Proof of Theorem 4. We first check that \((\mathsf Z,d_{V_\alpha,\alpha})\) is Polish. Since \(\mathsf Z\) is closed in a finite-dimensional Euclidean space, \(V_\alpha\) is continuous, and \(V_\alpha\ge1\), the metric \(d_{V_\alpha,\alpha}\) dominates the augmented base norm 3 . Conversely, on every set \(K_R:=\{z\in\mathsf Z:\vert z\vert_\alpha\le R\},\) continuity of \(V_\alpha\) gives \(\sup_{K_R}V_\alpha<\infty\), so \(d_{V_\alpha,\alpha}\) and the augmented base metric are locally comparable on \(K_R\). Hence \((\mathsf Z,d_{V_\alpha,\alpha})\) is Polish by elementary general topology.

Let \(c_0\) be as in Lemma 2, and choose \(\alpha_0 \in (0, L_{\mathsf G}^{-1}]\) small enough that the conclusions of Lemmas 2, 3, and 4 hold. Fix \(0<\alpha\le\alpha_0\). Equations 24 and 31 verify respectively the two conditions ?? and ?? of Proposition 11, with \((E,d)=(\mathsf Z,d_{V_\alpha,\alpha})\), \(r=1-c_0\alpha^{m-1}\), and \(z_\star\) as in Lemma 4. Proposition 11 therefore yields a unique invariant law \(\pi_\alpha\in\mathcal{P}_1(\mathsf Z,d_{V_\alpha,\alpha})\) and gives 8 with \(c = c_0\). ◻

Proof of Theorem 7. Fix \(R\ge1\). For all sufficiently small \(\alpha\), we have \(R\alpha^{1/m}\le1\), and Lemma 6 gives \[\begin{align} \mathbb{P}(\left\lVert Y_\alpha \right\rVert>R) &= \mathbb{P}(\left\lVert X_\alpha \right\rVert>R\alpha^{1/m})\\ &\le \mathbb{P}(R\alpha^{1/m}<\left\lVert X_\alpha \right\rVert\le1) +\mathbb{P}(\left\lVert X_\alpha \right\rVert>1)\\ &\le \frac{ \mathbb{E}[\left\lVert X_\alpha \right\rVert^m \mathbf{1}_{\{\left\lVert X_\alpha \right\rVert\le1\}}] }{R^m\alpha} +\mathbb{P}(\left\lVert X_\alpha \right\rVert>1)\\ &\le CR^{-m}+C\alpha, \end{align}\] where the last line uses 36 and 37 . Hence \[\limsup_{\alpha\downarrow0} \mathbb{P}(\left\lVert Y_\alpha \right\rVert>R) \le CR^{-m}.\] Letting \(R\to\infty\) shows that the family \(\{Y_\alpha\}_{\alpha>0}\) is tight as \(\alpha\downarrow0\).

We next identify all possible subsequential limits. Let \((\alpha_k)_{k\ge1}\) be an arbitrary sequence satisfying \(\alpha_k\downarrow0\). By tightness, there exist integers \(1\le k_1<k_2<\cdots\) and a probability law \(\nu\) on \(\mathbb{R}^d\) such that \(Y_{\alpha_{k_j}}\Rightarrow\nu\) as \(j\to\infty\). Since \(\alpha_{k_j}\downarrow0\), we may apply Lemma 8 to the sequence \((\alpha_{k_j})_{j\ge1}\). It follows that \(\nu\) satisfies the stationary generator equation 39 . Therefore, Lemma 9 implies that \(\nu=\nu_\infty,\) where \(\nu_\infty\) is the unique invariant law of the limiting diffusion 12 . Thus, \(Y_{\alpha_{k_j}}\Rightarrow\nu_\infty.\)

The same argument applies to every subsequence of the original sequence \((\alpha_k)_{k\ge1}\): every such subsequence has a further subsequence along which the corresponding random variables converge weakly to \(\nu_\infty\). As a result, the entire sequence satisfies \(Y_{\alpha_k}\Rightarrow\nu_\infty\) as \(k\to\infty\). Because the sequence \((\alpha_k)_{k\ge1}\) was arbitrary, we conclude that \(Y_\alpha=\alpha^{-1/m}X_\alpha \Rightarrow Y_\infty\) as \(\alpha\downarrow0\) and \(Y_\infty\sim\nu_\infty\). ◻

Proof of Proposition 11. The assumptions imply that \(P\) maps \(\mathcal{P}_1(E,d)\) into itself, since \(\mathbb{E}d(F_U(z),z_\star)\le r d(z,z_\star)+\mathbb{E}d(F_U(z_\star),z_\star)\). The one-step estimate ?? iterates under the synchronous coupling and gives ?? . Let \(\mu_0=\delta_{z_\star}\) and \(\mu_n=\mu_0P^n\). By ?? , \(W_1^d(\mu_1,\mu_0)<\infty\), and hence \(W_1^d(\mu_{k+1},\mu_k)\le r^kW_1^d(\mu_1,\mu_0)\). Thus \((\mu_n)\) is Cauchy in \((\mathcal{P}_1(E,d),W_1^d)\). Let its limit be \(\pi\). Applying the contraction to \(\mu_n\) and \(\pi\) shows \(\mu_{n+1}\to\pi P\), while \(\mu_{n+1}\to\pi\), so \(\pi P=\pi\).

For fixed \(z\in E\), taking \(\mu=\delta_z\) and \(\nu=\pi\) in ?? gives \(W_1^d(\delta_zP^n,\pi)\le r^nW_1^d(\delta_z,\pi)\to0\); the right-hand side is finite because \(\pi\in\mathcal{P}_1(E,d)\). Consequently, every bounded \(d\)-Lipschitz \(\varphi\) satisfies \(|P^n\varphi(z)-\pi(\varphi)|\le \mathop{\mathrm{Lip}}_d(\varphi)W_1^d(\delta_zP^n,\pi)\to0\).

Let \(\widetilde{\pi}\) be any Borel invariant probability law. By the pointwise convergence, invariance, and dominated convergence, \(\widetilde{\pi}(\varphi)=\lim_n\widetilde{\pi}(P^n\varphi)=\pi(\varphi)\). Bounded Lipschitz functions determine Borel probability measures on the Polish space \((E,d)\), so \(\widetilde{\pi}=\pi\). ◻

Proof of Lemma 1. Put \(A_{\rm s}(x,\xi):=(A(x,\xi)+A(x,\xi)^\top)/2\). Fix \(\xi\in\Xi\). Applying Cauchy–Schwarz to 6 gives \(\left\lVert \mathsf G(x,\xi)-\mathsf G(y,\xi) \right\rVert \le L_{\mathsf G}\left\lVert x-y \right\rVert\) for \(x,y\in\mathbb{R}^d\); the conclusion is immediate when the left-hand difference vanishes and otherwise follows by division by its norm. Since \(A(x,\xi)=\nabla_x\mathsf G(x,\xi)\), this proves the first bound in 25 .

Next fix \(x,v\in\mathbb{R}^d\). Apply 6 to \(x+tv\) and \(x\), divide by \(t^2\), and let \(t\to0\). Then \[\left\lVert A(x,\xi)v \right\rVert^2 \le L_{\mathsf G}\left\langle v,A(x,\xi)v\right\rangle =L_{\mathsf G}\left\langle v,A_{\rm s}(x,\xi)v\right\rangle, \qquad 0\preceq A_{\rm s}(x,\xi)\preceq L_{\mathsf G}I_d.\] Here the first inequality implies \(A_{\rm s}(x,\xi)\succeq0\), and the second also uses \(\left\lVert A_{\rm s} \right\rVert_{\mathrm{op}}\le\left\lVert A \right\rVert_{\mathrm{op}}\).

Since \(\mathsf G=h+g\), 7 implies \[\left\langle x-y,\mathbb{E}[\mathsf G(x,\Phi(\xi,U_1))- \mathsf G(y,\Phi(\xi,U_1))]\right\rangle \ge(1-\theta)\left\langle x-y,h(x)-h(y)\right\rangle.\] Apply this inequality to \(x+tv\) and \(x\), divide by \(t^2\), and let \(t\to0\). For \(t\ne0\), the preceding Lipschitz bound gives \(\left\lVert [\mathsf G(x+tv,\Phi(\xi,U_1))- \mathsf G(x,\Phi(\xi,U_1))]/t \right\rVert\le L_{\mathsf G}\left\lVert v \right\rVert\), so dominated convergence yields \[\mathbb{E}[A_{\rm s}(x,\Phi(\xi,U_1))]\succeq(1-\theta)\nabla^2H(x), \qquad 0\preceq\nabla^2H(x)\preceq\frac{L_{\mathsf G}}{1-\theta}I_d.\] Since \(\nabla_xg(x,\xi)=A(x,\xi)-\nabla^2H(x)\), the triangle inequality proves the second bound in 25 . Integrating the two Jacobian bounds along line segments yields the stated global Lipschitz estimates. Finally, 26 is the Baillon–Haddad inequality for the convex \(L_{\mathsf G}/(1-\theta)\)-smooth function \(H\), applied with \(h(0)=0\).

It remains to prove the two contraction bounds. Write \(A=A(x,\xi)\) and \(A_{\rm s}=A_{\rm s}(x,\xi)\). Since \(\alpha L_{\mathsf G}\le1\), \[\begin{align} \left\lVert (I_d-\alpha A)v \right\rVert^2 &=\left\lVert v \right\rVert^2-2\alpha\left\langle v,A_{\rm s}v\right\rangle+\alpha^2\left\lVert Av \right\rVert^2\\ &\le\left\lVert v \right\rVert^2-(2\alpha-\alpha^2L_{\mathsf G})\left\langle v,A_{\rm s}v\right\rangle \le\left\lVert v \right\rVert^2-\alpha\left\langle v,A_{\rm s}v\right\rangle\le\left\lVert v \right\rVert^2. \end{align}\] This proves 27 . If \(v\ne0\), then \[\begin{align} 0&\le\alpha\frac{\left\langle v,A_{\rm s}v\right\rangle}{\left\lVert v \right\rVert^2} \le\alpha\left\lVert A_{\rm s} \right\rVert_{\mathrm{op}}\le\alpha\left\lVert A \right\rVert_{\mathrm{op}}\le1,\\ \left\lVert (I_d-\alpha A)v \right\rVert &\le\left(1-\frac{\alpha}{2} \frac{\left\langle v,A_{\rm s}v\right\rangle}{\left\lVert v \right\rVert^2}\right)\left\lVert v \right\rVert, \end{align}\] where the second inequality uses \(\sqrt{1-t}\le1-t/2\). Replacing \(\xi\) by \(\Phi(\xi,U_1)\), taking expectation with \(v\) fixed, and using the averaged matrix inequality above proves 28 . ◻

Proof. This is the second inequality in Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}). ◻

Proof. Assumption 2([[itm:N6]](#itm:N6){reference-type=“ref” reference=“itm:N6”}) gives \(\inf_{\xi\in\Xi}\mathbb{P}(G_\xi\ge\varepsilon_g)\ge p_g\). For \(0<\alpha\le\varepsilon_g^{-1}\), \(\min\{\alpha G_\xi,1\}\ge \alpha\varepsilon_g\mathbf{1}_{\{G_\xi\ge\varepsilon_g\}}\), and therefore \(\bar a_\alpha\ge\alpha\varepsilon_gp_g\), proving 40 . Similarly, \(\min\{(\alpha G_\xi)^p,1\}\ge (\alpha\varepsilon_g)^p\mathbf{1}_{\{G_\xi\ge\varepsilon_g\}}\), so \(\bar a_{\alpha,p}\ge\alpha^p\varepsilon_g^pp_g\).

It remains to compare \(\alpha^p\) with \(\bar a_\alpha^p\). Let \(\xi_\star\) be the reference point in Assumption 2([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}). If \(\beta=2\), then \(\mathbb{E}G_{\xi_\star}<\infty\) by ([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}); if \(\beta<2\), Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) at \(x=0\) and \(\xi=\xi_\star\) gives the same conclusion. Thus \(\bar a_\alpha\le\mathbb{E}\min(\alpha G_{\xi_\star},1) \le\alpha\mathbb{E}G_{\xi_\star}\). Combining this with the previous lower bound on \(\bar a_{\alpha,p}\) proves 41 ; the same inequality gives 42 . Together with 40 , this proves \(\bar a_\alpha=\Theta(\alpha)\). ◻

Proof. For (i), set \(X=\langle u,Z\rangle\), so \(\mathbb{E}X=0\) and \(|X|\le\left\lVert Z \right\rVert\). For \(|t|\le\lambda/2\), \[\begin{align} e^{tX}&\le 1+tX+\frac{t^2\left\lVert Z \right\rVert^2}{2}e^{|t|\left\lVert Z \right\rVert},& r^2e^{(\lambda/2)r}&\le \frac{16}{\lambda^2}e^{\lambda r},\\ \mathbb{E}e^{tX}&\le 1+\frac{8M}{\lambda^2}t^2 \le \exp\!\left(\frac{8M}{\lambda^2}t^2\right). \end{align}\] For (ii), \(e^{tY}-1\le tYe^{tY}\le tYe^{(\lambda/2)Y}\) and \(re^{(\lambda/2)r}\le(2/\lambda)e^{\lambda r}\), so \(\mathbb{E}e^{tY}\le1+(2M/\lambda)t\). For (iii), apply Taylor’s theorem to \(\psi(w)=\left\lVert w \right\rVert^{2-\beta}\) on \(\mathbb{R}^d\setminus\{0\}\). Along \(y-sz\), \(s\in[0,1]\), the condition \(\left\lVert z \right\rVert\le\left\lVert y \right\rVert/2\) gives \(\left\lVert y-sz \right\rVert\ge\left\lVert y \right\rVert/2\); since \(\left\lVert \nabla^2\psi(w) \right\rVert_{\mathrm{op}}\le C_\beta\left\lVert w \right\rVert^{-\beta}\), the claimed inequality follows. ◻

Proof. The case \(\beta=2\) has \(\mathcal{T}_\beta\equiv1\), so 45 is immediate. Suppose throughout the rest of the proof that \(1\le\beta<2\). Since \(h\) is continuous, it is bounded on the compact set \(\{\left\lVert x \right\rVert\le R_H\}\). Also \(1+\left\lVert x \right\rVert^{\beta-1}\) is bounded on this compact set, so Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) implies that, for some \(\lambda_K,M_K>0\), \[\label{eq:compact-g-exp-moment} \sup_{\left\lVert x \right\rVert\le R_H}\sup_{\xi\in\Xi} \mathbb{E}\exp\{\lambda_K\left\lVert g(x,\xi^+) \right\rVert\}\le M_K.\tag{48}\]

First assume \(\left\lVert x \right\rVert\le \frac{R_H}{2}\). If \(\mathcal{T}_\beta(f_{\xi^+}(x))>1\), then \(\left\lVert f_{\xi^+}(x) \right\rVert>\frac{3R_H}{4}\). Since \(\frac{3R_H}{4}>\frac{R_H}{2}\) and \(\left\lVert h(x) \right\rVert\le C\), this implies, after decreasing the stepsize threshold, \(\alpha\left\lVert g(x,\xi^+) \right\rVert\ge c\). Moreover, \[\bigl(\left\lVert f_{\xi^+}(x) \right\rVert^{2-\beta}-(\tfrac{3R_H}{4})^{2-\beta}\bigr)_+ \le C\{1+\alpha\left\lVert g(x,\xi^+) \right\rVert\},\] because \(2-\beta\le1\). Taking \(\kappa\) small enough that \(C\kappa\alpha\le \lambda_K/2\), 48 gives \[\begin{align} \mathbb{E}\bigl[(\mathcal{T}_\beta(f_{\xi^+}(x))-1)_+\bigr] &\le C\mathbb{E}\!\bigl[ e^{C\kappa\alpha\left\lVert g(x,\xi^+) \right\rVert} \mathbf{1}_{\{\left\lVert g(x,\xi^+) \right\rVert\ge c/\alpha\}} \bigr] \\ &\le C e^{-c/\alpha}, \end{align}\] which proves 45 .

It remains to prove 46 . Fix \(\left\lVert x \right\rVert\le R_H\) and set \(G=\left\lVert g(x,\xi^+) \right\rVert\). Split according to \(B_\alpha:=\{G\le \alpha^{-1/2}\}\). On \(B_\alpha\), \(\left\lVert f_{\xi^+}(x)-x \right\rVert\le C\alpha(1+G) \le C\alpha^{1/2},\) so both \(x\) and \(f_{\xi^+}(x)\) lie in a fixed compact enlargement of \(\{\left\lVert y \right\rVert\le R_H\}\). On this enlargement, \(\mathcal{T}_\beta\) has Lipschitz constant at most \(C\kappa\), for \(0<\kappa\le1\). Hence \[\begin{align} \mathbb{E}\bigl[(\mathcal{T}_\beta(f_{\xi^+}(x))-\mathcal{T}_\beta(x))_+\mathbf{1}_{B_\alpha}\bigr] &\le C\kappa\alpha\mathbb{E}(1+G) \le C\kappa\alpha. \end{align}\] On \(B_\alpha^c\), use \(e^a-e^b\le (a-b)_+e^a\) for \(a\ge b\), the bound \(\bigl(\left\lVert f_{\xi^+}(x) \right\rVert^{2-\beta}-(\tfrac{3R_H}{4})^{2-\beta}\bigr)_+\le C\{1+(\alpha G)^{2-\beta}\}\), and 48 . Decreasing \(\alpha\) if necessary, \[\begin{align} &\mathbb{E}\bigl[(\mathcal{T}_\beta(f_{\xi^+}(x))-\mathcal{T}_\beta(x))_+\mathbf{1}_{B_\alpha^c}\bigr] \\ &\qquad\le C\kappa\mathbb{E}\!\bigl[(1+(\alpha G)^{2-\beta}) e^{C\kappa(1+(\alpha G)^{2-\beta})} \mathbf{1}_{\{G>\alpha^{-1/2}\}}\bigr] \le C\kappa e^{-c\alpha^{-1/2}} \le C\kappa\alpha. \end{align}\] Therefore \[\label{eq:compact-U-positive-increment-expanded} \mathbb{E}\bigl[(\mathcal{T}_\beta(f_{\xi^+}(x))-\mathcal{T}_\beta(x))_+\bigr] \le C\kappa\alpha.\tag{49}\] The \(\omega\)-part satisfies \(0\le\omega\le1\), and hence \[\delta_\alpha(\omega(f_{\xi^+}(x))-\omega(x))_+ \le \delta_\alpha\le C\kappa_0\alpha.\] Since \(V_\alpha(x)\ge1\), 46 follows from 49 . Finally, 47 follows from \(V_\alpha(f)\le V_\alpha(x)+(V_\alpha(f)-V_\alpha(x))_+\). ◻

Proof. Let \(r=\left\lVert x \right\rVert\), \(G=g(x,\xi^+)\), and \(\zeta=G-\bar g(x,\xi)\). Then \(\mathbb{E}[\zeta\mid\xi]=0\) and \(f_{\xi^+}(x)=x-\alpha\{h(x)+\bar g(x,\xi)+\zeta\}.\) By Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}), \(\left\lVert \bar g(x,\xi) \right\rVert\le C(1+\left\lVert x \right\rVert^{\beta-1})\), uniformly in \((x,\xi)\). Fix \(\eta_0\in(0,1/2)\), put \[N_x:=\frac{\left\lVert g(x,\xi^+) \right\rVert}{1+\left\lVert x \right\rVert^{\beta-1}}, \qquad E_x:=\{N_x\le\alpha^{-\eta_0}\}.\] On \(E_x\), for sufficiently small \(\alpha\), \[\alpha\left\lVert h(x)+G \right\rVert \le C\alpha^{1-\eta_0}r^{\beta-1} \le \varepsilon_* r, \qquad r\ge R_H,\] with \(\varepsilon_*>0\) chosen so that the segment from \(x\) to \(f_{\xi^+}(x)\) remains in the tail region where the cutoff in \(\mathcal{T}_\beta\) is inactive. Lemma 12(iii), applied to \(z=\alpha\{h(x)+G\}\), gives on \(E_x\) \[\begin{align} \left\lVert f_{\xi^+}(x) \right\rVert^{2-\beta} &\le r^{2-\beta} -(2-\beta)\alpha r^{-\beta}\left\langle x,h(x)+\bar g(x,\xi)\right\rangle -(2-\beta)\alpha r^{-\beta}\left\langle x,\zeta\right\rangle +C\alpha^{2-2\eta_0}. \end{align}\] By the tail dissipativity condition ([[itm:N4]](#itm:N4){reference-type=“ref” reference=“itm:N4”}), \(r^{-\beta}\left\langle x,h(x)+\bar g(x,\xi)\right\rangle\ge c_{\rm diss}\). Therefore, conditionally on \(\xi\), \[\mathcal{T}_\beta(f_{\xi^+}(x))\mathbf{1}_{E_x} \le \mathcal{T}_\beta(x) \exp\{-\kappa(2-\beta)c_{\rm diss}\alpha -\kappa(2-\beta)\alpha r^{-\beta}\left\langle x,\zeta\right\rangle +C\alpha^{2-2\eta_0}\}.\] Writing \(e=x/r\), the centered term is \[r^{-\beta}\left\langle x,\zeta\right\rangle =r^{1-\beta}(1+\left\lVert x \right\rVert^{\beta-1}) \left\langle e, \frac{g(x,\xi^+)}{1+\left\lVert x \right\rVert^{\beta-1}} -\mathbb{E}\!\left[\frac{g(x,\xi^+)}{1+\left\lVert x \right\rVert^{\beta-1}}\,\middle|\,\xi\right]\right\rangle.\] Since \(r^{1-\beta}(1+\left\lVert x \right\rVert^{\beta-1})\) is uniformly bounded for \(r\ge R_H\), the coefficient of the centered normalized noise is \(O(\alpha)\). The conditional exponential moment in ([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) also gives a uniform exponential moment for this centered normalized noise. Lemma 12(i) therefore implies \[\mathbb{E}\left[ \exp\{-\kappa(2-\beta)\alpha r^{-\beta}\left\langle x,\zeta\right\rangle\}\mid \xi \right] \le e^{C\alpha^2}.\] Thus \[\label{eq:tail-good-event-bound} \mathbb{E}[\mathcal{T}_\beta(f_{\xi^+}(x))\mathbf{1}_{E_x}] \le e^{-c\alpha+C\alpha^{2-2\eta_0}}\mathcal{T}_\beta(x).\tag{51}\]

It remains to control \(E_x^c\). Because \(2-\beta\in(0,1]\) and \(s\mapsto s^{2-\beta}\) is concave, \[\begin{align} \left\lVert x-\alpha(h(x)+G) \right\rVert^{2-\beta}-r^{2-\beta} &\le (r+\alpha\left\lVert h(x)+G \right\rVert)^{2-\beta}-r^{2-\beta} \\ &\le (2-\beta)r^{1-\beta}\alpha\left\lVert h(x)+G \right\rVert. \end{align}\] For \(r\ge R_H\), Assumption 1([[itm:H3b]](#itm:H3b){reference-type=“ref” reference=“itm:H3b”}) and the definition of \(N_x\) give \(\left\lVert h(x)+G \right\rVert\le C r^{\beta-1}(1+N_x).\) The powers of \(r\) therefore cancel, and we obtain the deterministic bound \(\left\lVert f_{\xi^+}(x) \right\rVert^{2-\beta}-r^{2-\beta} \le C\alpha(1+N_x).\) Consequently, \[\mathcal{T}_\beta(f_{\xi^+}(x)) \le \mathcal{T}_\beta(x)\exp\{C\kappa\alpha(1+N_x)\}.\] Choosing \(\alpha\) so that \(C\kappa\alpha\le\lambda_0/2\), Assumption ([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) yields \[\label{eq:tail-bad-event-bound} \mathbb{E}[\mathcal{T}_\beta(f_{\xi^+}(x))\mathbf{1}_{E_x^c}] \le Ce^{-c\alpha^{-\eta_0}}\mathcal{T}_\beta(x).\tag{52}\] Combining 51 and 52 , using \(2-2\eta_0>1\) and the fact that \(e^{-c\alpha^{-\eta_0}}=o(\alpha)\), gives 50 after reducing the stepsize threshold. ◻

Proof. For \(\left\lVert x \right\rVert\le R_H\), 53 is 46 . Suppose \(\left\lVert x \right\rVert\ge R_H\). If \(\beta=2\), then \(\mathcal{T}_\beta\equiv1\) and \(\omega(x)=0\), so \[\bigl(V_\alpha(f_{\Phi(\xi,U_1)}(x))-V_\alpha(x)\bigr)_+ \le\delta_\alpha\le C\alpha V_\alpha(x).\] If \(1\le\beta<2\), use the normalized noise \(N_x:=\frac{\left\lVert g(x,\Phi(\xi,U_1)) \right\rVert}{1+\left\lVert x \right\rVert^{\beta-1}}.\) The deterministic tail comparison used in the proof of Lemma 14 gives \[\mathcal{T}_\beta(f_{\Phi(\xi,U_1)}(x)) \le\mathcal{T}_\beta(x)e^{C\kappa\alpha(1+N_x)}.\] Since \(e^t-1\le te^t\) for \(t\ge0\), Assumption 2 ([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) yields, uniformly in \(x,\xi\), \[\begin{align} \mathbb{E}\bigl[ (\mathcal{T}_\beta(f_{\Phi(\xi,U_1)}(x)) -\mathcal{T}_\beta(x))_+\bigr] &\le C\alpha\mathcal{T}_\beta(x) \mathbb{E}[(1+N_x)e^{C\kappa\alpha(1+N_x)}]\\ &\le C\alpha V_\alpha(x). \end{align}\] The positive increment of \(\delta_\alpha\omega\) is at most \(\delta_\alpha\le C\alpha V_\alpha(x)\), which proves 53 . Finally, \(0\le L_\Phi\le1\) and \(\mathbb{E}L_\Phi(U_1)=\rho_\Xi\) give \[\begin{align} \mathbb{E}[L_\Phi(U_1)V_\alpha(f_{\Phi(\xi,U_1)}(x))] &\le \rho_\Xi V_\alpha(x) +\mathbb{E}[(V_\alpha(f_{\Phi(\xi,U_1)}(x))-V_\alpha(x))_+], \end{align}\] and 54 follows. ◻

Proof. Fix \(x\) with \(r=\left\lVert x \right\rVert\le \frac{R_H}{2}\), and fix \(e\in\mathbb{S}^{d-1}\). Since \(\frac{3R_H}{4}>\frac{R_H}{2}\), \(\mathcal{T}_\beta(x)=1\). Put \(\mathcal{R}(u)=\mathcal{T}_\beta(u)-1\) when \(\beta<2\), and \(\mathcal{R}\equiv0\) when \(\beta=2\). Then \[V_\alpha(u)=1+\delta_\alpha\omega(u)+\mathcal{R}(u), \qquad V_\alpha(x)=1+\delta_\alpha\omega(x).\] Using 43 , \[\label{eq:fixed-core-pre} \begin{align} (\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e)-V_\alpha(x) &\le \mathbb{E}[\mathcal{D}_{\xi,\alpha}(x,e)]-1 +\delta_\alpha\{\mathbb{E}\omega(f_{\xi^+}(x))-\omega(x)\} \\ &\qquad +\mathbb{E}\mathcal{R}(f_{\xi^+}(x)). \end{align}\tag{55}\]

Put \(a_{\rm in}:=(1-\theta)c_{\rm in}/2\). If \(m=2\), Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) gives \(\left\langle e,\nabla^2H(x)e\right\rangle\ge c_{\rm in}\) for \(\left\lVert x \right\rVert\le \frac{R_H}{2}\). Hence \[\mathbb{E}\mathcal{D}_{\xi,\alpha}(x,e) \le 1-a_{\rm in}\alpha.\] Since \(\omega\equiv0\) when \(m=2\), 55 and 45 give the claim after absorbing the exponentially small term into \(\alpha\).

Assume now \(m>2\), and put \(s_m:=(m-2)\wedge1\). Then \[\omega(y) = \left(1-\left(\frac{2\left\lVert y \right\rVert}{R_H}\right)^{s_m}\right)_+, \qquad y\in\mathbb{R}^d.\] By Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) and 44 , \(\mathbb{E}\mathcal{D}_{\xi,\alpha}(x,e) \le 1-a_{\rm in}\alpha r^{m-2}.\) We claim that \[\label{eq:fixed-core-omega-drop-new} \mathbb{E}\omega(f_{\xi^+}(x))-\omega(x) \le -c_\omega\mathbb{E}\min\{(\alpha\left\lVert g(0,\xi^+) \right\rVert)^{s_m},1\} +C_\omega r^{s_m}.\tag{56}\] Indeed, write \[f_{\xi^+}(x)=-\alpha g(0,\xi^+)+v, \qquad v=x-\alpha h(x)-\alpha\{g(x,\xi^+)-g(0,\xi^+)\}.\] Lemma 10 gives \(\left\lVert h \right\rVert(x)\le Cr^{m-1}\) for \(\left\lVert x \right\rVert\le \frac{R_H}{2}\), while 25 gives \(\left\lVert g(x,\xi^+)-g(0,\xi^+) \right\rVert \le (2-\theta)L_{\mathsf G}r/(1-\theta)\). Hence, after reducing the stepsize threshold, \(\left\lVert v \right\rVert\le Cr\). The function \(y\mapsto\omega(y)\) is \(s_m\)-Hölder, because \((1-s^{s_m})_+\) is \(s_m\)-Hölder on \([0,\infty)\). Therefore \[\omega(f_{\xi^+}(x)) \le \omega(-\alpha g(0,\xi^+))+C\left\lVert v \right\rVert^{s_m} \le \left(1-\left(\frac{2\alpha\left\lVert g(0,\xi^+) \right\rVert}{R_H}\right)^{s_m}\right)_+ +Cr^{s_m}.\] Since \[\left(1-\left(\frac{2z}{R_H}\right)^{s_m}\right)_+ \le 1-c\min\{z^{s_m},1\}, \qquad z\ge0,\] and \(\omega(x)=1-(\frac{2r}{R_H})^{s_m}\) for \(r\le \frac{R_H}{2}\), 56 follows.

Combining 55 , 56 , 45 , and Lemma 11, we get \[\label{eq:core-combined-before-absorption} (\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e)-V_\alpha(x) \le -a_{\rm in}\alpha r^{m-2} -c\delta_\alpha\bar a_{\alpha,s_m} +C\delta_\alpha r^{s_m} +Ce^{-c/\alpha}.\tag{57}\] If \(2<m\le3\), then \(s_m=m-2\) and \(\delta_\alpha=\kappa_0\alpha\). Using \(\bar a_{\alpha,s_m}\ge c\bar a_\alpha^{m-2}\), 57 becomes \[(\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e)-V_\alpha(x) \le -(a_{\rm in}-C\kappa_0)\alpha r^{m-2} -c\kappa_0\alpha\bar a_\alpha^{m-2} +Ce^{-c/\alpha}.\] Choose \(\kappa_0\) so small that \(a_{\rm in}-C\kappa_0\ge a_{\rm in}/2\). Then \(\bar a_\alpha\ge c_1\alpha\), and the exponentially small term is negligible compared with \(\alpha^{m-1}\). Hence \[\label{eq:core-additive-drop-flat-case} (\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e)-V_\alpha(x) \le -c\alpha^{m-1}.\tag{58}\]

If \(m>3\), then \(s_m=1\), \(\delta_\alpha=\kappa_0\alpha^{m-2}\), and \(\bar a_{\alpha,1}=\bar a_\alpha\). Factoring out \(\alpha\) in 57 gives \[\begin{align} (\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e)-V_\alpha(x) \le \alpha\bigl[ -a_{\rm in}r^{m-2} -c\kappa_0\alpha^{m-3}\bar a_\alpha +C\kappa_0\alpha^{m-3}r \bigr] +Ce^{-c/\alpha}. \end{align}\] Put \(q:=(m-2)/(m-3)>1\). Young’s inequality yields \[C\kappa_0\alpha^{m-3}r \le \frac{a_{\rm in}}{2}r^{m-2} +C_*\kappa_0^q\alpha^{m-2}.\] Since \(\bar a_\alpha\ge c_1\alpha\), the negative noise term inside the brackets is at most \(-cc_1\kappa_0\alpha^{m-2}\). Choose \(\kappa_0\) so small that \(C_*\kappa_0^{q-1}\le cc_1/2\). After restoring the outer factor \(\alpha\), the Young remainder is absorbed by the negative noise term. Finally, \(e^{-c/\alpha}=o(\alpha^{m-1})\), so 58 also holds for \(m>3\).

On the near-minimizer region, \(1\le V_\alpha(x)\le C\) uniformly in \(\alpha\). Since 58 gives an additive decrease of at least \(c\alpha^{m-1}\), it implies \[(\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e) \le (1-c_{\rm core}\alpha^{m-1})V_\alpha(x),\] after decreasing \(c_{\rm core}\). Taking the supremum over \(e\) completes the proof. ◻

Proof. Fix \(e\in\mathbb{S}^{d-1}\). By Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}), for \(\frac{R_H}{2}\le\left\lVert x \right\rVert\le R_H\), \(\left\langle e,\nabla^2H(x)e\right\rangle \ge c_{\rm in}\left(\frac{R_H}{2}\right)^{m-2}.\) Together with 44 , this gives \(\mathbb{E}\mathcal{D}_{\xi,\alpha}(x,e) \le 1-c\alpha\) for some \(c>0\). Since \(\mathcal{D}_{\xi,\alpha}(x,e)\le1\), \[\begin{align} (\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e) &\le V_\alpha(x)\mathbb{E}\mathcal{D}_{\xi,\alpha}(x,e) +\mathbb{E}[(V_\alpha(f_{\xi^+}(x))-V_\alpha(x))_+] \\ &\le (1-c\alpha+C(\kappa+\kappa_0)\alpha)V_\alpha(x), \end{align}\] where the last line uses 46 . Choose \(\kappa\) and \(\kappa_0\) sufficiently small to obtain a factor \(1-c\alpha\). Since \(0<\alpha\le1\) and \(m\ge2\), \(\alpha\ge\alpha^{m-1}\), which gives the stated estimate after decreasing the constant. ◻

Proof. First suppose \(\beta=2\). Then \(\mathcal{T}_\beta\equiv1\), and \(\omega(x)=0\) for \(\left\lVert x \right\rVert\ge R_H\), so \(V_\alpha(x)=1\). By Assumption 1([[itm:H3a]](#itm:H3a){reference-type=“ref” reference=“itm:H3a”}) and 44 , \(\mathbb{E}\mathcal{D}_{\xi,\alpha}(x,e) \le 1-c\alpha\) for some \(c>0\). Since \(0\le\omega\le1\), \((\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e) \le 1-c\alpha+\delta_\alpha.\) Since \(\delta_\alpha\le\kappa_0\alpha\), choosing \(\kappa_0\) small gives a factor \(1-c\alpha\).

Now suppose \(1\le\beta<2\). On \(\left\lVert x \right\rVert\ge R_H\), \(\omega(x)=0\), so \(V_\alpha(x)=\mathcal{T}_\beta(x)\). By 43 and Lemma 14, \[(\mathcal{K}_{\xi,\alpha}V_\alpha)(x;e) \le \mathbb{E}\mathcal{T}_\beta(f_{\xi^+}(x))+\delta_\alpha \le (1-c\alpha)\mathcal{T}_\beta(x)+\delta_\alpha.\] Since \(\mathcal{T}_\beta(x)\ge \exp\{\kappa(R_H^{2-\beta}-(\tfrac{3R_H}{4})^{2-\beta})\}>1\) on the far region and \(\delta_\alpha\le C\kappa_0\alpha\), choosing \(\kappa_0\) small absorbs the last term and again gives a factor \(1-c\alpha\). In both tail cases, \(\alpha\ge\alpha^{m-1}\); taking the supremum over \(e\) and decreasing the constant proves the stated estimate. ◻

Proof of Lemma 2. The estimates in Lemmas 16, 17, and 18 give 29 after taking \(c_0:=\min\{c_{\rm core},c_{\rm br},c_{\rm far}\}\) and reducing the common stepsize threshold.

It remains to prove the growth estimate 30 . On \(\left\lVert x \right\rVert\le R_H\), it is exactly 47 . If \(\left\lVert x \right\rVert\ge R_H\) and \(\beta=2\), then \(V_\alpha(x)=1\) and \(V_\alpha(f_{\xi^+}(x))\le1+\delta_\alpha\le(1+C\alpha)V_\alpha(x)\). If \(\left\lVert x \right\rVert\ge R_H\) and \(1\le\beta<2\), then Lemma 14 and \(0\le\omega\le1\) give \[\mathbb{E}V_\alpha(f_{\xi^+}(x)) \le (1-c\alpha)\mathcal{T}_\beta(x)+\delta_\alpha \le (1+C\alpha)V_\alpha(x),\] again after reducing the threshold and increasing \(C\). This proves Lemma 2. ◻

Proof of Lemma 3. Let \(C_V\) be as in Lemma 2. Let \(C_+\) be as in Lemma 15. Choose \(\alpha_0\in(0,1]\) small enough that the conclusions of Lemmas 2 and 15 hold for \(0<\alpha\le\alpha_0\) and \[\label{eq:augmented-xi-dominance-new} \rho_\Xi+C_+\alpha +\alpha^2L_{g,\Phi}(1+C_V\alpha) \le 1-c_0\alpha^{m-1}, \qquad 0<\alpha\le\alpha_0.\tag{59}\] This is possible because \(\rho_\Xi<1\) and \(\alpha^{m-1}\to0\).

Fix \(z=(x,\xi)\), \(z'=(y,\eta)\), and an absolutely continuous path \(\gamma(t)=(x(t),\xi(t))\) from \(z\) to \(z'\). For fixed \(u\), write \[\xi^+(t):=\Phi(\xi(t),u), \qquad X^+(t):=x(t)-\alpha\{h(x(t))+g(x(t),\xi^+(t))\}.\] The curve \(F_u\circ\gamma=(X^+(\cdot),\xi^+(\cdot))\) is absolutely continuous. Indeed, \(\Phi(\cdot,u)\) is Lipschitz by Assumption 2([[itm:N1]](#itm:N1){reference-type=“ref” reference=“itm:N1”}), the composed dependence \(g(x,\Phi(\xi,u))\) is Lipschitz in \(\xi\) by Assumption 2([[itm:N2]](#itm:N2){reference-type=“ref” reference=“itm:N2”}), and the map \(x\mapsto h(x)+g(x,\zeta)\) has derivative \(A(x,\zeta)\), which is uniformly bounded by 25 .

Let \(\operatorname{md}_\alpha\) denote the metric derivative with respect to the augmented base norm \(\vert\cdot\vert_\alpha\). For a.e. \(t\), \[\label{eq:augmented-metric-derivative-bound} \begin{align} \operatorname{md}_\alpha(F_u\circ\gamma)(t) &\le \left\lVert (I_d-\alpha A(x(t),\xi^+(t)))\dot{x}(t) \right\rVert \\ &\qquad +(L_\Phi(u)+\alpha^2L_{g,\Phi})\alpha^{-1}\left\lVert \dot{\xi}(t) \right\rVert. \end{align}\tag{60}\] To prove 60 , compare \(F_u(\gamma(s))\) and \(F_u(\gamma(t))\). In the \(x\)-coordinate, first freeze the noise argument at \(\xi^+(t)\); division by \(|s-t|\) and passage to the a.e.differentiability point gives the derivative \((I_d-\alpha A(x(t),\xi^+(t)))\dot{x}(t)\). The remaining change in the \(x\)-coordinate is bounded by \(\alpha L_{g,\Phi}\left\lVert \xi(s)-\xi(t) \right\rVert\). In the noise coordinate, Assumption 2([[itm:N1]](#itm:N1){reference-type=“ref” reference=“itm:N1”}) gives \(\left\lVert \Phi(\xi(s),u)-\Phi(\xi(t),u) \right\rVert \le L_\Phi(u)\left\lVert \xi(s)-\xi(t) \right\rVert\). Dividing by \(|s-t|\) and using the \(\alpha^{-1}\)-weight in the augmented norm gives 60 .

If \(\dot{x}(t)\ne0\), set \(e(t)=\dot{x}(t)/\left\lVert \dot{x}(t) \right\rVert\); otherwise choose any measurable unit vector \(e(t)\). By the definition of induced length, 60 , Lemmas 2 and 15, and 59 , \[\begin{align} &\mathbb{E}\, d_{V_\alpha,\alpha}\bigl(F_{U_1}(z),F_{U_1}(z')\bigr) \\ &\quad\le \int_0^1 \mathbb{E}\!\left[ V_\alpha(X^+(t)) \left\lVert (I_d-\alpha A(x(t),\xi^+(t)))\dot{x}(t) \right\rVert \right]dt \\ &\qquad + \int_0^1 \mathbb{E}\!\left[ (L_\Phi(U_1)+\alpha^2L_{g,\Phi})V_\alpha(X^+(t)) \right] \alpha^{-1}\left\lVert \dot{\xi}(t) \right\rVert\,dt \\ &\quad\le (1-c_0\alpha^{m-1}) \int_0^1 V_\alpha(x(t)) \left(\left\lVert \dot{x}(t) \right\rVert+\alpha^{-1}\left\lVert \dot{\xi}(t) \right\rVert\right)dt. \end{align}\] Taking the infimum over all admissible paths \(\gamma\) gives 24 . ◻

Proof of Lemma 4. Let \(z_\star=(0,\xi_\star)\), where \(\xi_\star\) is the point from Assumption 2([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}). Write \[\xi_\star^+:=\Phi(\xi_\star,U_1), \qquad x_{\rm ref}^+:=-\alpha g(0,\xi_\star^+).\] Move from \(z_\star=(0,\xi_\star)\) to \((0,\xi_\star^+)\), and then from \((0,\xi_\star^+)\) to \((x_{\rm ref}^+,\xi_\star^+)\). Since \(V_\alpha\) depends only on the \(x\)-coordinate, \[d_{V_\alpha,\alpha}\bigl(F_{U_1}(z_\star),z_\star\bigr) \le V_\alpha(0)\alpha^{-1}\left\lVert \xi_\star^+-\xi_\star \right\rVert + d_{V_\alpha}^{X}(x_{\rm ref}^+,0).\] The first term is integrable by Assumption 2([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}).

If \(\beta=2\), then \(\mathcal{T}_\beta\equiv1\) and \(V_\alpha\le1+\delta_\alpha\). Hence \[d_{V_\alpha}^{X}(x_{\rm ref}^+,0) \le (1+\delta_\alpha)\left\lVert x_{\rm ref}^+ \right\rVert = (1+\delta_\alpha)\alpha\left\lVert g(0,\xi_\star^+) \right\rVert,\] whose expectation is finite by Assumption 2([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}).

If \(\beta<2\), Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) at \(x=0\) and \(\xi=\xi_\star\) gives an exponential moment of \(g(0,\xi_\star^+)\). Along the straight line from \(0\) to \(x_{\rm ref}^+\), \[d_{V_\alpha}^{X}(x_{\rm ref}^+,0) \le \alpha\left\lVert g(0,\xi_\star^+) \right\rVert \left[ 1+\delta_\alpha+ \exp\!\left\{\kappa\bigl(\alpha^{2-\beta}\left\lVert g(0,\xi_\star^+) \right\rVert^{2-\beta}-(\tfrac{3R_H}{4})^{2-\beta}\bigr)_+\right\} \right].\] Since \(2-\beta\in(0,1]\), \(r^{2-\beta}\le1+r\) for \(r\ge0\), we may choose \(\alpha_0\in(0,1]\) such that, for \(0<\alpha\le\alpha_0\), the exponential factor above is dominated by \(C e^{\lambda\left\lVert g(0,\xi_\star^+) \right\rVert}\) for some \(\lambda>0\) below the exponential-moment threshold supplied by ([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}). Thus \(\mathbb{E}d_{V_\alpha}^{X}(x_{\rm ref}^+,0)<\infty.\) This proves 31 . ◻

Proof of Corollary 1. Since \(V_\alpha\ge1\), the augmented induced metric dominates the Euclidean distance in the \(X\)-coordinate: \(d_{V_\alpha,\alpha}\bigl((x,\xi),(y,\eta)\bigr) \ge \left\lVert x-y \right\rVert.\) Hence, for probability laws \(\mu,\nu\) on \(\mathsf Z\), \[W_1(\mu_X,\nu_X)\le W_{V_\alpha,\alpha}(\mu,\nu).\] Applying this with \(\nu=\pi_\alpha\), and then using Theorem 4, gives \[W_1\bigl((\mu P_\alpha^n)_X,(\pi_\alpha)_X\bigr) \le W_{V_\alpha,\alpha}(\mu P_\alpha^n,\pi_\alpha) \le (1-c\alpha^{m-1})^n W_{V_\alpha,\alpha}(\mu,\pi_\alpha).\] Since \(\pi_\alpha\in\mathcal{P}_1(\mathsf Z,d_{V_\alpha,\alpha})\), the final quantity is finite if and only if \(\mu\in\mathcal{P}_1(\mathsf Z,d_{V_\alpha,\alpha})\).

It remains to characterize \(\mathcal{P}_1(\mathsf Z,d_{V_\alpha,\alpha})\). Since \(\vert(x,\xi)\vert_\alpha = \left\lVert x \right\rVert+\alpha^{-1}\left\lVert \xi \right\rVert,\) and since \(V_\alpha\ge1\), projection of paths gives the lower bounds \[d_{V_\alpha,\alpha}\bigl((x,\xi),(0,\xi_\star)\bigr) \ge d_{V_\alpha}^{X}(x,0), \qquad d_{V_\alpha,\alpha}\bigl((x,\xi),(0,\xi_\star)\bigr) \ge \alpha^{-1}\left\lVert \xi-\xi_\star \right\rVert.\] Conversely, moving first in the \(x\)-coordinate and then in the \(\xi\)-coordinate gives \[d_{V_\alpha,\alpha}\bigl((x,\xi),(0,\xi_\star)\bigr) \le d_{V_\alpha}^{X}(x,0) + V_\alpha(0)\alpha^{-1}\left\lVert \xi-\xi_\star \right\rVert.\] Thus \(d_{V_\alpha,\alpha}((x,\xi),(0,\xi_\star))\)-integrability is equivalent to integrability of \(d_{V_\alpha}^{X}(x,0) + \alpha^{-1}\left\lVert \xi-\xi_\star \right\rVert.\)

If \(\beta=2\), then \(\mathcal{T}_\beta\equiv1\) and \(1\le V_\alpha\le1+\delta_\alpha\). Hence \(\left\lVert x-y \right\rVert \le d_{V_\alpha}^{X}(x,y) \le (1+\delta_\alpha)\left\lVert x-y \right\rVert,\) so \(d_{V_\alpha}^{X}(x,0)\)-integrability is equivalent to \(\left\lVert x \right\rVert\)-integrability.

If \(1\le\beta<2\), let \(R_H\) be as in Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) and define \[\Psi_\kappa(r) := \int_0^r \exp\{\kappa\bigl(s^{2-\beta}-(\tfrac{3R_H}{4})^{2-\beta}\bigr)_+\}\,ds.\] Because the correction \(\delta_\alpha\omega\) is compactly supported and \(0\le\omega\le1\), \[\Psi_\kappa(\left\lVert x \right\rVert) \le d_{V_\alpha}^{X}(x,0) \le \Psi_\kappa(\left\lVert x \right\rVert)+\frac{\delta_\alpha R_H}{2}.\] Moreover, \[\Psi_\kappa(r) \sim \frac{e^{-\kappa(\frac{3R_H}{4})^{2-\beta}}}{\kappa(2-\beta)} r^{\beta-1} e^{\kappa r^{2-\beta}}, \qquad r\to\infty.\] Therefore \(d_{V_\alpha}^{X}(x,0)\)-integrability is equivalent to \[\int (1+\left\lVert x \right\rVert)^{\beta-1} \exp\{\kappa\left\lVert x \right\rVert^{2-\beta}\}\,\mu(dx,d\xi) <\infty.\] Since \(2-\beta\in(0,1]\), for \(r\ge0\), \(r^{2-\beta}\le(1+r)^{2-\beta}\le1+r^{2-\beta},\) and hence \[e^{-\kappa}e^{\kappa r^{2-\beta}} \le \exp\!\left\{\kappa\bigl((1+r)^{2-\beta}-1\bigr)\right\} \le e^{\kappa r^{2-\beta}}.\] Thus the preceding integrability condition is equivalent to integrability of \(\Gamma_\beta(\left\lVert x \right\rVert)\). Combining this with the \(\xi\)-coordinate term gives the displayed description of \(\mathcal{P}_1(\mathsf Z,d_{V_\alpha,\alpha})\). ◻

Proof. For the lower bound near the minimizer, use \(h(x)=\int_0^1\nabla^2H(tx)x\,dt.\) Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) gives \[\left\langle x,h(x)\right\rangle = \int_0^1\left\langle x,\nabla^2H(tx)x\right\rangle\,dt \ge c_{\rm in}\left\lVert x \right\rVert^m\int_0^1t^{m-2}\,dt = \frac{c_{\rm in}}{m-1}\left\lVert x \right\rVert^m.\]

If \(1\le\beta<2\), the tail condition \(\left\langle x,h(x)\right\rangle\ge c_{\rm out}\left\lVert x \right\rVert^\beta\) on \(\{\left\lVert x \right\rVert\ge R_H\}\), together with boundedness on \(\{\left\lVert x \right\rVert<R_H\}\), gives \(\left\lVert x \right\rVert^{\beta} \le C\bigl(1+\left\langle x,h(x)\right\rangle\bigr).\) If \(\beta=2\), then for \(\left\lVert x \right\rVert\ge2R_H\), convexity and minimality at zero imply \(\left\langle x,h(x/2)\right\rangle\ge0.\) Using Assumption 1([[itm:H3a]](#itm:H3a){reference-type=“ref” reference=“itm:H3a”}) on the segment \(\{tx:1/2\le t\le1\}\), \[\begin{align} \left\langle x,h(x)\right\rangle &\ge \left\langle x,h(x)-h(x/2)\right\rangle \\ &= \int_{1/2}^1\left\langle x,\nabla^2H(tx)x\right\rangle\,dt \ge \frac{c_{\rm out}}{2}\left\lVert x \right\rVert^2. \end{align}\] The region \(\{\left\lVert x \right\rVert<2R_H\}\) is absorbed into the additive constant.

Finally, write \(x=ru\), where \(r=\left\lVert x \right\rVert\) and \(u\in\mathbb{S}^{d-1}\). Convexity of \(H\) and minimality at zero imply that \(s\mapsto\left\langle u,h(su)\right\rangle\) is nondecreasing on \([0,\infty)\). For \(r\ge r_0\), \[\left\langle x,h(x)\right\rangle = r\left\langle u,h(ru)\right\rangle \ge r_0\left\langle u,h(r_0u)\right\rangle = \left\langle r_0u,h(r_0u)\right\rangle \ge \frac{c_{\rm in}}{m-1}r_0^m.\] This proves item (iii). For item (iv), if \(\beta=2\), then \((1+\left\lVert x \right\rVert)(1+\left\lVert x \right\rVert^{\beta-1}) \le C(1+\left\lVert x \right\rVert^2),\) and the quadratic-tail estimate in item (ii) gives the result. If \(1\le\beta<2\), then \((1+\left\lVert x \right\rVert)(1+\left\lVert x \right\rVert^{\beta-1}) \le C(1+\left\lVert x \right\rVert^\beta),\) and the subquadratic-tail estimate in item (ii) gives the result. This proves the lemma. ◻

Proof. Let \(\xi_\star\) be as in Assumption 2([[itm:N3]](#itm:N3){reference-type=“ref” reference=“itm:N3”}). If \(\eta\sim\mu\in\mathcal{P}_1(\Xi)\), then Assumption 2([[itm:N1]](#itm:N1){reference-type=“ref” reference=“itm:N1”}) gives \[\begin{align} \mathbb{E}\left\lVert \Phi(\eta,U_1)-\xi_\star \right\rVert &\le \mathbb{E}\left\lVert \Phi(\eta,U_1)-\Phi(\xi_\star,U_1) \right\rVert + \mathbb{E}\left\lVert \Phi(\xi_\star,U_1)-\xi_\star \right\rVert \\ &\le \rho_\Xi\mathbb{E}\left\lVert \eta-\xi_\star \right\rVert + \mathbb{E}\left\lVert \Phi(\xi_\star,U_1)-\xi_\star \right\rVert. \end{align}\] Thus \(Q\) maps \(\mathcal{P}_1(\Xi)\) into itself.

For any coupling \((\eta,\widetilde{\eta})\) of \((\mu,\nu)\), drive both chains by the same innovation \(U_1\). Then \[\mathbb{E}\left\lVert \Phi(\eta,U_1)-\Phi(\widetilde{\eta},U_1) \right\rVert \le \rho_\Xi\mathbb{E}\left\lVert \eta-\widetilde{\eta} \right\rVert.\] Iteration with independent innovations gives the stated synchronous \(n\)-step estimate. Taking the infimum over couplings yields \(W_1(\mu Q,\nu Q)\le\rho_\Xi W_1(\mu,\nu).\) Since \(\Xi\) is closed in a finite-dimensional Euclidean space, \((\mathcal{P}_1(\Xi),W_1)\) is complete. Banach’s fixed-point theorem gives a unique invariant law \(\pi_\Xi\in\mathcal{P}_1(\Xi)\).

To prove uniqueness among all Borel invariant laws, let \(\rho\) be such a law on \(\Xi\). For each fixed \(\xi\), the contraction estimate gives \(\delta_\xi Q^n\to\pi_\Xi\) in \(W_1\). Hence, for every bounded Lipschitz \(\varphi\), \(Q^n\varphi(\xi)\to\pi_\Xi(\varphi).\) By invariance and dominated convergence, \(\rho(\varphi) = \rho(Q^n\varphi) \to \pi_\Xi(\varphi).\) Bounded Lipschitz functions determine Borel probability laws on the Polish space \(\Xi\), so \(\rho=\pi_\Xi\).

Finally, if \((X,\xi)\sim\pi_\alpha\), invariance of the augmented chain implies that \(\Phi(\xi,U_1)\) has the same law as \(\xi\). Thus the \(\Xi\)-marginal of \(\pi_\alpha\) is invariant for \(Q\) and equals \(\pi_\Xi\). ◻

Proof. Let \(\xi_k^\xi\) be the driving chain started from \(\xi\). Let \(\eta_0\sim\pi_\Xi\) and drive the stationary chain \((\eta_k)\) by the same innovations. Since \(f\) is centered under \(\pi_\Xi\), \(Q^kf(\xi) = \mathbb{E}[f(\xi_k^\xi)-f(\eta_k)].\) Therefore \(\left\lvert Q^kf(\xi) \right\rvert \le \min\left\{ 2\left\lVert f \right\rVert_\infty,\, \mathop{\mathrm{Lip}}(f)\rho_\Xi^k\bigl(\left\lVert \xi \right\rVert+\mathbb{E}_{\pi_\Xi}\left\lVert \eta \right\rVert\bigr) \right\}.\) This bound proves absolute convergence. Similarly, synchronous coupling of two chains started from \(\xi\) and \(\eta\) gives \(\left\lvert Q^kf(\xi)-Q^kf(\eta) \right\rvert \le \mathop{\mathrm{Lip}}(f)\rho_\Xi^k\left\lVert \xi-\eta \right\rVert,\) and summing over \(k\) proves the Lipschitz bound for \(u\).

The first bound also implies logarithmic growth: \(\left\lvert u(\xi) \right\rvert \le C_f\bigl(1+\log(1+\left\lVert \xi \right\rVert)\bigr).\) Since \(\pi_\Xi\in\mathcal{P}_1(\Xi)\) and \(\log^2(1+r)\le C(1+r)\), we get \(u\in L^2(\pi_\Xi)\). Finally, absolute convergence allows termwise application of \(Q\), giving \[Qu=\sum_{k=1}^{\infty}Q^kf, \qquad u-Qu=f.\] ◻

Proof of Lemma 5. Let \(\xi_k^\xi\) and \(\xi_k^\eta\) be two copies of the driving chain, started from \(\xi\) and \(\eta\) and coupled through the same innovations. By Assumption 2([[itm:N1]](#itm:N1){reference-type=“ref” reference=“itm:N1”}), \(\mathbb{E}\left\lVert \xi_k^\xi-\xi_k^\eta \right\rVert \le \rho_\Xi^k\left\lVert \xi-\eta \right\rVert.\) For \(k\ge1\), Assumption 2([[itm:N2]](#itm:N2){reference-type=“ref” reference=“itm:N2”}) gives \[\left\lVert Q^kg_x(\xi)-Q^kg_x(\eta) \right\rVert \le L_{g,\Phi}\rho_\Xi^{k-1}\left\lVert \xi-\eta \right\rVert.\] Let \(\eta\sim\pi_\Xi\), independent of the driving innovations. Since \(\pi_\Xi g_x=0\) by 9 , \[Q^kg_x(\xi)=\mathbb{E}[g_x(\xi_k^\xi)-g_x(\xi_k^\eta)], \qquad k\ge1.\] Consequently \[\left\lVert Q^kg_x(\xi) \right\rVert \le L_{g,\Phi}\rho_\Xi^{k-1} \bigl(\left\lVert \xi-\xi_\star \right\rVert +\mathbb{E}_{\pi_\Xi}\left\lVert \eta-\xi_\star \right\rVert\bigr),\] and the series defining \(\chi_x\) converges absolutely for every \(\xi\). Applying \(Q\) termwise gives \(\chi_x-Q\chi_x=Qg_x\). The same estimate gives \[\label{eq:poisson-solution-growth} \left\lVert \chi_x(\xi) \right\rVert \le C_\chi R_\chi(\xi), \qquad R_\chi(\xi):=1+\left\lVert \xi-\xi_\star \right\rVert,\tag{61}\] with \(R_\chi\in L^2(\pi_\Xi)\) by 10 . Centering of \(\chi_x\) follows from invariance of \(\pi_\Xi\): \(\int_\Xi \chi_x(\xi)\,\pi_\Xi(d\xi)=0,\) because \(\int Q^kg_x\,d\pi_\Xi=\int g_x\,d\pi_\Xi=0\) for every \(k\ge1\).

It remains to control the dependence on \(x\). The uniform derivative bound 25 shows that, for some \(L_x>0\), \[\left\lVert g_x(\zeta)-g_y(\zeta) \right\rVert \le L_x\left\lVert x-y \right\rVert, \qquad x,y\in\mathbb{R}^d,\;\zeta\in\Xi.\] Fix \(p\in(0,1)\), and put \(f_{x,y}:=g_x-g_y\). By 9 , \(\pi_\Xi f_{x,y}=0\). On the one hand, \[\left\lVert Q^kf_{x,y}(\xi) \right\rVert \le L_x\left\lVert x-y \right\rVert, \qquad k\ge1.\] On the other hand, comparison with the stationary chain used above and Assumption 2([[itm:N2]](#itm:N2){reference-type=“ref” reference=“itm:N2”}), applied separately to \(g_x\) and \(g_y\), give \[\begin{align} \left\lVert Q^kf_{x,y}(\xi) \right\rVert &\le 2L_{g,\Phi}\rho_\Xi^{k-1} \bigl(\left\lVert \xi-\xi_\star \right\rVert +\mathbb{E}_{\pi_\Xi}\left\lVert \eta-\xi_\star \right\rVert\bigr)\\ &\le C\rho_\Xi^{k-1}R_\chi(\xi). \end{align}\] Since \(\min\{a,b\}\le a^pb^{1-p}\) for \(a,b\ge0\), summing these two bounds over \(k\ge1\) yields the Hölder estimate \[\label{eq:poisson-solution-x-holder} \left\lVert \chi_x(\xi)-\chi_y(\xi) \right\rVert \le C_p\left\lVert x-y \right\rVert^pR_\chi(\xi)^{1-p}, \qquad x,y\in\mathbb{R}^d,\;\xi\in\Xi.\tag{62}\] The martingale-difference property 33 is immediate from 32 .

Finally, let \(f(\xi)=g(0,\xi)\). The geometric estimate used to prove 61 , with \(x=0\), gives \[\left\lVert Q^kf(\xi) \right\rVert \le C\rho_\Xi^{k-1}(1+\left\lVert \xi-\xi_\star \right\rVert), \qquad k\ge1.\] Together with 10 , this implies absolute summability of the lag covariances. Write the stationary chain as \((\xi_n)_{n\in\mathbb{Z}}\), set \(f_n=f(\xi_n)\) and \(\chi_n=\chi_0(\xi_n)\), and define \(D_n:=f_{n+1}+\chi_{n+1}-\chi_n.\) Then \((D_n)\) is a martingale-difference sequence and \(f_{n+1}=D_n+\chi_n-\chi_{n+1}.\) Therefore, for \(N\ge1\), \(\sum_{n=1}^N f_n = \sum_{n=0}^{N-1}D_n+\chi_0-\chi_N.\) Since \(\chi_0\in L^2(\pi_\Xi)\), the telescoping term is negligible after division by \(N\) in the covariance of the partial sums. Hence \[\lim_{N\to\infty} \frac{1}{N}\mathop{\mathrm{Var}}\!\left(\sum_{n=1}^N f_n\right) = \mathbb{E}[D_0D_0^\top] = \Sigma.\] On the other hand, stationarity and absolute summability of the lag covariances give \[\lim_{N\to\infty} \frac{1}{N}\mathop{\mathrm{Var}}\!\left(\sum_{n=1}^N f_n\right) = \Gamma_0+\sum_{k\ge1}(\Gamma_k+\Gamma_k^\top).\] This proves 11 and completes the proof of the lemma. ◻

Proof. By Lemma 20, \(\xi_\alpha^+\sim\pi_\Xi\).

If \(\beta=2\), 25 gives a uniform \(x\)-Lipschitz constant \((2-\theta)L_{\mathsf G}/(1-\theta)\) for \(g\). Hence \[\left\lVert g(x,\zeta) \right\rVert^2 \le 2\left\lVert g(0,\zeta) \right\rVert^2 +2\left(\frac{(2-\theta)L_{\mathsf G}}{1-\theta}\right)^2\left\lVert x \right\rVert^2.\] The stationary second-moment condition 10 gives \(\mathbb{E}\left\lVert g(0,\xi_\alpha^+) \right\rVert^2<\infty,\) and Lemma 19 gives \(\left\lVert x \right\rVert^2\le C(1+\left\langle x,h(x)\right\rangle).\) Taking expectations proves the claim in the quadratic-tail case.

If \(1\le\beta<2\), Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) gives a uniform second moment for \(\frac{g(x,\Phi(\xi,U_1))}{1+\left\lVert x \right\rVert^{\beta-1}}.\) Therefore, conditionally on \((X_\alpha,\xi_\alpha)\), \[\mathbb{E}\bigl[\left\lVert g(X_\alpha,\xi_\alpha^+) \right\rVert^2\mid X_\alpha,\xi_\alpha\bigr] \le C(1+\left\lVert X_\alpha \right\rVert^{\beta-1})^2 \le C(1+\left\lVert X_\alpha \right\rVert^{\beta}).\] Using again Lemma 19, \(\left\lVert x \right\rVert^{\beta}\le C(1+\left\langle x,h(x)\right\rangle),\) and taking expectations proves the claim. ◻

Proof. We first justify the second moments. If \(1\le\beta<2\), the induced first-moment bound in Theorem 4 implies that \(X_\alpha\) has finite moments of every polynomial order. Lemma 22 and the global Lipschitz bound on \(h\) from Lemma 1 then give \(\mathbb{E}\left\lVert h(X_\alpha)+g(X_\alpha,\xi_\alpha^+) \right\rVert^2<\infty.\)

For \(\beta=2\), Lemma 19 gives the global coercivity bound \[\label{eq:h-quadratic-coercive-for-L2} \left\langle x,h(x)\right\rangle\ge c\left\lVert x \right\rVert^2-C, \qquad x\in\mathbb{R}^d.\tag{64}\] Indeed, outside a sufficiently large ball this follows from the quadratic-tail lower Hessian bound, while the remaining bounded region is absorbed into the additive constant.

Start the augmented chain from \(X_0=0\) and \(\xi_0\sim\pi_\Xi\). Then \(\xi_n\sim\pi_\Xi\) for all \(n\). Define the martingale-corrected quadratic Lyapunov function \(J_\alpha^0(x,\xi) :=\left\lVert x \right\rVert^2-2\alpha\left\langle x,\chi_0(\xi)\right\rangle.\) For any law whose \(\Xi\)-marginal is \(\pi_\Xi\), 61 , Young’s inequality, and \(R_\chi\in L^2(\pi_\Xi)\) give \[\label{eq:Lalpha-comparable-L2} \frac{1}{2}\mathbb{E}\left\lVert X \right\rVert^2-C\alpha^2 \le \mathbb{E}J_\alpha^0(X,\xi) \le \frac{3}{2}\mathbb{E}\left\lVert X \right\rVert^2+C\alpha^2.\tag{65}\] For finite \(n\), the second moments below are finite by induction from \(X_0=0\), the finite variance of \(g(0,\xi)\) under \(\pi_\Xi\), and the global Lipschitz bounds in Lemma 1. Write \[\mathsf G_n:=\mathsf G(X_n,\xi_{n+1}), \qquad X_{n+1}=X_n-\alpha\mathsf G_n.\] The mean-perturbation condition 7 , with \(y=0\), implies \[\begin{align} &\mathbb{E}\!\left[ \left\langle X_n,\mathsf G(X_n,\xi_{n+1})-\mathsf G(0,\xi_{n+1})\right\rangle \,\middle|\,X_n,\xi_n\right]\\ &\qquad= \left\langle X_n, h(X_n)+\bar g(X_n,\xi_n)-\bar g(0,\xi_n)\right\rangle \ge (1-\theta)\left\langle X_n,h(X_n)\right\rangle. \end{align}\] The Poisson identity at the minimizer gives \(\mathbb{E}\left\langle X_n,g(0,\xi_{n+1})\right\rangle = \mathbb{E}\left\langle X_n,\chi_0(\xi_n)-\chi_0(\xi_{n+1})\right\rangle.\) Expanding \(J_\alpha^0\), using \(g(X_n,\xi_{n+1})=g(0,\xi_{n+1})+ \mathsf G(X_n,\xi_{n+1})-\mathsf G(0,\xi_{n+1})-h(X_n)\), and applying the preceding cancellation yield \[\begin{align} &\mathbb{E}[J_\alpha^0(X_{n+1},\xi_{n+1})- J_\alpha^0(X_n,\xi_n)]\\ &\quad=-2\alpha\mathbb{E}\left\langle X_n, \mathsf G(X_n,\xi_{n+1})-\mathsf G(0,\xi_{n+1})\right\rangle \\ &\quad+\alpha^2\mathbb{E}\left\lVert \mathsf G_n \right\rVert^2 +2\alpha^2\mathbb{E}\left\langle \mathsf G_n,\chi_0(\xi_{n+1})\right\rangle. \end{align}\] The global Lipschitz bounds in Lemma 1, together with \(\mathbb{E}_{\pi_\Xi}\left\lVert g(0,\xi) \right\rVert^2<\infty\), imply \(\mathbb{E}\left\lVert \mathsf G_n \right\rVert^2\le C\bigl(1+\mathbb{E}\left\lVert X_n \right\rVert^2\bigr).\) Moreover, Cauchy–Schwarz and \(\chi_0\in L^2(\pi_\Xi)\) give \[\left\lvert \mathbb{E}\left\langle \mathsf G_n,\chi_0(\xi_{n+1})\right\rangle \right\rvert \le \bigl(\mathbb{E}\left\lVert \mathsf G_n \right\rVert^2\bigr)^{1/2} \bigl(\mathbb{E}_{\pi_\Xi}\left\lVert \chi_0 \right\rVert^2\bigr)^{1/2} \le C\bigl(1+\mathbb{E}\left\lVert X_n \right\rVert^2\bigr).\] Combining these bounds with 64 yields \[\mathbb{E}J_\alpha^0(X_{n+1},\xi_{n+1}) \le \mathbb{E}J_\alpha^0(X_n,\xi_n)-c\alpha\mathbb{E}\left\lVert X_n \right\rVert^2+C\alpha +C\alpha^2\bigl(1+\mathbb{E}\left\lVert X_n \right\rVert^2\bigr).\] For sufficiently small \(\alpha\), 65 absorbs the \(C\alpha^2\mathbb{E}\left\lVert X_n \right\rVert^2\) term and gives \[\label{eq:Lalpha-geometric-bound} \mathbb{E}J_\alpha^0(X_{n+1},\xi_{n+1}) \le (1-c\alpha)\mathbb{E}J_\alpha^0(X_n,\xi_n)+C\alpha.\tag{66}\] Iterating 66 and using \(X_0=0\) gives \(\sup_n\mathbb{E}\left\lVert X_n \right\rVert^2<\infty\). By Theorem 4, the finite-time laws converge weakly to \(\pi_\alpha\). Lower semicontinuity of \(x\mapsto\left\lVert x \right\rVert^2\) then gives \(\mathbb{E}_{\pi_\alpha}\left\lVert X_\alpha \right\rVert^2<\infty\). Lemma 22 and the global Lipschitz bound in Lemma 1 give the second moment of \(h(X_\alpha)+g(X_\alpha,\xi_\alpha^+)\).

We now prove 63 . Lemma 22 and 26 imply \[\mathbb{E}\left\lVert h(X_\alpha)+g(X_\alpha,\xi_\alpha^+) \right\rVert^2 \le C\left(1+\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\right).\] By stationarity and the Poisson identity at the minimizer, \[\begin{align} \mathbb{E}\left\langle X_\alpha,g(0,\xi_\alpha^+)\right\rangle &= \mathbb{E}\left\langle X_\alpha,\chi_0(\xi_\alpha)-\chi_0(\xi_\alpha^+)\right\rangle\\ &= \mathbb{E}\left\langle X_\alpha^+-X_\alpha,\chi_0(\xi_\alpha^+)\right\rangle\\ &= -\alpha\mathbb{E}\left\langle h(X_\alpha)+g(X_\alpha,\xi_\alpha^+),\chi_0(\xi_\alpha^+)\right\rangle. \end{align}\] Thus Cauchy–Schwarz gives \[\mathbb{E}\left\langle X_\alpha,g(0,\xi_\alpha^+)\right\rangle \ge -C\alpha \left(\mathbb{E}\left\lVert h(X_\alpha)+g(X_\alpha,\xi_\alpha^+) \right\rVert^2\right)^{1/2} \ge -C\alpha\left(1+\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\right).\] Finally, the mean-perturbation condition with \(y=0\) gives \[\begin{align} &\mathbb{E}\left\langle X_\alpha, h(X_\alpha)+g(X_\alpha,\xi_\alpha^+)-g(0,\xi_\alpha^+)\right\rangle\\ &\qquad\ge(1-\theta)\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle. \end{align}\] Combining the last two displays and subtracting \(\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\) proves 63 . This proves the lemma. ◻

Proof of Lemma 6. By Lemma 23, the stationary square identity is justified. Since \(X_\alpha^+\stackrel d=X_\alpha\), \[0 = -2\alpha \mathbb{E}\left\langle X_\alpha,h(X_\alpha)+g(X_\alpha,\xi_\alpha^+)\right\rangle + \alpha^2 \mathbb{E}\left\lVert h(X_\alpha)+g(X_\alpha,\xi_\alpha^+) \right\rVert^2.\] Using 63 , Lemma 22, and 26 , we obtain \[(1-\theta)\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle \le C\alpha +C\alpha\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle.\] For small \(\alpha\), the last term is absorbed into the left-hand side, giving \(\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\le C\alpha.\) This proves 35 , and Lemma 22 then gives 38 .

The local moment and outside-probability estimates now follow from the same drift lower bounds. Let \(r_0:=\min\{1,\frac{R_H}{2}\}\). Lemma 19 gives \[\left\lVert x \right\rVert^m\le C\left\langle x,h(x)\right\rangle, \qquad \left\lVert x \right\rVert\le r_0, \qquad \left\langle x,h(x)\right\rangle\ge c_*, \qquad \left\lVert x \right\rVert\ge r_0.\] Therefore \[\begin{align} \mathbb{E}[\left\lVert X_\alpha \right\rVert^m\mathbf{1}_{\{\left\lVert X_\alpha \right\rVert\le1\}}] &\le C\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle +\mathbb{P}(\left\lVert X_\alpha \right\rVert\ge r_0) \\ &\le C\alpha, \end{align}\] and \[\mathbb{P}(\left\lVert X_\alpha \right\rVert\ge1) \le \mathbb{P}(\left\lVert X_\alpha \right\rVert\ge r_0) \le c_*^{-1}\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle \le C\alpha.\] This proves all estimates in Lemma 6. ◻

Proof. We first prove the local uniform integrability statement. On the event \(\{\left\lVert Y_\alpha \right\rVert\le R\}\), the point \(x=\alpha^{1/m}Y_\alpha\) remains in a fixed compact set. If \(\beta=2\), the uniform \(x\)-Lipschitz bound 25 gives \(\left\lVert g(x,\xi_\alpha^+) \right\rVert^2 \le C_R\{1+\left\lVert g(0,\xi_\alpha^+) \right\rVert^2\},\) and \(\xi_\alpha^+\sim\pi_\Xi\), so 10 gives uniform integrability. If \(1\le\beta<2\), Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) gives a uniform exponential moment for the ratio \(\frac{\left\lVert g(x,\xi_\alpha^+) \right\rVert}{1+\left\lVert x \right\rVert^{\beta-1}},\) while \(1+\left\lVert x \right\rVert^{\beta-1}\) is bounded on the same compact set. This again gives uniform integrability.

For 67 , write \[Y_\alpha^+-Y_\alpha =-\alpha^{1-1/m} \{h(X_\alpha)+g(X_\alpha,\xi_\alpha^+)\}.\] It is enough to prove \[\label{eq:scaled-Falpha-lindeberg-proof} \mathbb{E}\left[ \left\lVert \mathsf G_\alpha \right\rVert^2 \mathbf{1}_{\{\alpha^{1-1/m}\left\lVert \mathsf G_\alpha \right\rVert>\eta\}} \right]\to0, \qquad \mathsf G_\alpha:=\mathsf G(X_\alpha,\xi_\alpha^+).\tag{68}\]

Consider first \(\beta=2\). The global Lipschitz bounds in Lemma 1 imply \(\left\lVert \mathsf G_\alpha \right\rVert \le C\left\lVert X_\alpha \right\rVert+\left\lVert g(0,\xi_\alpha^+) \right\rVert.\) Using \((a+b)^2\mathbf{1}_{\{a+b>t\}}\le4a^2\mathbf{1}_{\{a>t/2\}}+4b^2\mathbf{1}_{\{b>t/2\}}\), the contribution of \(g(0,\xi_\alpha^+)\) to 68 tends to zero by dominated convergence under \(\pi_\Xi\). For the \(X_\alpha\)-term, the threshold \(c\eta\alpha^{1/m-1}\) tends to infinity. For all sufficiently small \(\alpha\), it lies in the quadratic-tail region, where \(\left\lVert x \right\rVert^2\le C\left\langle x,h(x)\right\rangle\). Therefore \[\mathbb{E}\!\bigl[\left\lVert X_\alpha \right\rVert^2 \mathbf{1}_{\{\alpha^{1-1/m}C\left\lVert X_\alpha \right\rVert>\eta\}}\bigr] \le C\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\to0\] by 35 . This verifies the Lindeberg condition in the quadratic-tail case.

Now assume \(1\le\beta<2\), and set \(w_\alpha:=1+\left\lVert X_\alpha \right\rVert^{\beta-1}.\) Since \(\left\lVert h(x) \right\rVert\le C(1+\left\lVert x \right\rVert^{\beta-1})\) globally, conditionally on \((X_\alpha,\xi_\alpha)\) we have \[\left\lVert \mathsf G_\alpha \right\rVert \le C w_\alpha\left\{1+ \frac{\left\lVert g(X_\alpha,\xi_\alpha^+) \right\rVert}{w_\alpha}\right\}.\] The exponential moment in Assumption 2([[itm:N5]](#itm:N5){reference-type=“ref” reference=“itm:N5”}) gives a uniform conditional second moment and a uniform conditional exponential tail for the normalized factor. Split according to \(w_\alpha\le \alpha^{-(m-1)/(2m)}\). On this event, the normalized factor must exceed a multiple of \(\alpha^{-(m-1)/(2m)}\), so the contribution is bounded by \(C\mathbb{E}w_\alpha^2 e^{-c\alpha^{-(m-1)/(2m)}}\to0.\) On the complementary event, the conditional second-moment bound gives a contribution at most \(C\mathbb{E}\bigl[w_\alpha^2 \mathbf{1}_{\{w_\alpha>\alpha^{-(m-1)/(2m)}\}}\bigr].\) To show that this expectation vanishes, fix a sufficiently large \(R_0\). The event \(\{w_\alpha>\alpha^{-(m-1)/(2m)}\}\) is eventually contained in \(\{\left\lVert X_\alpha \right\rVert\ge R_0\}\). On this region, Lemma 19 gives \((1+\left\lVert x \right\rVert^{\beta-1})^2\le C(1+\left\lVert x \right\rVert^\beta)\le C\left\langle x,h(x)\right\rangle\), after increasing \(R_0\) if necessary. Hence \[\mathbb{E}\bigl[w_\alpha^2 \mathbf{1}_{\{w_\alpha>\alpha^{-(m-1)/(2m)}\}}\bigr] \le C\mathbb{E}\left\langle X_\alpha,h(X_\alpha)\right\rangle\to0.\] This proves 68 , and hence 67 . ◻

Proof of Lemma 7. Let \(u\) be the Poisson solution from Lemma 21; thus \[u-Qu=f, \qquad u\in L^2(\pi_\Xi).\] Since \(\psi(Y_\alpha)\) is measurable with respect to \((X_\alpha,\xi_\alpha)\), \[\begin{align} \mathbb{E}[\psi(Y_\alpha)f(\xi_\alpha)] &= \mathbb{E}[\psi(Y_\alpha)u(\xi_\alpha)] - \mathbb{E}[\psi(Y_\alpha)Qu(\xi_\alpha)] \\ &= \mathbb{E}[\psi(Y_\alpha)u(\xi_\alpha)] - \mathbb{E}[\psi(Y_\alpha)u(\xi_\alpha^+)]. \end{align}\] Add and subtract \(\psi(Y_\alpha^+)u(\xi_\alpha^+)\). Since \((Y_\alpha^+,\xi_\alpha^+)\stackrel d=(Y_\alpha,\xi_\alpha),\) the first and third terms cancel, giving \[\mathbb{E}[\psi(Y_\alpha)f(\xi_\alpha)] = \mathbb{E}[(\psi(Y_\alpha^+)-\psi(Y_\alpha))u(\xi_\alpha^+)].\] Therefore, by Cauchy–Schwarz, \[\left\lvert \mathbb{E}[\psi(Y_\alpha)f(\xi_\alpha)] \right\rvert \le \mathop{\mathrm{Lip}}(\psi) \bigl(\mathbb{E}\left\lVert Y_\alpha^+-Y_\alpha \right\rVert^2\bigr)^{1/2} \bigl(\mathbb{E}\left\lvert u(\xi_\alpha^+) \right\rvert^2\bigr)^{1/2}.\] The second factor is finite and independent of \(\alpha\), because \(\xi_\alpha^+\sim\pi_\Xi\). For the first factor, \[Y_\alpha^+-Y_\alpha = -\alpha^{1-1/m} \{h(X_\alpha)+g(X_\alpha,\xi_\alpha^+)\}.\] Lemma 6 gives \(\mathbb{E}\left\lVert Y_\alpha^+-Y_\alpha \right\rVert^2 \le C\alpha^{2-2/m}.\) This proves the desired bound. ◻

Proof. The uniform \(x\)-Lipschitz bound 25 gives \(\left\lVert g_x(\zeta)-g_0(\zeta) \right\rVert\le C\left\lVert x \right\rVert.\) Fix \(p\in(0,1)\). The estimates for the solution of the Poisson equation in 61 and 62 give \[\left\lVert \chi_x(\zeta) \right\rVert+\left\lVert \chi_0(\zeta) \right\rVert\le C R_\chi(\zeta), \qquad \left\lVert \chi_x(\zeta)-\chi_0(\zeta) \right\rVert \le C_p\left\lVert x \right\rVert^pR_\chi(\zeta)^{1-p}.\] Expanding \(M_x-M_0\), for \(\left\lVert x \right\rVert\le1\) we obtain \(\left\lVert M_x(\zeta)-M_0(\zeta) \right\rVert \le C_p(\left\lVert x \right\rVert+\left\lVert x \right\rVert^p)F(\zeta),\) where \[F(\zeta):= 1+\left\lVert g_0(\zeta) \right\rVert+R_\chi(\zeta) +\left\lVert g_0(\zeta) \right\rVert R_\chi(\zeta).\] The stationary second moments in 10 and Cauchy–Schwarz imply \(F\in L^1(\pi_\Xi)\). In particular, \(M_0\in L^1(\pi_\Xi)\), and invariance gives \(QM_0\in L^1(\pi_\Xi)\). If \(K\) is compact, then, for all sufficiently small \(\alpha\), \(\alpha^{1/m}\sup_{y\in K}\left\lVert y \right\rVert\le1\). On \(\{Y_\alpha\in K\}\), the preceding pointwise bound, invariance of \(\pi_\Xi\), and \(\xi_\alpha\sim\pi_\Xi\) give \[\begin{align} &\mathbb{E}\left[ \left\lVert QM_{\alpha^{1/m}Y_\alpha}(\xi_\alpha)-QM_0(\xi_\alpha) \right\rVert \mathbf{1}_{\{Y_\alpha\in K\}}\right]\\ &\qquad\le C_{p,K}(\alpha^{1/m}+\alpha^{p/m})\mathbb{E}QF(\xi_\alpha) = C_{p,K}(\alpha^{1/m}+\alpha^{p/m})\int_\Xi F\,d\pi_\Xi \longrightarrow0. \end{align}\] This proves 70 . ◻

Proof. The constant part of \(\Psi\) is handled by weak convergence of \(Y_{\alpha_k}\), because the \(\Xi\)-marginal of \(\pi_{\alpha_k}\) is \(\pi_\Xi\) by Lemma 20. It remains to consider centered \(\Psi\), i.e. \(\pi_\Xi(\Psi)=0\). Bounded Lipschitz functions are dense in \(L^1(\pi_\Xi)\) on the closed Euclidean set \(\Xi\). Thus, for every \(\varepsilon>0\), choose a bounded Lipschitz \(f\) such that \(\left\lVert \Psi-f \right\rVert_{L^1(\pi_\Xi)}<\varepsilon\). Replacing \(f\) by \(f-\pi_\Xi(f)\) increases this error by at most another \(\varepsilon\), so we may assume \(\pi_\Xi(f)=0\). Since \(\xi_{\alpha_k}\sim\pi_\Xi\), \[\left\lvert \mathbb{E}[\psi(Y_{\alpha_k})(\Psi-f)(\xi_{\alpha_k})] \right\rvert \le 2\left\lVert \psi \right\rVert_\infty\varepsilon.\] Lemma 7 gives \(\mathbb{E}[\psi(Y_{\alpha_k})f(\xi_{\alpha_k})]\to0\). Letting \(\varepsilon\downarrow0\) proves the centered case and hence the lemma. ◻

Proof. Write \[G_\alpha:=g_{X_\alpha}(\xi_\alpha^+), \qquad \mathsf G_\alpha:=\mathsf G(X_\alpha,\xi_\alpha^+).\] Then \[Y_\alpha^+-Y_\alpha =-\alpha^{2-2/m}h_\alpha(Y_\alpha)-\alpha^{1-1/m}G_\alpha =-\alpha^{1-1/m}\mathsf G_\alpha.\] Let \(K_0:=\mathop{\mathrm{supp}}(\nabla\varphi)\), and let \(K_1:=\{y\in\mathbb{R}^d:\operatorname{dist}(y,K_0)\le1\}.\) If \(K_0=\varnothing\), the conclusion is immediate, so assume otherwise. Define \[\widetilde{\chi}_{\alpha,y}(\xi) := \begin{cases} \chi_{\alpha^{1/m}y}(\xi),&y\in K_1,\\ 0,&y\notin K_1. \end{cases}\] Set \(A_\alpha(y,\xi) :=\left\langle \nabla\varphi(y),\widetilde{\chi}_{\alpha,y}(\xi)\right\rangle.\) This is well defined because \(\nabla\varphi=0\) outside \(K_0\), and \(\left\lvert A_\alpha(y,\xi) \right\rvert\le C\left\lVert \nabla\varphi \right\rVert_\infty R_\chi(\xi)\). Introduce the localized perturbed test function \[\label{eq:perturbed-test-function-GK} \varphi_\alpha(y,\xi) := \varphi(y)-\alpha^{1-1/m}A_\alpha(y,\xi).\tag{72}\] The integrability required below follows from 38 , \(R_\chi\in L^2(\pi_\Xi)\), and 61 . Stationarity gives \[\label{eq:perturbed-stationary-identity} \mathbb{E}[\varphi_\alpha(Y_\alpha^+,\xi_\alpha^+)-\varphi_\alpha(Y_\alpha,\xi_\alpha)]=0.\tag{73}\]

First consider the uncorrected part. Taylor’s formula and \(Y_\alpha^+-Y_\alpha=-\alpha^{2-2/m}h_\alpha(Y_\alpha)-\alpha^{1-1/m}G_\alpha\) give \[\label{eq:uncorrected-taylor-generator} \begin{align} \frac{\varphi(Y_\alpha^+)-\varphi(Y_\alpha)}{\alpha^{2-2/m}} &=-\left\langle h_\alpha(Y_\alpha),\nabla\varphi(Y_\alpha)\right\rangle -\alpha^{1/m-1}\left\langle \nabla\varphi(Y_\alpha),G_\alpha\right\rangle \\ &\quad+ \frac{1}{2}\mathop{\mathrm{tr}}\!\left(G_\alpha G_\alpha^\top\nabla^2\varphi(Y_\alpha)\right) +r_\alpha^{(1)}, \end{align}\tag{74}\] with \(\mathbb{E}|r_\alpha^{(1)}|\to0\). The terms in the quadratic expansion containing \(h_\alpha\) are negligible because \[\alpha^{2-2/m}\left\lVert h_\alpha(Y_\alpha) \right\rVert^2=\left\lVert h(X_\alpha) \right\rVert^2, \qquad \alpha^{1-1/m}\left\lVert h_\alpha(Y_\alpha) \right\rVert\left\lVert G_\alpha \right\rVert =\left\lVert h(X_\alpha) \right\rVert\left\lVert G_\alpha \right\rVert,\] and 69 makes both expectations tend to zero. The third-order Taylor remainder is controlled by the uniform continuity of \(\nabla^2\varphi\): for every \(\eta>0\), after division by \(\alpha^{2-2/m}\) its absolute value is bounded by \[\eta\frac{\left\lVert Y_\alpha^+-Y_\alpha \right\rVert^2}{\alpha^{2-2/m}} +C_\eta \frac{\left\lVert Y_\alpha^+-Y_\alpha \right\rVert^2}{\alpha^{2-2/m}} \mathbf{1}_{\{\left\lVert Y_\alpha^+-Y_\alpha \right\rVert>\eta\}}.\] The first term is bounded in expectation uniformly in \(\alpha\) by 38 ; the second tends to zero by 67 . Sending \(\eta\downarrow0\) proves \(\mathbb{E}|r_\alpha^{(1)}|\to0\).

Now expand the term involving the solution of the Poisson equation in 72 . In \[-\alpha^{1/m-1}\mathbb{E}\left[ A_\alpha(Y_\alpha^+,\xi_\alpha^+) -A_\alpha(Y_\alpha,\xi_\alpha) \right],\] add and subtract \(A_\alpha(Y_\alpha,\xi_\alpha^+)\). Since \(A_\alpha(Y_\alpha,\cdot)\) can be nonzero only when \(Y_\alpha\in K_0\), the first resulting increment is \[\begin{align} -\alpha^{1/m-1} \mathbb{E}\left[A_\alpha(Y_\alpha,\xi_\alpha^+) -A_\alpha(Y_\alpha,\xi_\alpha)\right] &=\alpha^{1/m-1} \mathbb{E}\left\langle \nabla\varphi(Y_\alpha),Qg_{X_\alpha}(\xi_\alpha)\right\rangle. \end{align}\] Here the solution of the Poisson equation is evaluated at \(X_\alpha=\alpha^{1/m}Y_\alpha\) with \(Y_\alpha\in K_0\). The last display follows from \(\chi_x-Q\chi_x=Qg_x\) and cancels the conditional mean of the singular term in 74 , since \(\mathbb{E}[G_\alpha\mid Y_\alpha,\xi_\alpha]=Qg_{X_\alpha}(\xi_\alpha)\).

It remains to control \[-\alpha^{1/m-1}\mathbb{E}\left[ A_\alpha(Y_\alpha^+,\xi_\alpha^+) -A_\alpha(Y_\alpha,\xi_\alpha^+)\right].\] Let \(E_\alpha:=\{\left\lVert Y_\alpha^+-Y_\alpha \right\rVert\le1/2\}\). By 67 , \[\mathbb{P}(E_\alpha^c) \le4\mathbb{E}\left[\left\lVert Y_\alpha^+-Y_\alpha \right\rVert^2\mathbf{1}_{E_\alpha^c}\right] =o(\alpha^{2-2/m}).\] The bound on \(A_\alpha\), Cauchy–Schwarz, and \(\xi_\alpha^+\sim\pi_\Xi\) therefore give \[\label{eq:localized-poisson-large-jump} \begin{align} &\alpha^{1/m-1}\mathbb{E}\left[ \left|A_\alpha(Y_\alpha^+,\xi_\alpha^+) -A_\alpha(Y_\alpha,\xi_\alpha^+)\right| \mathbf{1}_{E_\alpha^c}\right]\\ &\qquad\le C\alpha^{1/m-1} \bigl(\mathbb{E}_{\pi_\Xi}R_\chi^2\bigr)^{1/2} \mathbb{P}(E_\alpha^c)^{1/2}=o(1). \end{align}\tag{75}\]

On \(E_\alpha\), if either term involving \(A_\alpha\) is nonzero, then one of \(Y_\alpha,Y_\alpha^+\) belongs to \(K_0\), and both belong to \(K_1\). Hence \[\begin{align} &A_\alpha(Y_\alpha^+,\xi_\alpha^+) -A_\alpha(Y_\alpha,\xi_\alpha^+)\\ &\quad= \left\langle \nabla\varphi(Y_\alpha^+)-\nabla\varphi(Y_\alpha),\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)\right\rangle +\left\langle \nabla\varphi(Y_\alpha^+),\widetilde{\chi}_{\alpha,Y_\alpha^+}(\xi_\alpha^+) -\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)\right\rangle. \end{align}\]

For the first term on \(E_\alpha\), Taylor’s formula gives \[\begin{align} &-\alpha^{1/m-1} \mathbb{E}\left\langle \nabla\varphi(Y_\alpha^+)-\nabla\varphi(Y_\alpha),\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)\right\rangle\mathbf{1}_{E_\alpha} \\ &\qquad= \mathbb{E}\mathop{\mathrm{tr}}\!\left(G_\alpha \widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)^\top \nabla^2\varphi(Y_\alpha)\right)+o(1). \end{align}\] The contribution of \(E_\alpha^c\) to the displayed leading term tends to zero. Indeed, \[\mathbb{E}[\left\lVert G_\alpha \right\rVert^2\mathbf{1}_{E_\alpha^c}] \le2\mathbb{E}[\left\lVert \mathsf G_\alpha \right\rVert^2\mathbf{1}_{E_\alpha^c}] +2\mathbb{E}\left\lVert h(X_\alpha) \right\rVert^2\longrightarrow0\] by 67 and 69 ; Cauchy–Schwarz with \(R_\chi\in L^2(\pi_\Xi)\) then applies. Also, \(\nabla^2\varphi(Y_\alpha)\ne0\) implies \(Y_\alpha\in K_0\), so the solution appearing in the leading term is well defined.

The contribution of \(-\alpha^{2-2/m}h_\alpha\) is \(O(\alpha^{1-1/m})\): the factor \(\nabla^2\varphi(Y_\alpha)\) restricts this term to a fixed compact set, where \(h_\alpha\) is locally uniformly bounded by the expansion in Assumption 3([[itm:H4]](#itm:H4){reference-type=“ref” reference=“itm:H4”}), while \(\mathbb{E}R_\chi(\xi_\alpha^+)<\infty\).

To control the Taylor remainder, let \(\omega_\varphi\) be a bounded modulus of continuity of \(\nabla^2\varphi\). Since \(Y_\alpha^+-Y_\alpha=-\alpha^{1-1/m}\mathsf G_\alpha\), the absolute expectation of the remainder after division by \(\alpha^{1-1/m}\) is at most \[C\mathbb{E}\!\left[ \omega_\varphi(\alpha^{1-1/m}\left\lVert \mathsf G_\alpha \right\rVert) \left\lVert \mathsf G_\alpha \right\rVert R_\chi(\xi_\alpha^+)\right].\] For every \(\eta>0\), the contribution of \(\{\alpha^{1-1/m}\left\lVert \mathsf G_\alpha \right\rVert\le\eta\}\) is bounded by \[C\omega_\varphi(\eta) \bigl(\mathbb{E}\left\lVert \mathsf G_\alpha \right\rVert^2\bigr)^{1/2} \bigl(\mathbb{E}_{\pi_\Xi}R_\chi^2\bigr)^{1/2} \le C\omega_\varphi(\eta).\] On the complementary event, Cauchy–Schwarz bounds the contribution by \[C\left( \mathbb{E}\!\left[\left\lVert \mathsf G_\alpha \right\rVert^2 \mathbf{1}_{\{\alpha^{1-1/m}\left\lVert \mathsf G_\alpha \right\rVert>\eta\}}\right] \right)^{1/2} \bigl(\mathbb{E}_{\pi_\Xi}R_\chi^2\bigr)^{1/2},\] which tends to zero by 67 . Sending \(\eta\downarrow0\) proves the \(o(1)\) remainder in the gradient increment.

For the second term on \(E_\alpha\), which changes the solution of the Poisson equation from \(\alpha^{1/m}Y_\alpha\) to \(\alpha^{1/m}Y_\alpha^+\), fix the exponent \(p\in(1-1/m,1)\) in 62 . That estimate gives \[\begin{align} &\alpha^{1/m-1} \mathbb{E}\left|\left\langle \nabla\varphi(Y_\alpha^+),\widetilde{\chi}_{\alpha,Y_\alpha^+}(\xi_\alpha^+) -\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)\right\rangle\right| \mathbf{1}_{E_\alpha} \\ &\qquad\le C_p\alpha^{p-(1-1/m)} \mathbb{E}[\left\lVert \mathsf G_\alpha \right\rVert^p R_\chi(\xi_\alpha^+)^{1-p}]\\ &\qquad\le C_p\alpha^{p-(1-1/m)} =o(1). \end{align}\] Here the second inequality follows from weighted AM–GM, \(a^pb^{1-p}\le pa+(1-p)b\), the uniform second moment of \(\mathsf G_\alpha\), and \(R_\chi\in L^2(\pi_\Xi)\).

Combining the preceding identities with 73 gives \[\begin{align} 0&=-\mathbb{E}\left\langle h_\alpha(Y_\alpha),\nabla\varphi(Y_\alpha)\right\rangle \\ &\quad+\frac{1}{2}\mathbb{E}\mathop{\mathrm{tr}}\!\left( \Bigl[G_\alpha G_\alpha^\top +G_\alpha\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)^\top +\widetilde{\chi}_{\alpha,Y_\alpha}(\xi_\alpha^+)G_\alpha^\top\Bigr] \nabla^2\varphi(Y_\alpha) \right)+o(1). \end{align}\] After multiplication by \(\nabla^2\varphi(Y_\alpha)\), conditioning on \((Y_\alpha,\xi_\alpha)\) identifies the conditional mean with \(QM_{\alpha^{1/m}Y_\alpha}(\xi_\alpha)\nabla^2\varphi(Y_\alpha)\). This proves 71 . ◻

Proof of Lemma 8. Let \(\varphi\in C_c^3(\mathbb{R}^d)\) and let \(K:=\mathop{\mathrm{supp}}(\nabla^2\varphi)\). Lemma 27 gives 71 . Along the subsequence \(\alpha_k\downarrow0\), local uniform convergence of \(h_\alpha\) and weak convergence give \[\label{eq:drift-limit-markov-new} \mathbb{E}\left\langle h_{\alpha_k}(Y_{\alpha_k}),\nabla\varphi(Y_{\alpha_k})\right\rangle \longrightarrow \int_{\mathbb{R}^d}\left\langle h_0(y),\nabla\varphi(y)\right\rangle\,\nu(dy).\tag{76}\] Here \(\nabla\varphi\) is compactly supported and \(h_\alpha\to h_0\) locally uniformly by Assumption 3([[itm:H4]](#itm:H4){reference-type=“ref” reference=“itm:H4”}).

For the covariance term, 70 replaces \(QM_{\alpha_k^{1/m}Y_{\alpha_k}}(\xi_{\alpha_k})\) by \(QM_0(\xi_{\alpha_k})\) on \(K\). Each entry of \(QM_0\) belongs to \(L^1(\pi_\Xi)\) by Lemma 25. Applying Lemma 26 entrywise, with the bounded Lipschitz entries of \(\nabla^2\varphi\), yields \[\label{eq:QM0-decorrelation-limit} \mathbb{E}\mathop{\mathrm{tr}}\!\left(QM_0(\xi_{\alpha_k})\nabla^2\varphi(Y_{\alpha_k})\right) \longrightarrow \int_{\mathbb{R}^d}\mathop{\mathrm{tr}}\!\left(\bar M\nabla^2\varphi(y)\right)\nu(dy),\tag{77}\] where \(\bar M:=\int_\Xi M_0(\xi)\,\pi_\Xi(d\xi).\) It remains only to identify \(\bar M\) with \(\Sigma\). Let \((\xi_0,\xi_1)\) be two successive values of the stationary driving chain and write \(f=g_0\), \(\chi=\chi_0\). Since \(\chi-Q\chi=Qf\), \(\mathbb{E}[f(\xi_1)+\chi(\xi_1)\mid\xi_0]=Qf(\xi_0)+Q\chi(\xi_0)=\chi(\xi_0).\) With \(D_0=f(\xi_1)+\chi(\xi_1)-\chi(\xi_0)\), expansion and the preceding conditional identity give \[\begin{align} \Sigma &=\mathbb{E}[D_0D_0^\top] \\ &=\mathbb{E}\bigl[f(\xi_1)f(\xi_1)^\top +f(\xi_1)\chi(\xi_1)^\top +\chi(\xi_1)f(\xi_1)^\top\bigr] \\ &=\int_\Xi M_0(\xi)\,\pi_\Xi(d\xi)=\bar M. \end{align}\] Combining 71 , 76 , and 77 proves 39 . ◻

Proof. The claim is trivial at \(y=0\). For \(y\ne0\), set \(x=\alpha^{1/m}y\). For all sufficiently small \(\alpha\), \(\left\lVert x \right\rVert\le \frac{R_H}{2}\). The identity \(h(x)=\int_0^1\nabla^2H(tx)x\,dt\) and Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) give \(\left\langle x,h(x)\right\rangle \ge \frac{c_{\rm in}}{m-1}\left\lVert x \right\rVert^m.\) Equivalently, \[\left\langle y,\alpha^{-(m-1)/m}h(\alpha^{1/m}y)\right\rangle = \alpha^{-1}\left\langle x,h(x)\right\rangle \ge \frac{c_{\rm in}}{m-1}\left\lVert y \right\rVert^m.\] Letting \(\alpha\downarrow0\) and using the expansion in Assumption 3([[itm:H4]](#itm:H4){reference-type=“ref” reference=“itm:H4”}) together with the homogeneity of \(h_0\) gives the result. ◻

Proof. The claim is trivial when \(y=z\). Set \(e:=y-z\ne0\). For sufficiently small \(\alpha\), the segment between \(\alpha^{1/m}z\) and \(\alpha^{1/m}y\) is contained in \(\{\left\lVert x \right\rVert\le \frac{R_H}{2}\}\). It can pass through the origin for at most one parameter value, which does not affect the integral below. Using Assumption 1([[itm:H2]](#itm:H2){reference-type=“ref” reference=“itm:H2”}) along the segment gives \[\begin{align} &\left\langle e,\alpha^{-(m-1)/m}\{h(\alpha^{1/m}y)-h(\alpha^{1/m}z)\}\right\rangle \\ &\qquad = \alpha^{-1}\left\langle \alpha^{1/m}e,h(\alpha^{1/m}y)-h(\alpha^{1/m}z)\right\rangle \\ &\qquad = \alpha^{-1}\int_0^1 \left\langle \alpha^{1/m}e,\nabla^2H(\alpha^{1/m}(z+te))\alpha^{1/m}e\right\rangle\,dt \\ &\qquad \ge c_{\rm in}\left\lVert e \right\rVert^2 \int_0^1\left\lVert z+te \right\rVert^{m-2}\,dt. \end{align}\] We use the elementary segment bound \[\label{eq:elementary-segment-bound-power} \inf_{\left\lVert v \right\rVert=1}\inf_{w\in\mathbb{R}^d} \int_0^1\left\lVert w+tv \right\rVert^{m-2}\,dt>0.\tag{78}\] To prove 78 , the case \(m=2\) is immediate. For \(m>2\), decompose \(w=a v+w_\perp\), with \(w_\perp\perp v\). Then \[\int_0^1\left\lVert w+tv \right\rVert^{m-2}\,dt \ge \int_0^1 |a+t|^{m-2}\,dt.\] The last integral is minimized over \(a\in\mathbb{R}\) when the interval \([a,a+1]\) is centered at the origin, and its minimum is \(2\int_0^{1/2}t^{m-2}\,dt>0\). Thus 78 holds and gives \(\int_0^1\left\lVert z+te \right\rVert^{m-2}\,dt \ge c\left\lVert e \right\rVert^{m-2}.\) Letting \(\alpha\downarrow0\) and using the expansion in Assumption 3([[itm:H4]](#itm:H4){reference-type=“ref” reference=“itm:H4”}) together with the homogeneity of \(h_0\) gives the result. ◻

Proof of Lemma 9. The drift \(-h_0\) is locally Lipschitz because \(H_0\in C^2\), and it has polynomial growth by homogeneity. Thus the SDE has a pathwise unique maximal strong solution up to its explosion time. Let \[W(y):=1+\left\lVert y \right\rVert^2, \qquad \mathcal{L}\varphi(y) :=-\left\langle h_0(y),\nabla\varphi(y)\right\rangle +\frac{1}{2}\mathop{\mathrm{tr}}(\Sigma\nabla^2\varphi(y)).\] By Lemma 28, \[\label{eq:limit-diffusion-LW} \mathcal{L}W(y) =-2\left\langle y,h_0(y)\right\rangle+\mathop{\mathrm{tr}}(\Sigma) \le C-c\left\lVert y \right\rVert^m.\tag{79}\] If \(\tau_R:=\inf\{t:\left\lVert Y_t \right\rVert\ge R\}\), Itô’s formula applied to \(W(Y_{t\wedge\tau_R})\) gives \(\mathbb{E}_y W(Y_{t\wedge\tau_R})\le W(y)+Ct.\) On \(\{\tau_R\le t\}\), the left-hand side is at least \(1+R^2\). Hence \(\mathbb{P}_y(\tau_R\le t)\le (W(y)+Ct)/(1+R^2)\to0\), which proves nonexplosion. Applying Itô’s formula without stopping and using 79 gives \(\frac{1}{T}\int_0^T \mathbb{E}_y\left\lVert Y_t \right\rVert^m\,dt \le C+\frac{W(y)}{cT}.\) The time-averaged laws \(T^{-1}\int_0^T P_t(y,\cdot)\,dt\) are therefore tight. Since the nonexplosive locally Lipschitz SDE is Feller, the Krylov–Bogoliubov argument gives at least one invariant probability law.

We next show that every invariant probability law has finite \(m\)-moment. Let \(\theta_n\in C^2([0,\infty))\) be nondecreasing and concave, with \(0\le\theta_n'\le1\), \(\theta_n'(s)=1\) for \(s\le n\), \(\theta_n'(s)=0\) for \(s\ge2n\), \(\theta_n''\le0\), and \(\theta_n'(s)\uparrow1\) for every fixed \(s\). Set \(W_n(y):=\theta_n(W(y))\). Then \(W_n\) is bounded with bounded first and second derivatives. If \(\mu\) is invariant, then \(\int \mathcal{L}W_n\,d\mu=0\). Moreover, \[\mathcal{L}W_n(y) =\theta_n'(W(y))\mathcal{L}W(y) +\frac{1}{2}\theta_n''(W(y)) \left\lVert \Sigma^{1/2}\nabla W(y) \right\rVert^2 \le \theta_n'(W(y))(C-c\left\lVert y \right\rVert^m),\] because \(\theta_n''\le0\) and \(\Sigma\succeq0\). Hence \[c\int \theta_n'(W(y))\left\lVert y \right\rVert^m\,\mu(dy) \le C\int \theta_n'(W(y))\,\mu(dy) \le C.\] Letting \(n\to\infty\) and using monotone convergence gives \(\int \left\lVert y \right\rVert^m\,\mu(dy)<\infty.\) In particular every invariant law has finite second moment.

Let \(\mu\) and \(\nu\) be invariant laws. Choose a coupling \((Y_0,\widetilde{Y}_0)\) with finite second moment and drive both solutions by the same Brownian motion. Since the diffusion coefficient is constant, \(\Delta_t:=Y_t-\widetilde{Y}_t\) satisfies, up to localization, \[\frac{d}{dt}\left\lVert \Delta_t \right\rVert^2 =-2\left\langle \Delta_t,h_0(Y_t)-h_0(\widetilde{Y}_t)\right\rangle \le -2c_m\left\lVert \Delta_t \right\rVert^m\] by Lemma 29. With \(D(t):=\mathbb{E}\left\lVert \Delta_t \right\rVert^2\), Fatou’s lemma and Jensen’s inequality give \(D'(t)\le -2c_m\mathbb{E}\left\lVert \Delta_t \right\rVert^m \le -2c_mD(t)^{m/2}.\) Thus \(D(t)\to0\), exponentially if \(m=2\) and polynomially if \(m>2\). The marginals remain \(\mu\) and \(\nu\), so \(W_2(\mu,\nu)^2\le D(t)\to0\). Hence \(\mu=\nu\). Denote the unique invariant law by \(\nu_\infty\).

Finally, suppose \(\rho\) satisfies the stationary weak equation for every \(\varphi\in C_c^3(\mathbb{R}^d)\). It then holds on \(C_c^\infty(\mathbb{R}^d)\). The martingale problem for \(\mathcal{L}\) is well posed because the SDE is pathwise unique and nonexplosive. The Echeverria criterion therefore implies that \(\rho\) is invariant; see [47]. By uniqueness, \(\rho=\nu_\infty\). ◻

Proof of Corollary 2. Write \(d_{i,\alpha}:=d_{V_{\alpha,i},\alpha}^{(i)}\). For each coordinate \(i\), the one-dimensional proof gives a constant \(c_i>0\) and, after taking the minimum of the finitely many stepsize thresholds, the synchronous one-step estimate \[\mathbb{E}\Bigl[d_{i,\alpha}\bigl(F_{U_1}^{(i)}(x_i,\xi),F_{U_1}^{(i)}(y_i,\eta)\bigr)\Bigr] \le (1-c_i\alpha^{m_i-1})\,d_{i,\alpha}\bigl((x_i,\xi),(y_i,\eta)\bigr)\] for the coordinate map \[F_u^{(i)}(x_i,\xi) := \Bigl(x_i-\alpha\bigl(h_i(x_i)+g_i(x_i,\Phi(\xi,u))\bigr),\Phi(\xi,u)\Bigr).\] The \(i\)th coordinate projection of the full separable map is \((x,\xi)\mapsto F_u^{(i)}(x_i,\xi).\) Since the same innovation \(U_1\) drives every coordinate, summing the coordinatewise estimates gives \[\mathbb{E}\Bigl[d_{{\rm sep},\alpha}\bigl(F_{U_1}(x,\xi),F_{U_1}(y,\eta)\bigr)\Bigr] \le (1-c_0\alpha^{m_{\max}-1})\,d_{{\rm sep},\alpha}\bigl((x,\xi),(y,\eta)\bigr),\] where \(c_0:=\min_{1\le i\le d}c_i>0\). Indeed, for \(0<\alpha\le1\), \[c_i\alpha^{m_i-1} \ge c_0\alpha^{m_{\max}-1}, \qquad 1\le i\le d.\] This is the required one-step contraction in the additive separable metric.

It remains to check the reference-point integrability. The one-dimensional hypotheses may use different reference points. Let \(\xi_\star^{(i)}\) be one for coordinate \(i\), and fix \(\xi^\circ\in\Xi\). By Assumption 2([[itm:N1]](#itm:N1){reference-type=“ref” reference=“itm:N1”}), \[\mathbb{E}\left\lVert \Phi(\xi^\circ,U_1)-\xi^\circ \right\rVert \le (1+\rho_\Xi)\left\lVert \xi^\circ-\xi_\star^{(i)} \right\rVert + \mathbb{E}\left\lVert \Phi(\xi_\star^{(i)},U_1)-\xi_\star^{(i)} \right\rVert<\infty.\] If \(\beta_i=2\), then Assumption 2([[itm:N2]](#itm:N2){reference-type=“ref” reference=“itm:N2”}) gives \[\mathbb{E}\left\lvert g_i(0,\Phi(\xi^\circ,U_1)) \right\rvert \le \mathbb{E}\left\lvert g_i(0,\Phi(\xi_\star^{(i)},U_1)) \right\rvert + L_{g,\Phi}\left\lVert \xi^\circ-\xi_\star^{(i)} \right\rVert<\infty.\] Thus \(\xi^\circ\) works for every coordinate. Each \(d_{i,\alpha}\) dominates the base norm \(\left\lvert x_i-y_i \right\rvert+\alpha^{-1}\left\lVert \xi-\eta \right\rVert\) and is locally comparable to it on bounded sets. Hence \(d_{{\rm sep},\alpha}\) is complete and separable on \(\mathbb{R}^d\times\Xi\) and has the usual Borel \(\sigma\)-field. Summing the coordinatewise reference-point costs at \((0,\xi^\circ)\) and applying Proposition 11 gives the invariant law, uniqueness, and 16 . ◻

Proof of Corollary 3. Write \(q:=\lvert\mathcal{I}_*\rvert\), use a subscript \(*\) to denote restriction to the coordinates in \(\mathcal{I}_*\), and set \(Y_{\alpha,*} :=\alpha^{-1/m_{\max}} X_{\infty,*}^{(\alpha),\rm sep}.\) The coordinatewise moment bounds from the proof of Theorem 7 imply that \((Y_{\alpha,*})_\alpha\) is tight in \(\mathbb{R}^q\). We first identify its subsequential limits. By separability, the update of the active block does not depend on the remaining iterate coordinates. On the scale \(\alpha^{1/m_{\max}}\), its stationary one-step increment is \[Y_{\alpha,*}^+-Y_{\alpha,*} =-\alpha^{2-2/m_{\max}}h_{\alpha,*}(Y_{\alpha,*}) -\alpha^{1-1/m_{\max}} g_*(\alpha^{1/m_{\max}}Y_{\alpha,*},\xi_\alpha^+),\] where \[h_{\alpha,*}(y) :=\bigl( \alpha^{-(m_{\max}-1)/m_{\max}} h_i(\alpha^{1/m_{\max}}y_i) \bigr)_{i\in\mathcal{I}_*},\] and \(g_*(x,\xi):=(g_i(x_i,\xi))_{i\in\mathcal{I}_*}\). The coordinatewise tangent assumptions give \[h_{\alpha,*}\longrightarrow h_{0,*}, \qquad h_{0,*}(y):=(h_{i,0}(y_i))_{i\in\mathcal{I}_*}\] locally uniformly.

For \(x\in\mathbb{R}^q\), let \(\chi_{*,x}(\xi) :=(\chi_{i,x_i}(\xi))_{i\in\mathcal{I}_*},\) where \(\chi_{i,x_i}\) is the one-dimensional solution of the Poisson equation for coordinate \(i\). At the minimizer, define \[D_i(\xi,U_1) :=g_i(0,\Phi(\xi,U_1)) +\chi_{i,0}(\Phi(\xi,U_1))-\chi_{i,0}(\xi), \qquad i\in\mathcal{I}_*.\] The vector \(D_*=(D_i)_{i\in\mathcal{I}_*}\) is conditionally centered and, by Lemma 5, \[\Sigma_{\rm act}:=\mathbb{E}[D_*D_*^\top] =(\Sigma_{\rm sep})_{\mathcal{I}_*\times\mathcal{I}_*}.\] Thus the off-diagonal asymptotic covariances created by the common driving chain are retained within the active block.

For \(\varphi\in C_c^3(\mathbb{R}^q)\), use the perturbed test function \[\varphi_\alpha(y,\xi) :=\varphi(y) -\alpha^{1-1/m_{\max}} \left\langle \nabla\varphi(y),\chi_{*,\alpha^{1/m_{\max}}y}(\xi)\right\rangle.\] The calculation in Lemma 8 applies componentwise to this block. For \[M_{*,x}(\xi) :=g_*(x,\xi)g_*(x,\xi)^\top +g_*(x,\xi)\chi_{*,x}(\xi)^\top +\chi_{*,x}(\xi)g_*(x,\xi)^\top,\] each entry is a finite sum of products of one-dimensional terms. The coordinatewise Lipschitz estimates for \(g_i\), the Hölder estimates 62 for \(\chi_i\), and the coordinatewise stationary second moments therefore give, as in Lemma 25, \[\mathbb{E}\!\left[ \left\lVert QM_{*,\alpha^{1/m_{\max}}Y_{\alpha,*}}(\xi_\alpha) -QM_{*,0}(\xi_\alpha) \right\rVert \mathbf{1}_{\{Y_{\alpha,*}\in K\}} \right]\longrightarrow0\] for every compact \(K\subset\mathbb{R}^q\). The entries of \(QM_{*,0}\) belong to \(L^1(\pi_\Xi)\). Applying Lemma 26 entrywise to \(\nabla^2\varphi(Y_{\alpha,*})\) gives the required product limits. Consequently, every subsequential weak limit \(\nu_*\) of \(Y_{\alpha,*}\) satisfies \[\int_{\mathbb{R}^q} \left[ -\left\langle \nabla\varphi(y),h_{0,*}(y)\right\rangle +\frac{1}{2}\mathop{\mathrm{tr}}\bigl(\Sigma_{\rm act}\nabla^2\varphi(y)\bigr) \right]\nu_*(dy)=0, \qquad \varphi\in C_c^3(\mathbb{R}^q).\] Because every coordinate in \(\mathcal{I}_*\) has exponent \(m_{\max}\), the active drift satisfies, for constants \(c,c'>0\), \[\left\langle y,h_{0,*}(y)\right\rangle \ge c\sum_{i\in\mathcal{I}_*}|y_i|^{m_{\max}} \ge c'\left\lVert y \right\rVert^{m_{\max}},\] and the same argument gives \(\left\langle y-z,h_{0,*}(y)-h_{0,*}(z)\right\rangle \ge c'\left\lVert y-z \right\rVert^{m_{\max}}.\) Lemma 9 therefore identifies a unique invariant law for the active block of 19 . Hence the entire family \(Y_{\alpha,*}\) converges to this law.

It remains to consider coordinates outside \(\mathcal{I}_*\). If \(i\notin\mathcal{I}_*\), then \[\frac{X_{\infty,i}^{(\alpha),\rm sep}}{\alpha^{1/m_{\max}}} =\alpha^{1/m_i-1/m_{\max}} \left( \frac{X_{\infty,i}^{(\alpha),\rm sep}}{\alpha^{1/m_i}} \right) \longrightarrow0 \qquad\text{in probability},\] because the bracketed family is tight by the coordinatewise limit stated before the corollary. In 19 , these inactive coordinates have no Brownian forcing and solve \(dY_{i,t}=-h_{i,0}(Y_{i,t})\,dt.\) The one-dimensional coercivity gives \(y_i h_{i,0}(y_i)\ge c|y_i|^{m_i}\), so every such deterministic flow converges to zero and has \(\delta_0\) as its unique invariant law. The active block has the unique invariant law identified above. Therefore 19 has a unique invariant law, with inactive coordinates equal to zero. Combining the active-block convergence with the inactive-coordinate convergence proves the asserted convergence to \(Y_\infty^{\rm sep}\). ◻

Proof of Remark 10. For each coordinate \(i\), the one-dimensional version of Theorem 7 gives \(\frac{X_{\infty,i}^{(\alpha)}}{\alpha^{1/m_i}}\Rightarrow Y_{\infty,i}.\) For each fixed \(\alpha\), the stationary law factorizes by Remark 9, so the rescaled coordinates are independent. Hence the joint characteristic function factorizes: \[\mathbb{E}\exp\left(i\sum_{i=1}^d t_i\frac{X_{\infty,i}^{(\alpha)}}{\alpha^{1/m_i}}\right) = \prod_{i=1}^d \mathbb{E}\exp\left(i t_i\frac{X_{\infty,i}^{(\alpha)}}{\alpha^{1/m_i}}\right) \longrightarrow \prod_{i=1}^d \mathbb{E}e^{it_iY_{\infty,i}}.\] Lévy’s continuity theorem gives the claimed joint convergence. ◻