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)\rangle]_{2,3}}\tag{116}\\ &\quad-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_t^\lambda)\rangle]_{2,4}}\tag{117}\\ &\quad-2\lambda\mathbb{E}[\langle V_t^\lambda-V_t^{\lambda,n},h_{MY,\epsilon}(Z_t^\lambda)-h_{MY,\epsilon}(\zeta_t^{\lambda,n})\rangle]_{2,5}}\tag{118}. \end{align}\] We proceed with establishing an upper bound for \(\mathsf{V}_{2,1}\). Applying Lemmas 7, 16, and Young’s inequality, we obtain \[\begin{align} \mathsf{V}_{2,1}&\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda\gamma^{-1}\mathbb{E}[|H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]\notag\\ &\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+18\lambda\gamma^{-3}\left(4K_H^2\mathbb{E}[(1+|X_0|)^{2\rho}]+m^2\right)(1+\mathbb{E}[|Z_{\lfloor t \rfloor}^\lambda|^{12r}])\notag\\ &\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+18\lambda\gamma^{-3}\left(4K_H^2\mathbb{E}[(1+|X_0|)^{2\rho}]+m^2\right)(1+\bar{C}_{12r}).\label{eq46V952441} \end{align}\tag{119}\] To establish an upper bound for \(\mathsf{V}_{2,2}\), we use 34 and ?? to obtain \[\begin{align} \mathsf{V}_{2,2}&=-2\lambda\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\rangle]_{2,2,1}}\\ &\quad+2\lambda^2\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t(V_{\lfloor s \rfloor}^\lambda-V_s^{\lambda,n})\,\mathrm{d}s,H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,2,2}}\\ &\quad+2\lambda^2\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t(H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})-h_{MY,\epsilon}(\zeta_s^{\lambda,n}))\,\mathrm{d}s,H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,2,3}}. \end{align}\] Since \(\mathcal{H}_t^\lambda=\mathcal{F}_t^\lambda\vee\mathcal{G}_{\lfloor t \rfloor}\vee\sigma(\theta_0,\nu_0)\) and \(X_{\lceil t \rceil}\) is independent of \(\mathcal{H}_{\lfloor t \rfloor}\), we have that \[\begin{align} \mathsf{V}_{2,2,1}&=-2\lambda\mathbb{E}[\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\rangle\mid\mathcal{H}_{\lfloor t \rfloor}]]\notag\\ &=-2\lambda\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},h(Z_{\lfloor t \rfloor}^\lambda)-h(Z_{\lfloor t \rfloor}^\lambda)\rangle]=0.\label{eq46V952442441} \end{align}\tag{120}\] Moreover, by 34 and ?? , we obtain \[\begin{align} \mathsf{V}_{2,2,2}&=2\lambda^2\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t (V_{\lfloor s \rfloor}^\lambda-V_{\lfloor s \rfloor}^{\lambda,n}) \,\mathrm{d}s, H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,2,2}^a}\\ &\quad+2\lambda^3\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s (\gamma V_r^{\lambda,n}+h_{MY,\epsilon}(\zeta_r^{\lambda,n}))\,\mathrm{d}r\, \mathrm{d}s, H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,2,2}^b}\\ &\quad-2\lambda^2\gamma\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s \sqrt{2\lambda\gamma\beta^{-1}}\,\mathrm{d}W_r^\lambda\, \mathrm{d}s, H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,2,2}^c}. \end{align}\] Since \(\int_{\lfloor t \rfloor}^t(V_{\lfloor s \rfloor}^\lambda-V_{\lfloor s \rfloor}^{\lambda,n}) \,\mathrm{d}s=(t-\lfloor t \rfloor)(V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n})\), it follows that \[\begin{align} \label{eq46V952442442441} \mathsf{V}_{2,2,2}^a=2\lambda^2\gamma(t-\lfloor t \rfloor)\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\rangle]=0, \end{align}\tag{121}\] where the second equality follows from the same argument as in the analysis of 120 . We note, by Remark 3 and Lemma 7, for any \(t\in(nT,(n+1)T]\), that \[\begin{align} &\mathbb{E}[|H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]\notag\\ &\leq 2K_H^2\mathbb{E}[(1+|X_{\lceil t \rceil}|)^{2\rho}(1+|Z_{\lfloor t \rfloor}^\lambda|^{4r})] +2L_h^2\mathbb{E}[(1+|Z_{\lfloor t \rfloor}^\lambda|)^{4r}]+2|h(0)|^2\notag\\ &\leq 2K_H^2\mathbb{E}[(1+|X_0|)^{2\rho}](1+\bar{C}_{4r}) + 2^{4r}L_h^2(1+\bar{C}_{4r})+2|h(0)|^2\leq c_A,\label{eq46cA} \end{align}\tag{122}\] where \(c_A\coloneq (2K_H^2\mathbb{E}[(1+|X_0|)^{2\rho}]+2^{4r}L_h^2)(1+\bar{C}_{4r})+2|h(0)|^2=\mathcal{O}((d/\beta)^{2r})\). By Young’s inequality, Cauchy-Schwarz inequality, Lemma 10, 90 and 122 , together with \(\gamma\geq\gamma_{\min}\geq\max\{1,K\}\), we have \[\begin{align} \mathsf{V}_{2,2,2}^b&\leq 2\lambda^3\gamma\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s \big(\gamma^2\mathbb{E}[|V_r^{\lambda,n}|^2]+\mathbb{E}[|h_{MY,\epsilon}(\zeta_r^{\lambda,n})|^2]\big)\,\mathrm{d}r\,\mathrm{d}s\notag\\ &\quad+2\lambda^3\gamma\big(\mathbb{E}[|H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]\big)\notag\\ &\leq \lambda^3\gamma^3\left(2C_V^\#+4C_\zeta^\#+4|h(0)|^2+2c_A\right).\label{eq46V952442442442} \end{align}\tag{123}\] Furthermore, we note that, an application of Itô’s formula yields \[(t-\lfloor t \rfloor)\int_{\lfloor t \rfloor}^t\,\mathrm{d}W_s^\lambda=\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s\,\mathrm{d}r\,\mathrm{d}W_s^\lambda+\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s\,\mathrm{d}W_r^\lambda\,\mathrm{d}s.\] Hence, by applying 122 , we obtain that \[\begin{align} \mathsf{V}_{2,2,2}^c&=-2\lambda^2\gamma\mathbb{E}\left[\left\langle(t-\lfloor t \rfloor)\sqrt{2\lambda\gamma\beta^{-1}}\int_{\lfloor t \rfloor}^t\,\mathrm{d}W_s^\lambda, H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]\notag\\ &\quad+2\lambda^2\gamma\mathbb{E}\left[\left\langle\sqrt{2\lambda\gamma\beta^{-1}}\int_{\lfloor t \rfloor}^t\int_{\lfloor s \rfloor}^s\,\mathrm{d}r\,\mathrm{d}W_s^\lambda, H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})-h(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]\notag\\ &\leq 2\lambda^3\gamma^2\beta^{-1}\mathbb{E}[|W_t^\lambda-W_{\lfloor t \rfloor}^\lambda|^2]+4\lambda^2\gamma\big(\mathbb{E}[|H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]\big)\notag\\ &\quad+2\lambda^3\gamma^2\beta^{-1}\mathbb{E}\left[\left|\int_{\lfloor t \rfloor}^t(s-\lfloor s \rfloor)\,\mathrm{d}W_s^\lambda\right|^2\right]\notag\\ &\leq 4\lambda^3\gamma^2 d/\beta+4\lambda^2\gamma c_A. \end{align}\] Moreover, using Young’s inequality, Lemmas 6 and 15, 90 and 122 , together with \(\gamma\geq\gamma_{\min}\geq\max\{1,K\}\), we obtain \[\begin{align} \mathsf{V}_{2,2,3}&\leq 2\lambda^2\gamma^{-1}\left(\mathbb{E}[|H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|h_{MY,\epsilon}(\zeta_t^{\lambda,n})|^2]\right)\notag\\ &\quad+2\lambda^2\gamma\left(\mathbb{E}[|H(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\mathbb{E}[|h(Z_{\lfloor t \rfloor})|^2]\right)\notag\\ &\leq 2\lambda^2\gamma^{-1}\big(2m^2\bar{C}_2+C_H\gamma+2K^2C_\zeta^\#+2|h(0)|^2\big)+2\lambda^2\gamma c_A\notag\\ &\leq \lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+4|h(0)|^2+2c_A\right).\label{eq46V952442443} \end{align}\tag{124}\] By using 120 , 121 , 123 124 and the condition \(\gamma\geq\gamma_{\min}\geq 1\), we obtain that, \[\begin{align} \label{eq46V952442} \begin{aligned} \mathsf{V}_{2,2}&\leq\lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+4|h(0)|^2+6c_A\right) \\ &\quad+\lambda^3\gamma^4\left(2C_V^\#+4C_\zeta^\#+4|h(0)|^2+2c_A+\frac{4d}{\beta}\right). \end{aligned} \end{align}\tag{125}\] Next, we consider establishing an upper bound for \(\mathsf{V}_{2,3}\). By 34 and ?? , we have \[\begin{align} \mathsf{V}_{2,3}&=-2\lambda\mathbb{E}[\langle V_{\lfloor t \rfloor}^\lambda-V_{\lfloor t \rfloor}^{\lambda,n},h(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)\rangle]_{2,3,1}}\\ &\quad+2\lambda^2\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t\gamma(V_{\lfloor s \rfloor}^\lambda-V_s^{\lambda,n})\,\mathrm{d}s,h(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,3,2}}\\ &\quad+2\lambda^2\mathbb{E}\left[\left\langle\int_{\lfloor t \rfloor}^t(H_\gamma(Z_{\lfloor s \rfloor}^\lambda,X_{\lceil s \rceil})-h_{MY,\epsilon}(\zeta_s^{\lambda,n}))\,\mathrm{d}s,h(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)\right\rangle\right]_{2,3,3}}. \end{align}\] Applying Young’s inequality together with Lemmas 3 and 7, we obtain \[\begin{align} \mathsf{V}_{2,3,1}&\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda\gamma^{-1}\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)|^2]\notag\\ &\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\cdot 2^{4r-2}\lambda\gamma^{-1}L_h^4\mathbb{E}[(1+|Z_{\lfloor t \rfloor}^\lambda|+R_0)^{8r-2}]\epsilon^2\notag\\ &\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+ C_4\lambda\gamma^{-1}\epsilon^2,\label{eq46V952443441} \end{align}\tag{126}\] where \(C_4\coloneq 9\cdot 2^{12r-5}L_h^4\big((1+R_0)^{8r-2}+\bar{C}_{8r-2}\big)=\mathcal{O}((d/\beta)^{4r-1})\). Similarly, using Young’s inequality, Remark 3, Lemmas 2, 6 and 8, and the assumption that \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\), we have that \[\begin{align} \mathsf{V}_{2,3,2}&\leq \frac{2\lambda^2\gamma^2}{9}\int_{\lfloor t \rfloor}^t\mathbb{E}[|V_s^\lambda-V_s^{\lambda,n}|^2]\,\mathrm{d}s+\frac{2\lambda^2\gamma^2}{9}\int_{\lfloor t \rfloor}^t\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda-V_s^\lambda|^2]\,\mathrm{d}s\notag\\ &\quad+18\lambda^2\big(\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]+\mathbb{E}[|h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)|^2]\big)\notag\\ &\leq \frac{2\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\frac{2}{9}\lambda^3\gamma^3 C_{1,v}+36\lambda^2\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]\notag\\ &\leq \frac{2\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\frac{2}{9}\lambda^3\gamma^3 C_{1,v}+\lambda^2 c_D, \end{align}\] where \(c_D\coloneq 36 (2^{4r}L_h^2(1+\bar{C}_{4r})+2|h(0)|^2)=\mathcal{O}((d/\beta)^{2r})\). Moreover, by using Young’s inequality, Lemmas 2, 6, 7, 10, and 15, together with Remark 3, 90 , and the condition \(\gamma\geq\gamma_{\min}\geq\max\{1,K\}\), we obtain that \[\begin{align} \mathsf{V}_{2,3,3}&\leq 2\lambda^2\gamma^{-1}\left(\mathbb{E}[|H_\gamma(Z_{\lfloor t \rfloor}^\lambda,X_{\lceil t \rceil})|^2]+\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|h_{MY,\epsilon}(\zeta_t^{\lambda,n})|^2]\right)\notag\\ &\quad+2\lambda^2\gamma\left(\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]+\mathbb{E}[|h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)|^2]\right)\notag\\ &\leq \lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+4|h(0)|^2\right)+4\lambda^2\gamma\mathbb{E}[|h(Z_{\lfloor t \rfloor}^\lambda)|^2]\notag\\ &\leq\lambda^2\gamma\left(4m^2\bar{C}_2+2C_H+4C_\zeta^\#+12|h(0)|^2+2^{4r+2}L_h^2(1+\bar{C}_{4r})\right).\label{eq46V952443443} \end{align}\tag{127}\] From 126 127 and \(\gamma\geq\gamma_{\min}\geq 1\), it follows that \[\begin{align} \label{eq46V952443} \begin{aligned} \mathsf{V}_{2,3}&\leq\frac{\lambda\gamma}{3}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+C_4\lambda\gamma^{-1}\epsilon^2+\frac{2}{9}\lambda^3\gamma^4 C_{1,v}\\ &\quad+\lambda^2\gamma\left(c_D+4m^2\bar{C}_2+2C_H+4C_\zeta^\#+12|h(0)|^2+2^{4r+2}L_h^2(1+\bar{C}_{4r})\right). \end{aligned} \end{align}\tag{128}\] To establish upper bounds for \(\mathsf{V}_{2,4}\) and \(\mathsf{V}_{2,5}\), by applying Young’s inequality, Lemma 8, Remark 9, and \(\gamma\geq\gamma_{\min}\geq K\), we obtain \[\begin{align} \mathsf{V}_{2,4}&\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda\gamma^{-1}\mathbb{E}[|h_{MY,\epsilon}(Z_{\lfloor t \rfloor}^\lambda)-h_{MY,\epsilon}(Z_t^\lambda)|^2]\notag\\ &\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda^2\gamma^{-1}K^2\bar{B}_2\notag\\ &\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda^2\gamma\bar{B}_2,\label{eq46V952444} \end{align}\tag{129}\] and \[\begin{align} \mathsf{V}_{2,5}&\leq \frac{\lambda\gamma}{9}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+9\lambda\gamma^{-1}K^2\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]\notag\\ &\leq \frac{\lambda\gamma}{9}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\frac{\lambda\gamma}{24}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2],\label{eq46V952445} \end{align}\tag{130}\] where the last inequality follows from the restriction that \(\gamma\geq\gamma_{\min}\geq 6\sqrt{6}K\). Combining 113 , 119 , 125 , 128 130 , we obtain \[\begin{align} \label{eq46differential32of32V} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]&\leq -2\lambda\gamma\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\lambda\gamma\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\\ &\quad+\lambda S_{\lambda,\gamma}+\frac{\lambda\gamma}{24}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2], \end{aligned} \end{align}\tag{131}\] where \[\begin{align} S_{\lambda,\gamma}&\coloneq C_1\lambda^2\gamma^4+C_2\lambda\gamma+C_3\gamma^{-3}+C_4\gamma^{-1}\epsilon^2,\label{eq46S95lambda44gamma}\\ C_1&\coloneq 4C_{1,v}/9+24d/\beta+18\bar{B}_2+36m^2\bar{C}_2+18C_H+2C_V^\#\\ &\quad+4C_\zeta^\#+4|h(0)|^2+2c_A=\mathcal{O}((d/\beta)^{2r}),\\ C_2&\coloneq 12m^2\bar{C}_2+6C_H+12C_\zeta^\#+20|h(0)|^2+6c_A+c_D\\ &\quad\;+2^{4r+2}L_h^2(1+\bar{C}_{4r})+9\bar{B}_2=\mathcal{O}((d/\beta)^{2r}),\\ C_3&\coloneq 18\big(4K_H^2\mathbb{E}[(1+|X_0|)^{2\rho}]+m^2\big)(1+\bar{C}_{12r})=\mathcal{O}((d/\beta)^{6r}),\\ C_4&\coloneq 9\cdot 2^{12r-5}L_h^4\big((1+R_0)^{8r-2}+\bar{C}_{8r-2}\big)=\mathcal{O}((d/\beta)^{4r-1}). \end{align}\tag{132}\] Combining Lemmas 6, 8, and 10 with Cauchy-Schwarz inequality, under the assumptions that \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\), we obtain \[\begin{align} \sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]&\leq 3\sup_{t\in(nT,(n+1)T]}\left(\mathbb{E}[|\zeta_t^{\lambda,n}|^2]+\mathbb{E}[|Z_{\lfloor t \rfloor}^\lambda|^2]+\mathbb{E}[|Z_t^\lambda-Z_{\lfloor t \rfloor}^\lambda|^2]\right)\leq C_J\\ \sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]&\leq 3\sup_{t\in(nT,(n+1)T]}\left(\mathbb{E}[|V_t^{\lambda,n}|^2]+\mathbb{E}[|V_{\lfloor t \rfloor}^\lambda|^2]+\mathbb{E}[|V_t^\lambda-V_{\lfloor t \rfloor}^\lambda|^2]\right)\leq C_M, \end{align}\] where \(C_J\coloneq 3C_\zeta^\#+3\bar{C}_2+3\bar{B}_2=\mathcal{O}(d/\beta)\) and \(C_M\coloneq 3C_V^\#+3\bar{B}_2+3C_{1,v}=\mathcal{O}(d/\beta)\). Consequently, 131 can be rewritten as \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]&\leq -2\lambda\gamma\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\\ &\quad+\lambda\gamma\left(\frac{S_{\lambda,\gamma}}{\gamma}+\frac{1}{24}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]\right.\\ &\quad\quad\quad\quad\;\left.+\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\right), \end{align}\] which implies, by applying integrating factor, that \[\begin{align} &\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\\ &\leq \frac{1}{2}\left(\frac{S_{\lambda,\gamma}}{\gamma}+\sup_{t\in(nT,(n+1)T]}\left(\frac{1}{24}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]+\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\right)\right)\left(1-e^{-2\lambda\gamma (t-nT)}\right)\\ &\leq \frac{1}{2}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+\frac{1}{2}\left(\frac{S_{\lambda,\gamma}}{\gamma}+\frac{1}{24}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]\right). \end{align}\] Taking the supremum over \(t\in(nT,(n+1)T]\) gives \[\label{eq46mse32of32momentum} \sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\leq \frac{S_{\lambda,\gamma}}{\gamma}+\frac{1}{24}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2].\tag{133}\]

To establish an upper bound for the second term on the right-hand side of 105 , we apply Itô’s formula to the expressions given in 34 and ?? to obtain that \[\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]=2\lambda\int_{nT}^t\mathbb{E}[\langle Z_s^\lambda-\zeta_s^{\lambda,n},V_{\lfloor s \rfloor}^\lambda-V_s^{\lambda,n}\rangle]\,\mathrm{d}s,\] which can be further rewritten as \[\begin{align} \mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]&=2\lambda\int_{nT}^t\mathbb{E}[\langle Z_s^\lambda-\zeta_s^{\lambda,n},V_{\lfloor s \rfloor}^\lambda-V_{\lfloor s \rfloor}^{\lambda,n}\rangle]\,\mathrm{d}s_{1}}\tag{134}\\ &\quad+2\lambda^2\int_{nT}^t\mathbb{E}\left[\left\langle Z_s^\lambda-\zeta_s^{\lambda,n},\int_{\lfloor s \rfloor}^s\left(\gamma V_r^{\lambda,n}+h_{MY,\epsilon}(\zeta_r^{\lambda,n})\right)\,\mathrm{d}r\right\rangle\right]\,\mathrm{d}s_{2}}\tag{135}\\ &\quad-2\lambda\int_{nT}^t\mathbb{E}\left[\left\langle Z_s^\lambda-\zeta_s^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}\int_{\lfloor s \rfloor}^s\,\mathrm{d}W_r^\lambda \right\rangle\right]\,\mathrm{d}s._{3}} \end{align}\] Applying Young’s inequality to \(\mathsf{Z}_1\) yields \[\begin{align} \label{eq46Z951} \mathsf{Z}_{1}\leq \frac{\lambda}{4}\int_{nT}^t\mathbb{E}[|Z_s^\lambda-\zeta_s^{\lambda,n}|^2]\,\mathrm{d}s+4\lambda\int_{nT}^t\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda-V_{\lfloor s \rfloor}^{\lambda,n}|^2]\,\mathrm{d}s. \end{align}\tag{136}\] Moreover, applying Young’s inequality and Cauchy-Schwarz inequality to \(\mathsf{Z}_2\) yields \[\label{eq46Z952} \mathsf{Z}_{2}\leq \frac{\lambda}{4}\int_{nT}^t\mathbb{E}[|Z_s^\lambda-\zeta_s^{\lambda,n}|^2]\,\mathrm{d}s+4\lambda^3\int_{nT}^t\mathbb{E}\left[\int_{\lfloor s \rfloor}^s|\gamma V_r^{\lambda,n}+h_{MY,\epsilon}(\zeta_r^{\lambda,n})|^2\,\mathrm{d}r\right]\,\mathrm{d}s.\tag{137}\] By Lemma 10, 90 , and \(\gamma\geq\gamma_{\min}\geq K\), we obtain \[\begin{align} &4\lambda^3\int_{nT}^t\mathbb{E}\left[\int_{\lfloor s \rfloor}^s|\gamma V_r^{\lambda,n}+h_{MY,\epsilon}(\zeta_r^{\lambda,n})|^2\,\mathrm{d}r\right]\,\mathrm{d}s\notag\\ &\leq 8 \lambda^3\gamma^2\int_{nT}^t\sup_{nT\leq r\leq s}\mathbb{E}[|V_r^{\lambda,n}|^2]\,\mathrm{d}s+16\lambda^3\int_{nT}^t\mathbb{E}\left[\int_{\lfloor s \rfloor}^s\left(K^2|\zeta_r^{\lambda,n}|^2+|h(0)|^2\right)\,\mathrm{d}r\right]\,\mathrm{d}s\notag\\ &\leq 8\lambda^2\gamma^2 C_V^\#+16\lambda^2 \gamma^2 C_\zeta^\#+16\lambda^2|h(0)|^2.\label{eq46Z952-1} \end{align}\tag{138}\] Thus, by substituting 138 into 137 , it holds that \[\label{eq46Z951442} \mathsf{Z}_{2}\leq \frac{\lambda}{4}\int_{nT}^t\mathbb{E}[|Z_s^\lambda-\zeta_s^{\lambda,n}|^2]\,\mathrm{d}s+8\lambda^2\gamma^2 C_V^\#+16\lambda^2 \gamma^2 C_\zeta^\#+16\lambda^2|h(0)|^2.\tag{139}\] Next, recall that \(\mathcal{H}_t^\lambda= \mathcal{F}_t^\lambda\vee\mathcal{G}_{\lfloor t \rfloor}\vee\sigma(\theta_0,\nu_0)\). We have, for any \(s\in(nT,t]\), that \[\begin{align} &\mathbb{E}\left[\left\langle Z_{\lfloor s \rfloor}^\lambda-\zeta_{\lfloor s \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_s^\lambda-W_{\lfloor s \rfloor}^\lambda) \right\rangle\right]\notag\\ &=\mathbb{E}\left[\mathbb{E}\left[\left\langle Z_{\lfloor s \rfloor}^\lambda-\zeta_{\lfloor s \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_s^\lambda-W_{\lfloor s \rfloor}^\lambda)\right\rangle\mid\mathcal{H}_{\lfloor s \rfloor}^\lambda\right] \right]\notag\\ &=\mathbb{E}\left[\left\langle Z_{\lfloor s \rfloor}^\lambda-\zeta_{\lfloor s \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}\mathbb{E}[W_s^\lambda-W_{\lfloor s \rfloor}^\lambda\mid\mathcal{H}_{\lfloor s \rfloor}^\lambda]\right\rangle\right]=0.\label{eq46Z953-1} \end{align}\tag{140}\] Moreover, notice that \[\label{eq46Z9531} -2\lambda\int_{nT}^t\mathbb{E}\left[\left\langle\lambda\int_{\lfloor s \rfloor}^s V_{\lfloor r \rfloor}^\lambda\,\mathrm{d}r,\sqrt{2\lambda\gamma\beta^{-1}}(W_s^\lambda-W_{\lfloor s \rfloor}^\lambda)\right\rangle\right]\,\mathrm{d}s=0.\tag{141}\] Thus, by Young’s inequality, Cauchy-Schwarz inequality, 140 , 141 and Lemma 8, we have \[\begin{align} \mathsf{Z}_{3}&=-2\lambda\int_{nT}^t\mathbb{E}\left[\left\langle Z_{\lfloor s \rfloor}^\lambda-\zeta_{\lfloor s \rfloor}^{\lambda,n},\sqrt{2\lambda\gamma\beta^{-1}}(W_s^\lambda-W_{\lfloor s \rfloor}^\lambda) \right\rangle\right]\,\mathrm{d}s\notag\\ &\quad-2\lambda\int_{nT}^t\mathbb{E}\left[\left\langle\lambda\int_{\lfloor s \rfloor}^s(V_{\lfloor r \rfloor}^\lambda-V_r^{\lambda,n})\,\mathrm{d}r,\sqrt{2\lambda\gamma\beta^{-1}}(W_s^\lambda-W_{\lfloor s \rfloor}^\lambda) \right\rangle\right]\,\mathrm{d}s\notag\\ &\leq \lambda\int_{nT}^t\left(\mathbb{E}\left[\lambda\int_{\lfloor s \rfloor}^s|V_{\lfloor r \rfloor}^\lambda-V_r^{\lambda,n}|^2\,\mathrm{d}r\right]\,\mathrm{d}s+2\lambda^2\gamma\beta^{-1}\mathbb{E}[|W_s^\lambda-W_{\lfloor s \rfloor}^\lambda|^2]\right)\,\mathrm{d}s\notag\\ &\leq \lambda^2\int_{nT}^t\mathbb{E}\left[\int_{\lfloor s \rfloor}^s|V_{\lfloor r \rfloor}^\lambda-V_r^{\lambda,n}|^2\,\mathrm{d}r\right]\,\mathrm{d}s+2\lambda^2\gamma d/\beta\notag\\ &\leq 2\lambda^2\int_{nT}^t\mathbb{E}\left[\int_{\lfloor s \rfloor}^s\big(|V_{\lfloor r \rfloor}^\lambda-V_r^{\lambda}|^2+|V_r^\lambda-V_r^{\lambda,n}|^2\big)\,\mathrm{d}r\right]\,\mathrm{d}s+2\lambda^2\gamma d/\beta\notag\\ &\leq 2\lambda^2\int_{nT}^t\int_{\lfloor s \rfloor}^s\mathbb{E}[|V_r^\lambda-V_r^{\lambda,n}|^2]\,\mathrm{d}r\,\mathrm{d}s+2\lambda^2\gamma C_{1,v}+2\lambda^2\gamma d/\beta.\label{eq46Z953} \end{align}\tag{142}\] Combining 136 , 139 and 142 , together with \(\gamma\geq\gamma_{\min}\geq 1\) and \(\lambda\leq\lambda_{\max,\gamma}\leq 1/(4\gamma)\), yields \[\begin{align} \mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]&\leq \frac{\lambda}{2}\int_{nT}^t\mathbb{E}[|Z_s^\lambda-\zeta_s^{\lambda,n}|^2]\,\mathrm{d}s+4\lambda\int_{nT}^t\mathbb{E}[|V_{\lfloor s \rfloor}^\lambda-V_{\lfloor s \rfloor}^{\lambda,n}|^2]\,\mathrm{d}s\\ &\quad+2\lambda^2\int_{nT}^t\int_{\lfloor s \rfloor}^s\mathbb{E}[|V_r^\lambda-V_r^{\lambda,n}|^2]\,\mathrm{d}r\,\mathrm{d}s\\ &\quad+8\lambda^2\gamma^2 C_V^\#+16\lambda^2 \gamma^2 C_\zeta^\#+16\lambda^2|h(0)|^2+2\lambda^2\gamma C_{1,v}+2\lambda^2\gamma d/\beta\\ &\leq \frac{\lambda}{2}\int_{nT}^t\mathbb{E}[|Z_s^\lambda-\zeta_s^{\lambda,n}|^2]\,\mathrm{d}s+6\lambda\int_{nT}^t\sup_{nT\leq r\leq s}\mathbb{E}[|V_r^\lambda-V_r^{\lambda,n}|^2]\,\mathrm{d}s\\ &\quad+8\lambda^2\gamma^2 C_V^\#+16\lambda^2 \gamma^2 C_\zeta^\#+16\lambda^2|h(0)|^2+2\lambda^2\gamma C_{1,v}+2\lambda^2\gamma d/\beta\\ &\leq\frac{1}{2}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]+6\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]+C_5\lambda^2\gamma^2, \end{align}\] where \(C_5\coloneq 8C_V^\#+16C_\zeta^\#+16|h(0)|^2+2C_{1,v}+2d/\beta=\mathcal{O}(d/\beta)\), which implies, by using 133 , that \[\begin{align} \mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]&\leq \frac{1}{2}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]+C_5\lambda^2\gamma^2\\ &\quad+6\left(\frac{S_{\lambda,\gamma}}{\gamma}+\frac{1}{24}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]\right)\\ &\leq \frac{3}{4}\sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]+\frac{6S_{\lambda,\gamma}}{\gamma}+C_5\lambda^2\gamma^2. \end{align}\] Taking the supremum over \(t\in(nT,(n+1)T]\) gives \[\label{eq46sup32Z-xi} \sup_{t\in(nT,(n+1)T]}\mathbb{E}[|Z_t^\lambda-\zeta_t^{\lambda,n}|^2]\leq\frac{24S_{\lambda,\gamma}}{\gamma}+4C_5\lambda^2\gamma^2.\tag{143}\] Substituting 143 back into 133 yields \[\label{eq46mse32of32velocity} \sup_{t\in(nT,(n+1)T]}\mathbb{E}[|V_t^\lambda-V_t^{\lambda,n}|^2]\leq \frac{2S_{\lambda,\gamma}}{\gamma}+\frac{C_5}{6}\lambda^2\gamma^2.\tag{144}\] Finally, in view of 105 , by using 143 and 144 , we conclude that \[W_2^2(\mathcal{L}(Z_t^\lambda,V_t^\lambda),\mathcal{L}(\zeta_t^{\lambda,n},V_t^{\lambda,n}))\leq C_0^2\left(\frac{S_{\lambda,\gamma}}{\gamma}+\lambda^2\gamma^2\right),\] where \(C_0\coloneq\sqrt{26(C_5+1)}=\mathcal{O}((d/\beta)^{1/2})\). This completes the proof. ◻

Proof of Theorem 3. The proof is adapted from [28], we provide details here for explicit constants. Let \((r_t,R_t)\) and \((r_t',R_t')\) be two solutions of the underdamped Langevin SDE 32 with initial conditions \((r_0,R_0)\) and \((r_0',R_0')\), and define \[\label{eq46y44z} y_t\coloneq (R_t-R_t')+\gamma(r_t-r_t'),\quad z_t\coloneq -(R_t-R_t')\tag{145}\] for all \(t\geq 0\). By Lemma 1, \(u_{MY,\epsilon}\) is twice continuously differentiable. Hence, by the fundamental theorem of calculus, \[\label{eq46taylor-MY} h_{MY,\epsilon}(r_t)-h_{MY,\epsilon}(r_t')=S_t(r_t-r_t'),\tag{146}\] where \[\label{eq46S95t} S_t\coloneq\int_0^1 \operatorname{Hess}(u_{MY,\epsilon})(r_t'+s(r_t-r_t'))\,\mathrm{d}s.\tag{147}\] By 145 , 146 and the expression of 32 , we have \[\begin{align} \frac{\mathrm{d}y_t}{\mathrm{d}t}&=-\gamma(R_t-R_t')-(h_{MY,\epsilon}(r_t)-h_{MY,\epsilon}(r_t'))+\gamma(R_t-R_t')\notag\\ &=-(h_{MY,\epsilon}(r_t)-h_{MY,\epsilon}(r_t'))=-S_t(r_t-r_t')=-\frac{S_t(y_t+z_t)}{\gamma},\label{eq46dy} \end{align}\tag{148}\] and \[\begin{align} \label{eq46dz} \begin{aligned} \frac{\mathrm{d}z_t}{\mathrm{d}t}&=\gamma(R_t-R_t')+(h_{MY,\epsilon}(r_t)-h_{MY,\epsilon}(r_t'))=-\gamma z_t+\frac{S_t(y_t+z_t)}{\gamma}. \end{aligned} \end{align}\tag{149}\] Combining 148 and 149 , we have \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}|(y_t,z_t)|^2&=2\left\langle y_t,\frac{\mathrm{d}y_t}{\mathrm{d}t}\right\rangle+2\left\langle z_t,\frac{\mathrm{d}z_t}{\mathrm{d}t}\right\rangle\notag\\ &=-\frac{2}{\gamma}\left\langle y_t, S_t(y_t+z_t)\right\rangle-2\gamma \left\langle z_t, z_t\right\rangle+\frac{2}{\gamma}\left\langle z_t, S_t(y_t+z_t)\right\rangle\notag\\ &=-\frac{2}{\gamma}\langle y_t,S_t y_t\rangle+\frac{2}{\gamma} \langle z_t,S_t z_t\rangle-2\gamma|z_t|^2.\label{eq46derivative32of32y44z} \end{align}\tag{150}\] Using Remark 9, we have, for all \(t\geq 0\), that \[\label{eq46quadratic32form32of32S95t} \frac{m}{2}I_d\preceq S_t\preceq KI_d.\tag{151}\] Substituting 151 into the right-hand side of 150 yields \[\begin{align} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}|(y_t,z_t)|^2&\leq -\frac{m}{\gamma}|y_t|^2+\frac{2K}{\gamma}|z_t|^2-2\gamma|z_t|^2 \leq \frac{\max\{-m,2(K-\gamma^2)\}}{\gamma}\,|(y_t,z_t)|^2. \end{aligned} \end{align}\] By integrating factor and the assumption that \(\gamma\geq\gamma_{\min}\geq \sqrt{K+m/2}\), we obtain, for all \(t\geq 0\), that \[|(y_t,z_t)|^2\leq \exp\left(\frac{\max\{-m,2(K-\gamma^2)\}}{\gamma} t\right)|(y_0,z_0)|^2\leq \exp\left(-\frac{m}{\gamma} t\right)|(y_0,z_0)|^2,\] and thus \[\begin{align} \label{eq46contraction32y44z} |(y_t,z_t)|\leq \exp\left(-\frac{m}{2\gamma} t\right)|(y_0,z_0)|. \end{align}\tag{152}\] Next, by the definition of \(y_t\) and \(z_t\) in 145 , we have \[|y_0|^2\leq 2|R_0-R_0'|^2+2\gamma^2|r_0-r_0'|^2,\quad |z_0|^2=|R_0-R_0'|^2,\] so that \[\label{eq46initial32y44z} |(y_0,z_0)|\leq\sqrt{3+2\gamma^2}\,|(r_0-r_0',R_0-R_0')|.\tag{153}\] Moreover, using the identity \(r_t-r_t'=(y_t+z_t)/\gamma\), along with 152 and 153 , and the condition \(\gamma\geq\gamma_{\min}\geq 1\), we have, for all \(t\geq 0\), that \[\begin{align} |r_t-r_t'|\leq\frac{\sqrt{2}}{\gamma}|(y_t,z_t)|&\leq \frac{\sqrt{2(3+2\gamma^2)}}{\gamma}\exp\left(-\frac{m}{2\gamma}t\right)|(r_0-r_0',R_0-R_0')|\\ &\leq 4\exp\left(-\frac{m}{2\gamma}t\right)|(r_0-r_0',R_0-R_0')|. \end{align}\] Taking expectations and then the infimum, we conclude that \[W_2(\mathcal{L}(r_t),\mathcal{L}(r_t'))\leq 4\exp\left(-\frac{m}{2\gamma}t\right)W_2(\mathcal{L}(r_0,R_0),\mathcal{L}(r_0',R_0')),\] which completes the proof. ◻

Proof of Proposition 5. For all \(n\in\mathbb{N}_0\), \(t\in(nT,(n+1)T]\), and \(\gamma\geq\gamma_{\min}\), by Theorem 3, Corollary 1, and Proposition 4, we have \[\begin{align} W_2\big(\mathcal{L}(\zeta_t^{\lambda,n}),\mathcal{L}(r_t^\lambda)\big)&\leq \sum_{k=1}^n W_2\big(\mathcal{L}(\zeta_t^{\lambda,k}),\mathcal{L}(\zeta_t^{\lambda,k-1})\big)\\ &=\sum_{k=1}^n W_2(\mathcal{L}\big(\widehat{\zeta}_t^{kT,Z_{kT}^\lambda,V_{kT}^\lambda,\lambda}),\mathcal{L}(\widehat{\zeta}_t^{kT,\zeta_{kT}^{\lambda,k-1},V_{kT}^{\lambda,k-1},\lambda})\big)\\ &\leq \sum_{k=1}^n 4 \exp\left(-\frac{m\lambda}{2\gamma}(t-kT)\right)W_2\big(\mathcal{L}(Z_{kT}^\lambda,V_{kT}^\lambda),\mathcal{L}(\zeta_{kT}^{\lambda,k-1},V_{kT}^{\lambda,k-1})\big)\\ &\leq 4C_0\sum_{k=1}^n\exp\left(-\frac{m}{4\gamma}(n-k)\right)\left(\lambda\gamma+\sqrt{\frac{S_{\lambda,\gamma}}{\gamma}}\right)\\ &\leq 4C_0\left(\lambda\gamma+\sqrt{\frac{S_{\lambda,\gamma}}{\gamma}}\right)\frac{1}{1-e^{-\frac{m}{4\gamma}}}\\ &\leq \frac{16C_0\gamma}{m}\exp\left(\frac{m}{4\gamma}\right)\left(\lambda\gamma+\sqrt{\frac{S_{\lambda,\gamma}}{\gamma}}\right)\\ &\leq \frac{16C_0e}{m}\left(\lambda\gamma^2+\sqrt{\gamma S_{\lambda,\gamma}}\right), \end{align}\] where \(C_0\) is given in Proposition 4, with explicitly expression given in Table [tab46constants], and where the third inequality holds due to \(T\geq 1/(2\lambda)\) for \(\lambda\leq\lambda_{\max,\gamma}\leq 1\), the fifth inequality holds due to \(1-e^{-x}\geq xe^{-x}\) for all \(x\geq 0\), and the last inequality holds due to \(\gamma\geq\gamma_{\min}\geq 14m\). ◻

Proof of Corollary 1. Let \((r_t,R_t)_{t\ge 0}\) be the solution of 32 with initial condition \((r_0,R_0)=(\theta_0,\nu_0)\). By definition of the scaled process 33 , we have \((r_t^\lambda,R_t^\lambda)=(r_{\lambda t},R_{\lambda t})\). Let \((r_t',R_t')_{t\ge 0}\) be another solution of 32 initialized according to the invariant measure \(\Pi_\beta^\epsilon\). Applying Theorem 3 at time \(\lambda t\) yields \[W_2\bigl(\mathcal{L}(r_{\lambda t}),\mathcal{L}(r_{\lambda t}')\bigr) \le 4\exp\left(-\frac{\lambda m}{2\gamma}t\right) W_2\bigl(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta^\epsilon\bigr).\] Since \(\Pi_\beta^\epsilon\) is invariant for 32 , its position marginal is \(\pi_\beta^\epsilon\), and hence \(\mathcal{L}(r_{\lambda t}')=\pi_\beta^\epsilon\). Therefore, \[W_2\bigl(\mathcal{L}(r_t^\lambda),\pi_\beta^\epsilon\bigr) \le 4\exp\left(-\frac{\lambda m}{2\gamma}t\right) W_2\bigl(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta^\epsilon\bigr),\] which proves the claim. ◻

Proof of Lemma 11. By Remark 3, we have, for all \(\theta\in\mathbb{R}^d\), that \[\begin{align} \label{eq46upper32bound32of32htheta} |h(\theta)|-|h(0)|\leq |h(\theta)-h(0)|\leq L_h(1+|\theta|)^{2r-1}|\theta|\leq L_h(1+|\theta|)^{2r}, \end{align}\tag{154}\] which implies, \[\begin{align} \label{eq46upper32bound32of32h} |h(\theta)|\leq |h(0)|+L_h(1+|\theta|)^{2r}\leq |h(0)|+L_h 2^{2r-1}(1+|\theta|^{2r}). \end{align}\tag{155}\] Let \((\Theta_n, \Theta)\) follow an optimal coupling of \(\mathcal{L}(\theta_n^\lambda)\) and \(\pi_\beta\), that is, \(\mathcal{L}(\Theta_n)=\mathcal{L}(\theta_n^\lambda)\), \(\mathcal{L}(\Theta)=\pi_\beta\) and \[\label{eq46coupling} \left(\mathbb{E}[|\Theta_n-\Theta|^2]\right)^{1/2}=W_2(\mathcal{L}(\theta_n^\lambda),\pi_\beta).\tag{156}\] Then, we have, for all \(\theta,\theta'\in\mathbb{R}^d\), that \[\begin{align} u(\theta)-u(\theta')=\int_0^1\langle h(\theta'+t(\theta-\theta')),\theta-\theta'\rangle\,\mathrm{d}t. \end{align}\] Therefore, using Cauchy-Schwarz inequality and 155 , we obtain \[\begin{align} |u(\theta)-u(\theta')|&\leq \left|\int_0^1\langle h(\theta'+t(\theta-\theta')),\theta-\theta'\rangle\,\mathrm{d}t\right|\\ &\leq \int_0^1|h(\theta'+t(\theta-\theta'))|\,|\theta-\theta'|\,\mathrm{d}t\\ &\leq \int_0^1\left(|h(0)|+L_h2^{2r-1}(1+|t\theta+(1-t)\theta'|^{2r})\right)|\theta-\theta'|\,\mathrm{d}t\\ &\leq \left(|h(0)|+L_h 2^{2r-1}+L_h 2^{4r-2}(|\theta|^{2r}+|\theta'|^{2r})\right)|\theta-\theta'|. \end{align}\] Taking expectations and applying the Cauchy-Schwarz inequality, the elementary bound \(|a+b+c|^2\le 3(|a|^2+|b|^2+|c|^2)\) for \(a,b,c\in\mathbb{R}\), together with 156 and Lemmas 4 and 7, we obtain \[\begin{align} &\mathbb{E}[u(\theta_n^\lambda)]-\mathbb{E}[u(\theta_\infty)]\\ &=\mathbb{E}[u(\Theta_n)]-\mathbb{E}[u(\Theta)]\\ &\leq \mathbb{E}\left[\left(|h(0)|+L_h2^{2r-1}+L_h 2^{4r-2}(|\Theta_n|^{2r}+|\Theta|^{2r})\right)|\Theta_n-\Theta|\right]\\ &\leq \left(\mathbb{E}\left[\left||h(0)|+L_h2^{2r-1}+L_h 2^{4r-2}(|\Theta_n|^{2r}+|\Theta|^{2r})\right|^2\right]\right)^{1/2}\left(\mathbb{E}[|\Theta_n-\Theta|^2]\right)^{1/2}\\ &\leq \left(3(|h(0)|+L_h2^{2r-1})^2+3L_h^2 2^{8r-4}\mathbb{E}[|\theta_n^\lambda|^{4r}]+3L_h^2 2^{8r-4}\mathbb{E}[|\theta_\infty|^{4r}]\right)^{1/2}W_2(\mathcal{L}(\theta_n^\lambda),\pi_\beta)\\ &\leq \left(3(|h(0)|+L_h2^{2r-1})^2+3L_h^2 2^{8r-4}\big(\bar C_{4r}+C_{\pi_\beta ,4r}\big)\right)^{1/2}W_2(\mathcal{L}(\theta_n^\lambda),\pi_\beta). \end{align}\] Finally, applying Theorem 1 yields \[\mathbb{E}[u(\theta_n^\lambda)]-\mathbb{E}[u(\theta_\infty)]\leq C'\left(\sqrt{\lambda}\gamma+\lambda\gamma^{5/2}+\gamma^{-1}+\exp\left(-\frac{\lambda m}{2\gamma}n\right)W_2(\mathcal{L}(\theta_0,\nu_0),\Pi_\beta)+\epsilon\right),\] where \(C'\coloneq\big(3(|h(0)|+L_h2^{2r-1})^2+3L_h^2 2^{8r-4}(\bar C_{4r}+C_{\pi_\beta ,4r})\big)^{1/2} \dot{C}=\mathcal{O}((d/\beta)^{4r+1/2})\) with \(\dot{C}\) given in Theorem 1. ◻

Proof of Lemma 12. By Remark 5, we have \(\theta^*\in\bar{B}(0,R_0)\) with \(R_0= \sqrt{2u(0)/m}\). We note that, for all \(\theta,\theta'\in\bar{B}(0,R_0+1)\), \[|h(\theta)-h(\theta')|\leq L_h(1+2R_0)^{2r-1}|\theta-\theta'|.\] Hence, by the convexity of \(u\) as shown in Remark 5 and the fact that \(h(\theta^*)=0\), we have \[\begin{align} \mathbb{E}[u(\theta_\infty)\mathbb{1}_{\bar{B}(0,R_0+1)}(\theta_\infty)]-u(\theta^*)&\leq \int_{\bar{B}(0,R_0+1)} (u(\theta)-u(\theta^*))\,\pi_\beta(\mathrm{d}\theta)\notag\\ &\leq \int_{\bar{B}(0,R_0+1)} \langle h(\theta)-h(\theta^*),\theta-\theta^*\rangle\,\pi_\beta(\mathrm{d}\theta)\notag\\ &\leq L_h(1+2R_0)^{2r-1}\int_{\bar{B}(0,R_0+1)}|\theta-\theta^*|^2\,\pi_\beta(\mathrm{d}\theta)\notag\\ &\leq L_h(1+2R_0)^{2r-1}\mathbb{E}[|\theta_\infty-\theta^*|^2\mathbb{1}_{\bar{B}(0,R_0+1)}]\notag\\ &\leq L_h(1+2R_0)^{2r-1}\mathbb{E}[|\theta_\infty-\theta^*|^2].\label{eq46first32part32of32u40invariant41} \end{align}\tag{157}\] Moreover, since \(\theta^*\in\bar{B}(0,R_0)\), we have \[\mathbb{P}(\theta_\infty\notin\bar{B}(0,R_0+1))\leq \mathbb{P}(|\theta_\infty-\theta^*|>1)\leq \mathbb{E}[|\theta_\infty-\theta^*|^2],\] where the last inequality follows from the Chebyshev’s inequality. Applying Cauchy-Schwarz inequality yields \[\begin{align} \mathbb{E}[u(\theta_\infty)\mathbb{1}_{{\bar{B}(0,R_0+1)}^c}(\theta_\infty)]&\leq \left(\mathbb{E}[u(\theta_\infty)^2]\right)^{1/2}\left(\mathbb{P}(\theta_\infty\in {\bar{B}(0,R_0+1)}^c)\right)^{1/2}\notag\\ &\leq \left(\mathbb{E}[u(\theta_\infty)^2]\right)^{1/2}\left(\mathbb{E}[|\theta_\infty-\theta^*|^2]\right)^{1/2}.\label{eq46second32part32of32u40invariant41} \end{align}\tag{158}\] Next, for any \(\theta\in\mathbb{R}^d\), by the fundamental theorem of calculus, Cauchy-Schwarz inequality, and 154 , we have that \[\begin{align} |u(\theta)-u(0)|=\left|\int_0^1\langle h(t\theta),\theta\rangle\,\mathrm{d}t\right|&\leq \int_0^1|h(t\theta)|\, |\theta|\,\mathrm{d}t\\ &\leq |\theta|\int_0^1 \left(|h(0)|+L_h(1+|t\theta|)^{2r}\right)\,\mathrm{d}t\\ &\leq (|h(0)|+L_h)(1+|\theta|)^{2r+1}. \end{align}\] Therefore, we obtain that \[\begin{align} |u(\theta)|^2&\leq 2|u(\theta)-u(0)|^2+2|u(0)|^2\\ &\leq 2(|h(0)|+L_h)^2(1+|\theta|)^{4r+2}+2|u(0)|^2\\ &\leq 2^{4r+3}(|h(0)|^2+L_h^2)(1+|\theta|^{4r+2})+2|u(0)|^2 \end{align}\] Therefore, by Lemma 4, the second moment of \(u(\theta_\infty)\) satisfies \[\label{eq46moment32bounds32of32u40invariant41} \mathbb{E}[|u(\theta_\infty)|^2]\leq 2^{4r+3}(|h(0)|^2+L_h^2)(1+C_{\pi_\beta ,4r+2})+2|u(0)|^2.\tag{159}\] Combining 157 , 158 and 159 , we conclude that \[\mathbb{E}[u(\theta_\infty)]-u(\theta^*)\leq C''\mathbb{E}[|\theta_\infty-\theta^*|^2],\] where \(C''\coloneq L_h(1+2R_0)^{2r-1}+(2^{4r+3}(|h(0)|^2+L_h^2)(1+C_{\pi_\beta ,4r+2})+2u(0)^2)^{1/2}=\mathcal{O}((d/\beta)^{r+1/2})\). ◻