Proof. Step 1: \(L^1(\pi^\varepsilon)\)-convergence. By 1(ii), \(u^\varepsilon(t,z)\to \bar u(t,\Phi(z))\) uniformly on compacts, and by 1(i) the family \((\pi^\varepsilon)\) is tight. Thus for any \(\delta>0\) we can pick a compact \(K\subset E\) with \(\limsup_{\varepsilon\to0}\pi^\varepsilon(K^c)\le\delta\), and then \[\mathcal{R}_{L_1}^\varepsilon:=\int_E |u^\varepsilon-\bar u|\,d\pi^\varepsilon \le \sup_{K}|u^\varepsilon-\bar u| + 2\|f\|_\infty\,\pi^\varepsilon(K^c).\] Letting \(\varepsilon\to0\) and then \(\delta\downarrow0\) yields \(\|u^\varepsilon-\bar u\|_{L^1(\pi^\varepsilon)}\to0\).

Step 2: Free Energy. Let \(\phi(x)=x\log x\) with \(0\log0:=0\) and \(M:=\|f\|_\infty\). By \(L^\infty\)-contractivity, \(0\le u^\varepsilon,\bar u\le M\), hence \(\phi\) is bounded and uniformly continuous on \([0,M]\) with modulus \(\omega(\eta):=\sup\{|\phi(a)-\phi(b)|:\;a,b\in[0,M],\,|a-b|\le\eta\}\). For any \(\eta>0\), \[\int_E |\phi(u^\varepsilon)-\phi(\bar u)|\,d\pi^\varepsilon \le \omega(\eta) + 2\|\phi\|_{L^\infty([0,M])}\,\pi^\varepsilon(|u^\varepsilon-\bar u|>\eta) \le \omega(\eta) + \frac{2\|\phi\|_{L^\infty([0,M])}}{\eta}\mathcal{R}_{L_1}^\varepsilon.\] First let \(\varepsilon\to0\) and then \(\eta\downarrow0\) to get \(\int_E |\phi(u^\varepsilon)-\phi(\bar u)|\,d\pi^\varepsilon\to0\).

Finally, \(\phi(\bar u)\) is bounded continuous, so by 1(i) (i.e.\(\pi^\varepsilon\rightharpoonup\Pi\) and \(\Phi_\#\Pi=\bar\pi\)), \[\int_E \phi(\bar u(t,\Phi(z))\,d\pi^\varepsilon\to \int_{\bar E}\phi(\bar u(t,x))\,d\bar\pi(x)=\bar\mathcal{F}(t).\] Combining the last two displays gives \(\mathcal{F}^\varepsilon(t)\to\bar\mathcal{F}(t)\). ◻

Proof. For \(f\in\mathcal{M}\) we have \(P_t^\varepsilon f\in C_b\). Under 2, integrating against \(\pi^\varepsilon\),the \(CD(-\kappa,\infty)\) gradient estimate gives \[\label{eq:I-mono} \int \Gamma^\varepsilon(P_t^\varepsilon f)\,d\pi^\varepsilon \;\le\; e^{2\kappa t}\int P_t^\varepsilon\Gamma^\varepsilon(f)\,d\pi^\varepsilon \;=e^{2\kappa t}\int \Gamma^\varepsilon(f)\,d\pi^\varepsilon \;<\;\infty.\tag{5}\] Hence \(\Gamma^\varepsilon(P_t^\varepsilon f)\in L^1(\pi^\varepsilon)\), so \(P_t^\varepsilon f\in\mathcal{M}\). ◻

Proof. (1) Monotonicity. Define \(\mathcal{G}^\varepsilon(t):=e^{-2\kappa t}\mathcal{I}^\varepsilon(t)\) for \(t>0\). Using 2, Eq. 5 gives \[e^{-2\kappa t}\mathcal{I}^\varepsilon(t+s)\le \mathcal{I}^\varepsilon(s) \qquad (s,t>0).\] Equivalently, \(\mathcal{G}^\varepsilon(t+s)\le \mathcal{G}^\varepsilon(s)\), hence \(t\mapsto \mathcal{G}^\varepsilon(t)\) is nonincreasing on \((0,\infty)\).

(2) Apply Lemma 6. Apply 6 with \[h^\varepsilon=\mathcal{F}^\varepsilon,\qquad h=\bar{\mathcal{F}}.\] By ?? , the weighted derivative in 6 is \[G^\varepsilon(t):=-e^{-2\kappa t}\frac{d}{dt}\mathcal{F}^\varepsilon(t) =e^{-2\kappa t}\mathcal{I}^\varepsilon(t)=\mathcal{G}^\varepsilon(t),\] which is nonincreasing by Step (1). Moreover, \(\mathcal{F}^\varepsilon(t)\to\bar{\mathcal{F}}(t)\) for every \(t>0\) by 2, and \(\bar{\mathcal{F}}\) is differentiable on \((0,\infty)\) with \(\bar{\mathcal{F}}'(t)=-\bar{\mathcal{I}}(t)\) by ?? . Therefore 6 yields, for each fixed \(t>0\), \[\mathcal{G}^\varepsilon(t)\to -e^{-2\kappa t}\bar{\mathcal{F}}'(t)=e^{-2\kappa t}\bar{\mathcal{I}}(t).\] Multiplying by \(e^{2\kappa t}\) gives \(\mathcal{I}^\varepsilon(t)\to \bar{\mathcal{I}}(t)\). ◻

Proof. Fix \(T>0\) and \(\tau\in(0,T)\). By 3 we have \(\mathcal{I}^\varepsilon(\tau)\to \bar\mathcal{I}(\tau)<\infty\), hence \(\sup_{\varepsilon\le \varepsilon_1}\mathcal{I}^\varepsilon(\tau)\le C\) for some \(\varepsilon_1>0\) and \(C<\infty\). By Step (1) in the proof of 3, the function \(\mathcal{G}^\varepsilon(t):=e^{-2\kappa t}\mathcal{I}^\varepsilon(t)\) is nonincreasing on \((0,\infty)\), so for all \(t\in[\tau,T]\), \[\mathcal{I}^\varepsilon(t)=e^{2\kappa t}\mathcal{G}^\varepsilon(t)\le e^{2\kappa t}\mathcal{G}^\varepsilon(\tau)=e^{2\kappa(t-\tau)}\mathcal{I}^\varepsilon(\tau) \le e^{2\kappa(T-\tau)}\,C=:L.\] Therefore, for any \(s,t\in[\tau,T]\), \[|\mathcal{F}^\varepsilon(t)-\mathcal{F}^\varepsilon(s)| =\Big|\int_s^t \mathcal{F}^{\varepsilon\,\prime}(r)\,dr\Big| =\int_s^t \mathcal{I}^\varepsilon(r)\,dr \le L\,|t-s|,\] so \(\{\mathcal{F}^\varepsilon\}_{\varepsilon\le\varepsilon_1}\) is equi-Lipschitz (hence equicontinuous) on \([\tau,T]\). Together with the pointwise convergence \(\mathcal{F}^\varepsilon(t)\to\bar\mathcal{F}(t)\) for every \(t>0\) from 2, this implies uniform convergence on \([\tau,T]\). ◻

Proof. Fix \(\psi\in C_b(\bar E;\mathbb{R}^n)\) and define \[\label{eq:tilde-sigma-hk} \tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi) :=\int_E u^\varepsilon(t,z)\Big( 2\big\langle D\Phi(z)\gamma^\varepsilon(z),\psi(\Phi(z))\big\rangle -\psi(\Phi(z))^\top\big(D\Phi(z)A^\varepsilon(z)D\Phi(z)^\top\big)\psi(\Phi(z)) \Big)\,d\pi^\varepsilon(z).\tag{6}\]

By completion of squares, for every \(\varepsilon>0\), \[\label{eq:cs-split} \sigma_{\mathrm{hk}}^\varepsilon(t) = \tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi) + \int_E u^\varepsilon(t,z)\Big\| \big(A^\varepsilon(z)\big)^{-1}\gamma^\varepsilon(z)-D\Phi(z)^\top\psi(\Phi(z)) \Big\|_{A^\varepsilon(z)}^2\,d\pi^\varepsilon(z),\tag{7}\] hence \[\label{eq:lsc-psi} \liminf_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t) \;\ge\;\liminf_{\varepsilon\to0}\tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi).\tag{8}\]

By 7, \[\tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi)\;\longrightarrow\; 2\!\int_{\bar E}\bar u(t,x)\,\langle \psi(x),\bar\gamma(x)\rangle\,d\bar\pi(x) -\!\int_{\bar E}\bar u(t,x)\,\psi(x)^\top\bar A(x)\psi(x)\,d\bar\pi(x).\] Rewriting the limit by completion of squares gives \[2\langle \psi,\bar\gamma\rangle-\psi^\top\bar A\psi = \bar\gamma^\top\bar A^{-1}\bar\gamma -\big\|\psi-\bar A^{-1}\bar\gamma\big\|_{\bar A}^2,\] thus, for every \(\psi\in C_b(\bar E;\mathbb{R}^n)\), \[\label{eq:lsc-ineq} \liminf_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t) \;\ge\;\bar\sigma_{\mathrm{hk}}(t) -\int_{\bar E}\bar u(t,x)\, \big\|\psi(x)-\bar A(x)^{-1}\bar\gamma(x)\big\|_{\bar A(x)}^2\,d\bar\pi(x).\tag{9}\]

By 9, there exists \((\psi_k)_{k\in\mathbb{N}}\subset C_b^\infty(\bar E;\mathbb{R}^n)\subset C_b(\bar E;\mathbb{R}^n)\) with \[\int_{\bar E}\bar u(t,x)\, \big\|\psi_k(x)-\bar A(x)^{-1}\bar\gamma(x)\big\|_{\bar A(x)}^2\,d\bar\pi(x)\;\longrightarrow\;0.\] Plugging \(\psi=\psi_k\) into 9 and letting \(k\to\infty\) yields the claim. ◻

Proof. For \(\psi\in C_b(\bar E;\mathbb{R}^n)\), completion of squares yields, \[\label{eq:sigma-split-lock} \sigma_{\mathrm{hk}}^\varepsilon(t)=\tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi)+R^\varepsilon(t;\psi),\qquad \forall \varepsilon>0\tag{10}\] where \[\label{eq:R-def} R^\varepsilon(t;\psi) :=\int_E u^\varepsilon(t,z)\, \Big\| \big(A^\varepsilon(z)\big)^{-1}\gamma^\varepsilon(z) - D\Phi(z)^\top \psi\big(\Phi(z)\big) \Big\|_{A^\varepsilon(z)}^2\,d\pi^\varepsilon(z)\;\ge\;0.\tag{11}\] Moreover, by 7, for each fixed \(\psi\in C_b(\bar E;\mathbb{R}^n)\), \[\label{eq:limit-tilde-short} \lim_{\varepsilon\to0}\tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi) =\bar\sigma_{\mathrm{hk}}(t) -\int_{\bar E}\bar u(t,x)\, \big\|\psi(x)-\bar A(x)^{-1}\bar\gamma(x)\big\|_{\bar A(x)}^2\,d\bar\pi(x),\tag{12}\]

[it:locking]\(\Rightarrow\)[it:hk-conv]. Let \((\psi_k)\) satisfy [it:locking]. Fix \(k\) and apply 10 with \(\psi=\psi_k\); taking \(\limsup_{\varepsilon\to0}\) and using 12 gives \[\limsup_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t) \le \bar\sigma_{\mathrm{hk}}(t) -\!\int_{\bar E}\bar u\,\big\|\psi_k-\bar A^{-1}\bar\gamma\big\|_{\bar A}^2\,d\bar\pi +\limsup_{\varepsilon\to0}R^\varepsilon(t;\psi_k).\] Let \(k\to\infty\) and use [def:recovery,ass:locking] to conclude \(\limsup_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t)\le \bar\sigma_{\mathrm{hk}}(t)\). Together with 4, this implies [it:hk-conv].

[it:hk-conv]\(\Rightarrow\)[it:locking]. Let \((\psi_k)\) be any recovery sequence at time \(t\) (cf.4) and set \[\delta_k:=\int_{\bar E}\bar u(t,x)\, \big\|\psi_k(x)-\bar A(x)^{-1}\bar\gamma(x)\big\|_{\bar A(x)}^2\,d\bar\pi(x)\xrightarrow[k\to\infty]{}0.\] By 12 , \(\tilde{\sigma}_{\mathrm{hk}}^\varepsilon(t;\psi_k)\to\bar\sigma_{\mathrm{hk}}(t)-\delta_k\). Assuming [it:hk-conv] and subtracting in 10 yields \(R^\varepsilon(t;\psi_k)\to\delta_k\), hence \(\limsup_{\varepsilon\to0}R^\varepsilon(t;\psi_k)=\delta_k\). Letting \(k\to\infty\) gives ?? . ◻

Proof. Fix \(t>0\). Using \(\sigma^\varepsilon(t)=\sigma_{\mathrm{hk}}^\varepsilon(t)+\mathcal{I}^\varepsilon(t)\) and \(\bar\sigma(t)=\bar\sigma_{\mathrm{hk}}(t)+\bar\mathcal{I}(t)\), together with \[\liminf_{\varepsilon\downarrow0}\sigma_{\mathrm{hk}}^\varepsilon(t)\ge\bar\sigma_{\mathrm{hk}}(t) \quad \qquad\text{and}\qquad \mathcal{I}^\varepsilon(t)\to\bar\mathcal{I}(t)\] proved in [thm:FI-conv,thm:hk-lsc] yields \(\liminf_{\varepsilon\downarrow0}\sigma^\varepsilon(t)\ge\bar\sigma(t)\).

By 8, \(\bar\sigma_{\mathrm{hk,ss}}<\infty\). Since \(\bar\sigma_{\mathrm{hk}}(t)\) differs from \(\bar\sigma_{\mathrm{hk,ss}}\) only by the additional weight \(\bar u(t,\cdot)\), which is bounded for \(t>0\) by the Markov property, we have \(\bar\sigma_{\mathrm{hk}}(t)<\infty\). Hence \[\sigma^\varepsilon(t)\to\bar\sigma(t)\quad\Longleftrightarrow\quad \sigma_{\mathrm{hk}}^\varepsilon(t)\to\bar\sigma_{\mathrm{hk}}(t),\] and the latter is equivalent to the locking condition at time \(t\) by 5. ◻

Proof. Clearly \(|(\mathcal{P}f)(y)|\le \|f\|_\infty\), so \(\mathcal{P}f\) is bounded. Let \(y_k\to y\) and fix \(\delta>0\). Choose \(R>0\) such that \(\mu^y(\{|x|>R\})\le \delta\). By weak continuity \(\mu^{y_k}\Rightarrow\mu^y\) and Portmanteau (for the closed set \(\{|x|\ge R\}\)), we have \(\mu^{y_k}(\{|x|>R\})\le \delta\) for all \(k\) large enough.

Decompose \[(\mathcal{P}f)(y_k)-(\mathcal{P}f)(y) =\int_{|x|\le R}\!\!\big(f(x,y_k)-f(x,y)\big)\,\mu^{y_k}(dx) +\int_{|x|\le R}\!\! f(x,y)\,(\mu^{y_k}-\mu^{y})(dx) +\mathrm{Tail}_k,\] where \(\mathrm{Tail}_k:=\int_{|x|>R} f(x,y_k)\,d\mu^{y_k}-\int_{|x|>R} f(x,y)\,d\mu^{y}\). The first term tends to \(0\), since \(f(\cdot,y_k)\to f(\cdot,y)\) uniformly on \(\{|x|\le R\}\).

For the second term, \(g_R(x):=f(x,y)\mathbf{1}_{\{|x|\le R\}}\) is bounded and is \(\mu^y\)–a.e.continuous because \(\mu^y(\{|x|=R\})=0\) (density assumption). Hence \(\int g_R\,d\mu^{y_k}\to\int g_R\,d\mu^y\).

Finally, \[|\mathrm{Tail}_k|\le \|f\|_\infty\big(\mu^{y_k}(\{|x|>R\})+\mu^y(\{|x|>R\})\big)\le 2\|f\|_\infty\,\delta\] for \(k\) large. Since \(\delta\) is arbitrary, \((\mathcal{P}f)(y_k)\to(\mathcal{P}f)(y)\). ◻

Proof. See 12. ◻

Proof. Let \(\pi(dx,dy)=\pi^y(dx)\,\tilde{\pi}(dy)\) be a regular conditional disintegration. Since \(\pi\) is invariant for \(\mathcal{L}^\varepsilon\), for \(\phi\in C_c^\infty(\mathbb{R}^{d_x})\), \(\psi\in C_c^\infty(\mathbb{R}^{d_y})\), \[0=\int \mathcal{L}^\varepsilon(\phi\psi)\,d\pi =\frac{1}{\varepsilon}\int \psi(y)\,\mathcal{L}_y^{\mathrm{fast}}\phi(x)\,d\pi +\int \phi(x)\,\mathcal{L}^{\mathrm{slow}}\psi(y)\,d\pi.\] Multiplying by \(\varepsilon\) and letting \(\varepsilon\downarrow0\) gives \[\int \psi(y)\,\mathcal{L}_y^{\mathrm{fast}}\phi(x)\,d\pi=0,\] hence, by disintegration and arbitrariness of \(\psi\), \[\int_{\mathbb{R}^{d_x}} \mathcal{L}_y^{\mathrm{fast}}\phi(x)\,\pi^y(dx)=0 \quad\text{for \tilde{\pi}-a.e.\;}y,\;\forall\,\phi\in C_c^\infty(\mathbb{R}^{d_x}).\] Thus \(\pi^y\) is \(\mathcal{L}_y^{\mathrm{fast}}\)-invariant for \(\tilde{\pi}\)-a.e.\(y\), and by uniqueness of the fast invariant law, \(\pi^y=\mu^y\) \(\tilde{\pi}\)-a.e., proving \(\pi(dx,dy)=\mu^y(dx)\,\tilde{\pi}(dy)\).

Finally, for \(\varphi\in C_c^\infty(\mathbb{R}^{d_y})\), invariance of \(\pi\) applied to \(\varphi\circ\Phi\) yields \[0=\int_E \mathcal{L}^\varepsilon(\varphi\circ\Phi)\,d\pi.\] Using \(\bar{\mathcal{L}}\varphi=\mathcal{P}(\mathcal{L}^\varepsilon(\varphi\circ\Phi))\) and the disintegration above, \[\int_{\mathbb{R}^{d_y}} \bar{\mathcal{L}}\varphi\,d\tilde{\pi} =\int_{\mathbb{R}^{d_y}}\Big(\int_{\mathbb{R}^{d_x}}\mathcal{L}^\varepsilon(\varphi\circ\Phi)(x,y)\,\mu^y(dx)\Big)\tilde{\pi}(dy) =\int_E \mathcal{L}^\varepsilon(\varphi\circ\Phi)\,d\pi =0.\] Hence \(\tilde{\pi}\) is invariant for \(\bar{\mathcal{L}}\), and uniqueness of the averaged invariant measure gives \(\tilde{\pi}=\bar\pi\). ◻

Proof. Since 21 yields \(\pi^\varepsilon\equiv\pi\) and \(\gamma^\varepsilon\equiv\gamma\), the static part of 1(i) reduces to identifying the \(\Phi\)–push-forward of \(\pi\) and its disintegration along the fibres. Under the uniqueness of the frozen fast invariant laws, 5 gives \(\pi(dx,dy)=\mu^y(dx)\,\bar\pi(dy)\) and thus 1(i). The definition of \(\mathcal{P}\) in terms of \(\mu^y\) together with weak continuity of \(y\mapsto\mu^y\) gives 1(ii). Finally, ?? is exactly 1(iii).

For 3, note that \(D\Phi\,\gamma^\varepsilon=\gamma_y\) and \(D\Phi A^\varepsilon D\Phi^\top=a_2\), hence \[J^\varepsilon=\Phi_\#(\gamma_y\,\pi)=\bar\gamma\bar\pi,\qquad Q^\varepsilon=a_2\,\pi,\] and the finiteness and weak convergence assertions follow from 7 and the disintegration \(\pi(dx,dy)=\mu^y(dx)\,\bar\pi(dy)\). ◻

Proof. In this subclass, work under the assumptions, fix \(R>0\). Splitting the integral over \(\{|(x,y)|\le R\}\) and its complement, we get \[\begin{align} &\int_E |u^\varepsilon(t,x,y)-\bar u(t,y)|\,\gamma_y^\top a_2^{-1}\gamma_y\,d\pi \\ &\le \sup_{|(x,y)|\le R}|u^\varepsilon(t,x,y)-\bar u(t,y)| \int_{|(x,y)|\le R}\gamma_y^\top a_2^{-1}\gamma_y\,d\pi +2\|f\|_\infty\int_{|(x,y)|> R}\gamma_y^\top a_2^{-1}\gamma_y\,d\pi. \end{align}\] The first term \(\to0\) as \(\varepsilon\to0\) by local uniform convergence, while the second term \(\to0\) as \(R\to\infty\) by 7. Hence \[\lim_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t) =\int_E \bar u(t,y)\,\gamma_y(x,y)^\top a_2(x,y)^{-1}\gamma_y(x,y)\,d\pi(x,y).\] Disintegrating \(\pi(dx,dy)=\mu^y(dx)\,\bar\pi(dy)\) and using \(\bar u(t,\Phi(x,y))=\bar u(t,y)\), \[\lim_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t) =\int \bar u(t,y)\Big(\int \gamma_y^\top a_2^{-1}\gamma_y\,\mu^y(dx)\Big)\,d\bar\pi(y).\] By fibrewise Jensen for the jointly convex map \((A,g)\mapsto g^\top A^{-1}g\), \[\int \gamma_y^\top a_2^{-1}\gamma_y\,\mu^y(dx) \ge \bar\gamma_y(y)^\top \bar A(y)^{-1}\bar\gamma_y(y)\quad\text{for \bar\pi--a.e.\;}y,\] hence \(\lim_{\varepsilon\to0}\sigma_{\mathrm{hk}}^\varepsilon(t)\ge \bar\sigma_{\mathrm{hk}}(t)\). Moreover, equality holds iff \(a_2(x,y)^{-1}\gamma_y(x,y)=\bar A(y)^{-1}\bar\gamma_y(y)\) holds \(\mu^y\)–a.s.for \(\bar\pi\)–a.e.\(y\), equivalently \(\pi\)–a.e.\((x,y)\). ◻

Proof. Let \(r:=x-Hy-b\). The fast flow \(\dot{z}=-\nabla U(z)\) satisfies \(\dot{r}=-(I+HH^\top)B\,r\), hence contracts exponentially onto \(M=\{r=0\}\). Moreover, \(y+H^\top x\) is conserved, which yields the global asymptotic phase map \(\Phi\) and projection \(\mathrm{proj}=\iota\circ\Phi\). Therefore the hypotheses of [20] apply to the large-drift SDE 23 with stable manifold \(M\), and imply weak convergence in path space to a diffusion constrained to \(M\). Transporting the constrained diffusion through the chart \(\Phi|_M:M\to\bar E\) gives the limiting dynamics Eqs. 25 ,26 . In particular, for each fixed \(t>0\) we have \(Z_t^\varepsilon \Rightarrow \iota(U_t)\) in distribution. Therefore, for every \(f\in C_b(E)\), \[P_t^\varepsilon f(z)=\mathbb{E}_z[f(Z_t^\varepsilon)] \to \mathbb{E}_{\Phi(z)}[f(\iota(U_t))]=\bar P_t(\mathcal{P}f)(\Phi(z)).\] ◻

Proof. The uniform \(CD(-\kappa,\infty)\) bound follows from 9 by the Ricci/Hessian criterion for reversible diffusions with constant diffusion matrix \(I\) (cf.9).

Fix \(t>0\) and \(f\in C_b^1(E)\). By the gradient commutation estimate under \(CD(-\kappa,\infty)\) (see, e.g., [29]), \[\Gamma(P_t^\varepsilon f)\le e^{2\kappa t}\,P_t^\varepsilon\Gamma(f)\le e^{2\kappa t}\,\|\nabla f\|_\infty^2,\] and since here \(\Gamma(g)=|\nabla g|^2\), we obtain the uniform Lipschitz bound \[\|\nabla P_t^\varepsilon f\|_\infty\le e^{\kappa t}\,\|\nabla f\|_\infty=:L,\] so \(\{P_t^\varepsilon f\}_{\varepsilon\in(0,\varepsilon_0]}\) is equi-Lipschitz on \(E\).

Let \(K\subset E\) be compact and set \(u_\varepsilon:=P_t^\varepsilon f\). By equi-Lipschitzness and pointwise convergence 27 , the limit \[F(z):=\bar P_t(\mathcal{P} f)\big(\Phi(z)\big)\] is \(L\)–Lipschitz on \(K\) (hence uniformly continuous). In particular, its modulus of continuity \(\omega_F(\delta):=\sup\{|F(z)-F(z')|:\;z,z'\in K,\;|z-z'|\le\delta\}\) satisfies \(\omega_F(\delta)\to0\) as \(\delta\downarrow0\).

Given \(\delta>0\), choose a finite \(\delta\)-net \(\{z_i\}_{i=1}^N\subset K\). For any \(z\in K\), pick \(i\) with \(|z-z_i|\le \delta\), and write \[\begin{align} |u_\varepsilon(z)-F(z)| &\le |u_\varepsilon(z)-u_\varepsilon(z_i)| + |u_\varepsilon(z_i)-F(z_i)| +|F(z_i)-F(z)| \\ &\le L\,\delta + \max_{1\le i\le N}|u_\varepsilon(z_i)-F(z_i)| + \omega_F(\delta). \end{align}\] Letting \(\varepsilon\downarrow0\) and using 27 at the finitely many points \(z_i\), and then letting \(\delta\downarrow0\), yields the claim. ◻

Proof. Fix \(t>0\) and \(g\in C_b^1(\bar E)\). Consider the lift \(f:=g\circ\Phi\in C_b^1(E)\). Invariance of \(\pi^\varepsilon\) gives \[\int_E P_t^\varepsilon (g\circ\Phi)\,d\pi^\varepsilon=\int_E (g\circ\Phi)\,d\pi^\varepsilon.\] Let \(\varepsilon\downarrow0\). By [prop:stiff-uniform] (applied to \(f=g\circ\Phi\)) we have \[P_t^\varepsilon(g\circ\Phi)(z)\to \bar P_t(\mathcal{P}(g\circ\Phi))\big(\Phi(z)\big)=\bar P_t g\big(\Phi(z)\big)\] uniformly on compacts, and by 28 the measures \(\pi^\varepsilon\) are tight. Passing to the limit yields \[\int_E \bar P_t g(\Phi(z))\,\Pi(dz)=\int_E g(\Phi(z))\,\Pi(dz),\] i.e.\(\Phi_\#\Pi\) is invariant for \(\bar P_t\) on the separating class \(C_b^1(\bar E)\). By uniqueness of the invariant probability measure for \((\bar P_t)\) (assumed in 2), we conclude \(\Phi_\#\Pi=\bar\pi\).

Finally, since \(\Pi\) is supported on \(M\) and \(\Phi|_M:M\to\bar E\) is a bijection with inverse \(\iota\), we have \[\Pi=\iota_\#(\Phi_\#\Pi)=\iota_\#\bar\pi.\] The last statement \(\bar\pi=\bar\pi_0\) follows by combining \(\Phi_\#\Pi=\bar\pi\) with 29 . ◻

Proof. Fix \(\varepsilon\in(0,\varepsilon_0]\), \(z\in E\), and \(u\in\mathbb{R}^N\). Let \(Z_t^{z,\varepsilon}\) denote the solution started from \(z\). By Kunita’s \(C^1\)-flow theory [31], the map \(z\mapsto Z_t^{z,\varepsilon}\) is differentiable and \[\delta Z_t(z;u):=\lim_{h\to0}\frac{Z_t^{z+hu,\varepsilon}-Z_t^{z,\varepsilon}}{h} =:J_t(z)u \quad\text{exists in }L^2(\Omega).\] Apply ?? with \((z_1,z_2)=(z,z+hu)\) and divide by \(h^2\). Letting \(h\to0\) and using Fatou’s lemma yields \[\label{eq:deriv-energy-G} \mathbb{E}\Big[ \delta Z_t(z;u)^\top \big(A^\varepsilon(Z_t^{z,\varepsilon})\big)^{-1} \delta Z_t(z;u) \Big] \;\le\; e^{2\rho t}\,u^\top \big(A^\varepsilon(z)\big)^{-1}u.\tag{32}\]

For \(f\in\mathcal{M}\) we use the gradient representation (see, e.g., [32]) \[\nabla P_t^\varepsilon f(z)=\mathbb{E}\big[J_t(z)^\top\nabla f(Z_t^{z,\varepsilon})\big],\] hence \[u^\top\nabla P_t^\varepsilon f(z)=\mathbb{E}\big[\langle \delta Z_t(z;u),\nabla f(Z_t^{z,\varepsilon})\rangle\big].\] Applying Cauchy–Schwarz inequality with the random inner product \(\langle\xi,\eta\rangle:=\xi^\top \big(A^\varepsilon(Z_t^{z,\varepsilon})\big)^{-1}\eta\), we obtain \[\big(u^\top\nabla P_t^\varepsilon f(z)\big)^2 \le \mathbb{E}\big[\delta Z_t(z;u)^\top \big(A^\varepsilon(Z_t^{z,\varepsilon})\big)^{-1}\delta Z_t(z;u)\big]\, \mathbb{E}\big[\nabla f(Z_t^{z,\varepsilon})^\top A^\varepsilon(Z_t^{z,\varepsilon})\nabla f(Z_t^{z,\varepsilon})\big].\] Combining this with 32 and the definition of \(\Gamma^\varepsilon\) yields \[\big(u^\top\nabla P_t^\varepsilon f(z)\big)^2 \le e^{2\rho t}\,u^\top \big(A^\varepsilon(z)\big)^{-1}u\, \big(P_t^\varepsilon\Gamma^\varepsilon(f)\big)(z).\]

Finally, recall the quadratic representation \[\Gamma^\varepsilon(P_t^\varepsilon f)(z) =\nabla P_t^\varepsilon f(z)^\top A^\varepsilon(z)\nabla P_t^\varepsilon f(z) =\sup_{u\neq0}\frac{\big(u^\top\nabla P_t^\varepsilon f(z)\big)^2}{u^\top \big(A^\varepsilon(z)\big)^{-1}u}.\] Taking the supremum over \(u\neq0\) in the previous bound gives ?? . ◻

Proof. Step 1: duality identity. Fix \(\xi\in C_c^1(\bar E;\mathbb{R}^n)\). Using \(\gamma^\varepsilon\pi^\varepsilon=b^\varepsilon\pi^\varepsilon-\nabla\!\cdot(A^\varepsilon\pi^\varepsilon)\) and integration by parts, \[\int_E\!\langle \xi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon =\int_E\!\langle \xi(\Phi),D\Phi b^\varepsilon\rangle\,d\pi^\varepsilon +\int_E (A^\varepsilon\pi^\varepsilon):\nabla\!\big(D\Phi^\top\xi(\Phi)\big)\,dz.\] Since \(\Phi\) is affine, \(\nabla\!\big(D\Phi^\top\xi(\Phi)\big)=D\Phi^\top(\nabla\xi)(\Phi)\,D\Phi\), so \[\langle J^\varepsilon,\xi\rangle =\langle B^\varepsilon,\xi\rangle +\int_E\mathrm{tr}\!\big((\nabla\xi)(\Phi(z))\,D\Phi A^\varepsilon(z)D\Phi^\top\big)\,d\pi^\varepsilon(z) =\langle B^\varepsilon,\xi\rangle+\big\langle \Phi_\# Q^\varepsilon,\nabla\xi\big\rangle,\] where \(\langle \Phi_\# Q^\varepsilon,\nabla\xi\rangle:=\int_{\bar E}\mathrm{tr}\big((\nabla\xi)(x)\,d(\Phi_\#Q^\varepsilon)(x)\big)\).

Step 2: passage to the limit on \(C_c^1\). By assumption, \(B^\varepsilon\rightharpoonup \bar B\) in \(\mathcal{M}(\bar E;\mathbb{R}^n)\), and 3(ii) implies \(\Phi_\#Q^\varepsilon\rightharpoonup \Phi_\#Q=\bar Q=\bar A\bar\pi\) in \(\mathcal{M}(\bar E;\mathbb{S}_+^n)\). Passing to the limit in the previous identity yields \(\langle J^\varepsilon,\xi\rangle\to \langle \bar J,\xi\rangle\) for all \(\xi\in C_c^1(\bar E;\mathbb{R}^n)\), where \[\langle \bar J,\xi\rangle:=\langle \bar B,\xi\rangle+\langle \bar Q,\nabla\xi\rangle .\]

Step 3: extension to \(C_b\). By 1 we have \(\nu^\varepsilon:=\Phi_\#\pi^\varepsilon\rightharpoonup\bar\pi\), hence \((\nu^\varepsilon)_\varepsilon\) is tight. Moreover, by Cauchy–Schwarz inequality and 3(iii), \[|J^\varepsilon|(A) =\int_{\Phi^{-1}(A)}|D\Phi\,\gamma^\varepsilon|\,d\pi^\varepsilon \le \Big(\int_E |D\Phi\,\gamma^\varepsilon|^2\,d\pi^\varepsilon\Big)^{1/2}\,\nu^\varepsilon(A)^{1/2} \le C\,\nu^\varepsilon(A)^{1/2},\] so \((J^\varepsilon)_\varepsilon\) is tight and \(\sup_\varepsilon|J^\varepsilon|(\bar E)<\infty\). Hence every subsequence admits a further weakly convergent subsequence in \(\mathcal{M}(\bar E;\mathbb{R}^n)\). Step 2 identifies the unique possible limit on \(C_c^1\), so \(J^\varepsilon\rightharpoonup \bar J\) in \(\mathcal{M}(\bar E;\mathbb{R}^n)\), i.e.3(i). ◻

Proof. Step 1: tail control. Let \(T_M\) be the radial truncation and \(\bar F_M:=T_M(\bar F)\in C_b(\bar E;\mathbb{R}^n)\). Since \(\bar F_M=\alpha_M\bar F\) with \(\alpha_M\in[0,1]\), for any symmetric \(Q\ge0\), \[\|\bar F-\bar F_M\|_{Q}^2 \le \bar F^\top Q\bar F-\bar F_M^\top Q\bar F_M .\] Hence \[\int_E u^\varepsilon\,\|\bar F(\Phi)-\bar F_M(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon \le \int_E u^\varepsilon\,\bar F(\Phi)^\top Q^\varepsilon\bar F(\Phi)\,d\pi^\varepsilon - \int_E u^\varepsilon\,\bar F_M(\Phi)^\top Q^\varepsilon\bar F_M(\Phi)\,d\pi^\varepsilon.\] Taking \(\limsup_{\varepsilon\to0}\), using Eq. 33 for the first term and Eq. ?? with \(\psi=\bar F_M\) for the second term yields \[\limsup_{\varepsilon\to0}\int_E u^\varepsilon\,\|\bar F(\Phi)-\bar F_M(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon \le \int_{\bar E}\bar u\,\big(\bar F^\top\bar A\bar F-\bar F_M^\top\bar A\bar F_M\big)\,d\bar\pi.\] Since \(\bar F_M^\top\bar A\bar F_M\uparrow \bar F^\top\bar A\bar F\) and the RHS of Eq. 33 is finite, monotone convergence gives \[\label{eq:tail-Q} \lim_{M\to\infty}\;\limsup_{\varepsilon\to0}\int_E u^\varepsilon(t,z)\,\|\bar F(\Phi(z))-\bar F_M(\Phi(z))\|_{Q^\varepsilon(z)}^2\,d\pi^\varepsilon(z)=0.\tag{35}\]

Step 2: Eq. eq. ¿eq:eq:canonical-locking? \(\Rightarrow\) 4. From the definition in Eq. 11 and the above estimate, \[R^\varepsilon(t;\bar F_M)\;\le\;2R^\varepsilon(t;\bar F) +2\int_E u^\varepsilon\,\|\bar F(\Phi)-\bar F_M(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon.\] Taking \(\limsup_{\varepsilon\to0}\) and then \(M\to\infty\), Eq. 34 and Eq. 35 imply \(\lim_{M\to\infty}\limsup_{\varepsilon\to0}R^\varepsilon(t;\bar F_M)=0\). Moreover, \((\bar F_M)\) is a recovery sequence in the sense of 4 since \(\int \bar u\,\|\bar F_M-\bar F\|_{\bar A}^2\,d\bar\pi\to0\) (monotone convergence). Thus Eq. ?? holds with \(\psi_k=\bar F_k\), i.e.4 holds.

Step 3: 4\(\Rightarrow\) Eq. eq. ¿eq:eq:canonical-locking? . Let \((\psi_k)\subset C_b\) be a recovery sequence such that Eq. ?? holds. Fix \(M\) and write again \(\bar F_M\). By repeated use of the elementary estimate, \[R^\varepsilon(t;\bar F) \;\le\;2R^\varepsilon(t;\psi_k) +4\int_E u^\varepsilon\,\|\psi_k(\Phi)-\bar F_M(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon +4\int_E u^\varepsilon\,\|\bar F_M(\Phi)-\bar F(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon.\] Take \(\limsup_{\varepsilon\to0}\). The first term vanishes as \(k\to\infty\) by Eq. ?? . The last term can be made arbitrarily small by Eq. 35 (choose \(M\) large). For the middle term, since \(\psi_k-\bar F_M\in C_b\), Eq. ?? yields \[\lim_{\varepsilon\to0}\int_E u^\varepsilon\,\|\psi_k(\Phi)-\bar F_M(\Phi)\|_{Q^\varepsilon}^2\,d\pi^\varepsilon = \int_{\bar E}\bar u\,\|\psi_k-\bar F_M\|_{\bar A}^2\,d\bar\pi.\] Using the recovery property from 4, \[\limsup_{k\to\infty}\int_{\bar E}\bar u\,\|\psi_k-\bar F_M\|_{\bar A}^2\,d\bar\pi \le 2\int_{\bar E}\bar u\,\|\bar F-\bar F_M\|_{\bar A}^2\,d\bar\pi,\] and the RHS tends to \(0\) as \(M\to\infty\) (monotone convergence). Letting \(k\to\infty\) and then \(M\to\infty\) yields Eq. 34 . ◻

Proof. Fix \(t_0>0\) and \(r>0\). Since \(h^{\varepsilon\,\prime}(t)=-e^{2\kappa t}G^\varepsilon(t)\) and \(G^\varepsilon\) is nonincreasing, \[G^\varepsilon(t_0+r)\le G^\varepsilon(t)\le G^\varepsilon(t_0)\qquad(t\in[t_0,t_0+r]).\] Integrating \(h^{\varepsilon\,\prime}\) over \([t_0,t_0+r]\) gives the squeeze \[-\,G^\varepsilon(t_0)\,A_\kappa(t_0,r) \;\le\; \frac{h^\varepsilon(t_0+r)-h^\varepsilon(t_0)}{r} \;\le\; -\,G^\varepsilon(t_0+r)\,A_\kappa(t_0,r),\] where \[A_\kappa(t_0,r):=\frac{1}{r}\int_{t_0}^{t_0+r}e^{2\kappa t}\,dt = \begin{cases} e^{2\kappa t_0}, & \kappa=0,\\[2mm] \dfrac{e^{2\kappa(t_0+r)}-e^{2\kappa t_0}}{2\kappa r}, & \kappa>0. \end{cases}\] Letting \(\varepsilon\to0\) and using \(h^\varepsilon\to h\) pointwise yields the same squeeze with \(h\) in place of \(h^\varepsilon\). Then letting \(r\downarrow0\) gives \(A_\kappa(t_0,r)\to e^{2\kappa t_0}\) and, if \(h\) is differentiable at \(t_0\), \[\frac{h(t_0+r)-h(t_0)}{r}\to h'(t_0).\] This forces \(\lim_{\varepsilon\to0}G^\varepsilon(t_0)=-e^{-2\kappa t_0}h'(t_0)\), hence \(h^{\varepsilon\,\prime}(t_0)=-e^{2\kappa t_0}G^\varepsilon(t_0)\to h'(t_0)\). ◻

Proof. Fix \(t>0\) and \(\psi\in C_b(\bar E;\mathbb{R}^n)\). Set \[r^\varepsilon(z):=u^\varepsilon(t,z)-\bar u(t,\Phi(z)),\qquad Q^\varepsilon(z):=D\Phi(z)A^\varepsilon(z)D\Phi(z)^\top .\] Then \(|r^\varepsilon|\le 2\|f\|_\infty\), and \(r^\varepsilon\to0\) locally uniformly by 1(iii). Moreover, define the finite measures \[\mu^\varepsilon(dz):=\mathrm{tr}(Q^\varepsilon(z))\,d\pi^\varepsilon(z).\] By 3, the family \((\mu^\varepsilon)_\varepsilon\) is tight and has uniformly bounded total mass.

Cross term. Decompose \[\int_E u^\varepsilon\,\langle \psi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon = \int_E \bar u(t,\Phi)\,\langle \psi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon +\int_E r^\varepsilon\,\langle \psi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon.\] For the first term, \(J^\varepsilon\rightharpoonup\bar J\) and \(\xi(x):=\bar u(t,x)\psi(x)\in C_b(\bar E;\mathbb{R}^n)\) yield \[\int_E \bar u(t,\Phi)\,\langle \psi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon \longrightarrow \int_{\bar E}\bar u(t,x)\,\langle \psi(x),\bar\gamma(x)\rangle\,d\bar\pi(x).\] For the remainder, the Cauchy–Schwarz inequality and the definition of \(\mathcal{J}_{\mathrm{hk,proj}}^\varepsilon\) give \[\Big|\int_E r^\varepsilon\,\langle \psi(\Phi),D\Phi\gamma^\varepsilon\rangle\,d\pi^\varepsilon\Big| \le \Big(\int_E |r^\varepsilon|^2\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon\Big)^{1/2} \big(\mathcal{J}_{\mathrm{hk,proj}}^\varepsilon\big)^{1/2}.\] Since \(\sup_\varepsilon\mathcal{J}_{\mathrm{hk,proj}}^\varepsilon<\infty\), it suffices to show \[S^\varepsilon:=\int_E |r^\varepsilon|^2\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon\longrightarrow 0.\] Using \(\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\le \|\psi\|_\infty^2\,\mathrm{tr}(Q^\varepsilon)\), we have \(S^\varepsilon\le \|\psi\|_\infty^2\int_E |r^\varepsilon|^2\,d\mu^\varepsilon\). Fix \(\delta>0\). By tightness of \((\mu^\varepsilon)_\varepsilon\), choose a compact \(K\subset E\) such that \(\sup_\varepsilon\mu^\varepsilon(K^c)\le\delta\). Then \[\int_E |r^\varepsilon|^2\,d\mu^\varepsilon =\int_K |r^\varepsilon|^2\,d\mu^\varepsilon+\int_{K^c} |r^\varepsilon|^2\,d\mu^\varepsilon \le (\sup_K|r^\varepsilon|^2)\mu^\varepsilon(K)+4\|f\|_\infty^2\,\mu^\varepsilon(K^c).\] As \(\varepsilon\to0\), \(\sup_K|r^\varepsilon|\to0\) and \(\sup_\varepsilon\mu^\varepsilon(K)<\infty\), hence the first term \(\to0\). Moreover the second term is bounded by \(4\|f\|_\infty^2\delta\). Therefore \[\limsup_{\varepsilon\to0} \int_E |r^\varepsilon|^2\,d\mu^\varepsilon\le 4\|f\|_\infty^2\delta,\] and since \(\delta\) is arbitrary, \(\int_E |r^\varepsilon|^2\,d\mu^\varepsilon\to0\), hence \(S^\varepsilon\to0\) and ?? follows.

Quadratic term. Similarly, \[\int_E u^\varepsilon\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon = \int_E \bar u(t,\Phi)\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon +\int_E r^\varepsilon\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon.\] The remainder satisfies \[\Big|\int_E r^\varepsilon\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon\Big| \le \|\psi\|_\infty^2 \int_E |r^\varepsilon|\,d\mu^\varepsilon,\] and the same compact/tail argument (with \(|r^\varepsilon|\le 2\|f\|_\infty\)) shows it tends to \(0\).

For the main term, set \[\eta_t(z):=\bar u(t,\Phi(z))\,\psi(\Phi(z))\psi(\Phi(z))^\top \in C_b(E;\mathbb{S}_+^n).\] Then \[\int_E \bar u(t,\Phi)\,\psi(\Phi)^\top Q^\varepsilon\psi(\Phi)\,d\pi^\varepsilon =\int_E \mathrm{tr}(\eta_t Q^\varepsilon)\,d\pi^\varepsilon\longrightarrow \int_E \mathrm{tr}(\eta_t Q)\,d\pi\] by \(Q^\varepsilon\rightharpoonup Q\). Using \(\Phi_\# Q=\bar A\,\bar\pi\) yields \[\int_E \mathrm{tr}(\eta_t Q)\,d\pi =\int_{\bar E}\bar u(t,x)\,\psi(x)^\top\bar A(x)\psi(x)\,d\bar\pi(x),\] which is ?? . ◻

Proof. For \(\psi\in C_b(\bar E;\mathbb{R}^n)\) define the steady variational functional \[\tilde{\sigma}_{\mathrm{hk,ss}}(\psi) :=2\int_{\bar E}\langle \psi(x),\bar\gamma(x)\rangle\,d\bar\pi(x) -\int_{\bar E}\psi(x)^\top\bar A(x)\psi(x)\,d\bar\pi(x),\] and its microscopic counterpart \[\tilde{\sigma}_{\mathrm{hk,ss}}^\varepsilon(\psi) :=2\int_E\!\big\langle \psi(\Phi(z)),D\Phi(z)\gamma^\varepsilon(z)\big\rangle\,d\pi^\varepsilon(z) -\int_E\!\psi(\Phi(z))^\top\!\big(D\Phi(z)A^\varepsilon(z)D\Phi(z)^\top\big)\psi(\Phi(z))\,d\pi^\varepsilon(z).\]

Step 1: \(\tilde{\sigma}_{\mathrm{hk,ss}}\) is uniformly bounded above on \(C_b\). By pointwise completion of squares, \[2\langle D\Phi\,\gamma^\varepsilon,\psi(\Phi)\rangle -\psi(\Phi)^\top(D\Phi A^\varepsilon D\Phi^\top)\psi(\Phi) \le (D\Phi\,\gamma^\varepsilon)^\top(D\Phi A^\varepsilon D\Phi^\top)^{-1}(D\Phi\,\gamma^\varepsilon),\] hence \(\tilde{\sigma}_{\mathrm{hk,ss}}^\varepsilon(\psi)\le \mathcal{J}_{\mathrm{hk,proj}}^\varepsilon\) for all \(\psi\in C_b(\bar E;\mathbb{R}^n)\). Moreover, by 7 applied with the constant observable \(f\equiv 1\) (so \(u^\varepsilon\equiv \bar u\equiv 1\)), we have \(\tilde{\sigma}_{\mathrm{hk,ss}}^\varepsilon(\psi)\to \tilde{\sigma}_{\mathrm{hk,ss}}(\psi)\) for each fixed \(\psi\in C_b(\bar E;\mathbb{R}^n)\). Therefore, with \(C:=\sup_{\varepsilon>0}\mathcal{J}_{\mathrm{hk,proj}}^\varepsilon<\infty\), \[\label{eq:sig95tilde95bounded} \tilde{\sigma}_{\mathrm{hk,ss}}(\psi)\le C\qquad \forall\,\psi\in C_b(\bar E;\mathbb{R}^n), \qquad\text{so}\qquad \sup_{\psi\in C_b}\tilde{\sigma}_{\mathrm{hk,ss}}(\psi)\le C<\infty.\tag{36}\]

Step 2: contradiction if \(\bar\sigma_{\mathrm{hk,ss}}=\infty\). Assume for contradiction that \[\bar\sigma_{\mathrm{hk,ss}} =\int_{\bar E}\bar\gamma^\top\bar A^{-1}\bar\gamma\,d\bar\pi =\int_{\bar E} \bar F^\top \bar A \bar F\,d\bar\pi =+\infty, \qquad \bar F:=\bar A^{-1}\bar\gamma.\] Fix \(L>0\). Since \(\bar F^\top\bar A\bar F\ge0\) and \((B_R\cap\{|\bar F|\le M\})_{R,M}\) increases to \(\bar E\), by monotone convergence we can choose \(R,M\) such that \[\label{eq:RM95choice95sig95tilde} \int_{B_R\cap\{|\bar F|\le M\}} \bar F^\top\bar A\bar F\,d\bar\pi\ge 4L.\tag{37}\] Let \(\chi\in C_c^\infty(\mathbb{R}^n)\) satisfy \(0\le\chi\le 1\), \(\chi\equiv 1\) on \(B_R\), \(\mathrm{supp}\chi\subset B_{R+1}\), and let \(T_M\) be the radial truncation. Set \[h:=\chi\,T_M(\bar F).\] Writing \(T_M(\bar F)=\alpha \bar F\) with \(\alpha\in[0,1]\), we have pointwise \[2\langle \bar\gamma,h\rangle-h^\top\bar A h =\chi(2\alpha-\chi\alpha^2)\,\bar F^\top\bar A \bar F \ge \chi\,\mathbf{1}_{\{|\bar F|\le M\}}\,\bar F^\top\bar A \bar F,\] hence \[\label{eq:sig95tilde95h95lower} \tilde{\sigma}_{\mathrm{hk,ss}}(h)\ge \int_{B_R\cap\{|\bar F|\le M\}} \bar F^\top\bar A \bar F\,d\bar\pi \ge 4L.\tag{38}\]

Step 3: smoothing. Let \(\rho_\epsilon\) be the standard Gaussian mollifier and set \(\tilde{h}_\epsilon:=\rho_\epsilon*h\). By [34], \(\tilde{h}_\epsilon\in C_b\) and \(\|\tilde{h}_\epsilon\|_\infty\le\|h\|_\infty\). Moreover, since \(h\in L^2(\mathbb{R}^n)\), [34] (with \(p=2\)) yields \(\|\tilde{h}_\epsilon-h\|_{L^2(\mathbb{R}^n)}\to0\) as \(\epsilon\downarrow0\). Choose \(\tilde{\chi}\in C_c^\infty\) with \(\tilde{\chi}\equiv 1\) on \(\mathrm{supp}\chi\) and set \(\psi_\epsilon:=\tilde{\chi}\,\tilde{h}_\epsilon\in C_b^\infty(\bar E;\mathbb{R}^n)\). By the standing local regularity assumptions on \((\bar\gamma,\bar A,\bar\pi)\), there exists \(C_{R,M}<\infty\) such that \[\big|\tilde{\sigma}_{\mathrm{hk,ss}}(\psi_\epsilon)-\tilde{\sigma}_{\mathrm{hk,ss}}(h)\big| \le C_{R,M}\Big(\|\tilde{h}_\epsilon-h\|_{L^2(\mathbb{R}^n)}+\|\tilde{h}_\epsilon-h\|_{L^2(\mathbb{R}^n)}^2\Big)\xrightarrow[\epsilon\downarrow0]{}0.\] Thus for \(\epsilon\) small enough, 38 gives \(\tilde{\sigma}_{\mathrm{hk,ss}}(\psi_\epsilon)\ge 2L\). Since \(L>0\) is arbitrary, we obtain \(\sup_{\psi\in C_b}\tilde{\sigma}_{\mathrm{hk,ss}}(\psi)=\infty\), contradicting 36 . Hence \(\bar\sigma_{\mathrm{hk,ss}}<\infty\). ◻

Proof. Since \(\bar u(t,\cdot)=\bar P_t\bar f\) with \(\bar f\in C_b\), the Markov property yields \(\|\bar u(t,\cdot)\|_{L^\infty}<\infty\). Moreover, by 8, \[\int_{\bar E}\bar F^\top\bar A\bar F\,d\bar\pi<\infty, \qquad\text{hence}\qquad \int_{\bar E}\bar u(t,\cdot)\,\bar F^\top\bar A\bar F\,d\bar\pi<\infty.\]

Step 1 (space and amplitude truncation). Let \(\chi_R\in C_c^\infty(\mathbb{R}^n)\) satisfy \(0\le\chi_R\le 1\), \(\chi_R\equiv 1\) on \(B_R\) and \(\mathrm{supp}\chi_R\subset B_{R+1}\). Let \(T_M\) be the radial truncation on \(\mathbb{R}^n\) and set \(h_{R,M}:=\chi_R\,T_M(\bar F)\) on \(\bar E\). Write \(\|v\|_{\bar A}^2:=v^\top\bar A v\). Using \[\bar F-h_{R,M}=(1-\chi_R)\bar F+\chi_R(\bar F-T_M(\bar F)),\qquad (a+b)^2\le 2a^2+2b^2,\] and \((1-\chi_R)^2\le \mathbf{1}_{B_R^c}\), we obtain \[\begin{align} \int_{\bar E}\bar u\,\|\bar F-h_{R,M}\|_{\bar A}^2\,d\bar\pi &\le 2\int_{\bar E\cap B_R^c}\bar u\,\|\bar F\|_{\bar A}^2\,d\bar\pi +2\int_{\bar E}\bar u\,\|\bar F-T_M(\bar F)\|_{\bar A}^2\,d\bar\pi. \end{align}\] The first term tends to \(0\) as \(R\to\infty\) by dominated convergence, since \(\bar u\in L^\infty\) and \(\bar u\,\|\bar F\|_{\bar A}^2\in L^1(d\bar\pi)\). For the second term, note that pointwise \[\bar F-T_M(\bar F)=\mathbf{1}_{\{|\bar F|>M\}}\Big(1-\frac{M}{|\bar F|}\Big)\bar F \quad\Longrightarrow\quad \|\bar F-T_M(\bar F)\|_{\bar A}^2\downarrow 0,\] and \(0\le \|\bar F-T_M(\bar F)\|_{\bar A}^2\le \|\bar F\|_{\bar A}^2\), hence it tends to \(0\) as \(M\to\infty\) by dominated convergence. In particular, choosing \(R=M=k\) gives \[\label{eq:step195diag} \int_{\bar E}\bar u\,\|\bar F-h_{k,k}\|_{\bar A}^2\,d\bar\pi\xrightarrow[k\to\infty]{}0.\tag{39}\]

Step 2 (Gaussian mollification). Extend \(h_{R,M}\) by \(0\) outside \(\bar E\) (still denoted by \(h_{R,M}\)) and let \(\rho_\epsilon\) be the standard Gaussian mollifier on \(\mathbb{R}^n\). Define \(\psi_{R,M,\epsilon}:=\rho_\epsilon*h_{R,M}\) on \(\mathbb{R}^n\) and then restrict it to \(\bar E\). By [34], \(\psi_{R,M,\epsilon}\in C_b^\infty(\bar E;\mathbb{R}^n)\) and \(\|\psi_{R,M,\epsilon}\|_{L^\infty}\le \|h_{R,M}\|_{L^\infty}\le M\). Since \(h_{R,M}\in L^2(\mathbb{R}^n)\) (bounded with compact support), [34] (with \(p=2\)) yields \(\|\psi_{R,M,\epsilon}-h_{R,M}\|_{L^2(\mathbb{R}^n)}\to0\) as \(\epsilon\downarrow0\). By the standing local regularity assumptions on \((\bar u,\bar A,\bar\pi)\) and \(\mathrm{supp}h_{R,M}\subset B_{R+1}\), there exists \(C_R<\infty\) such that for all \(\epsilon>0\), \[\label{eq:step295weighted} \int_{\bar E}\bar u\,\|\psi_{R,M,\epsilon}-h_{R,M}\|_{\bar A}^2\,d\bar\pi \le C_R\,\|\psi_{R,M,\epsilon}-h_{R,M}\|_{L^2(\mathbb{R}^n)}^2 \xrightarrow[\epsilon\downarrow0]{}0.\tag{40}\]

Step 3 (diagonal choice). Set \(R_k=M_k:=k\). Choose \(\epsilon_k\downarrow0\) such that the left-hand side of 40 (with \(R=R_k\), \(M=M_k\), \(\epsilon=\epsilon_k\)) is at most \(k^{-1}\), and define \(\psi_k:=\psi_{R_k,M_k,\epsilon_k}\). Then \(\psi_k\in C_b^\infty(\bar E;\mathbb{R}^n)\) and \(\|\psi_k\|_{L^\infty}\le k\). Finally, by \((a+b)^2\le 2a^2+2b^2\), \[\int_{\bar E}\bar u\,\|\psi_k-\bar F\|_{\bar A}^2\,d\bar\pi \le 2\int_{\bar E}\bar u\,\|\psi_k-h_{k,k}\|_{\bar A}^2\,d\bar\pi +2\int_{\bar E}\bar u\,\|\bar F-h_{k,k}\|_{\bar A}^2\,d\bar\pi \xrightarrow[k\to\infty]{}0\] by 39 and the choice of \(\epsilon_k\). This proves ?? and \(\|\psi_k\|_{L^\infty}\le k\). Hence the set of recovery sequences in 4 is nonempty. ◻

Proof. Fix \(\psi\in C_b(\bar E;\mathbb{R}^n)\). By expanding the square, \[\begin{align} &\sigma_{\mathrm{hk,ss}}^\varepsilon -\bigl\|(A^\varepsilon)^{-1}\gamma^\varepsilon-D\Phi^\top(\psi\circ\Phi)\bigr\|_{L^2(A^\varepsilon,\pi^\varepsilon)}^2\\ =&2\int_E (D\Phi^\top\psi\circ\Phi)^\top\gamma^\varepsilon\,d\pi^\varepsilon -\int_E \psi^\top(D\Phi A^\varepsilon D\Phi^\top)\psi\,d\pi^\varepsilon\\ &\xrightarrow[\varepsilon\to0]{}\; 2\int_{\bar E}\psi^\top\bar\gamma\,d\bar\pi-\int_{\bar E}\psi^\top\bar A\,\psi\,d\bar\pi\\ =&\bar\sigma_{\mathrm{hk,ss}}-\|\bar A^{-1}\bar\gamma-\psi\|_{L^2(\bar A,\bar\pi)}^2, \end{align}\] where we used 3. Hence \[\liminf_{\varepsilon\downarrow0}\sigma_{\mathrm{hk,ss}}^\varepsilon \;\ge\;\bar\sigma_{\mathrm{hk,ss}}-\|\bar A^{-1}\bar\gamma-\psi\|_{L^2(\bar A,\bar\pi)}^2.\] Taking \(\psi=\psi_k\) along a recovery sequence and \(k\to\infty\) gives the liminf inequality, with gap \[\inf_{\{\psi_k\}\;\text{recovery}} \;\lim_{k\to\infty}\;\limsup_{\varepsilon\downarrow0} \bigl\|(A^\varepsilon)^{-1}\gamma^\varepsilon-D\Phi^\top(\psi_k\circ\Phi)\bigr\|_{L^2(A^\varepsilon,\pi^\varepsilon)}^2.\] ◻

Proof. Since \(B_{11}\) is Hurwitz, there exist \(M,\alpha>0\) such that \(\|e^{-B_{11}r}\|\le Me^{-\alpha r}\) for all \(r\ge0\), hence \[\label{eq:fast-semigroup} \|e^{-(\varepsilon^{-1}B_{11})r}\|=\|e^{-B_{11}(r/\varepsilon)}\|\le M e^{-\alpha r/\varepsilon},\qquad r\ge0.\tag{47}\] Differentiating \(w_s^\varepsilon=x_s^\varepsilon+B_{11}^{-1}B_{12}y_s^\varepsilon\) and using \(\dot{m}_s^\varepsilon=-(I^\varepsilon B)m_s^\varepsilon\) yields the coupled system \[\label{eq:wYsys-short} \dot{w}_s^\varepsilon=-(\varepsilon^{-1}B_{11}+B_{11}^{-1}B_{12}B_{21})\,w_s^\varepsilon-B_{11}^{-1}B_{12}C\,y_s^\varepsilon, \qquad \dot{y}_s^\varepsilon=-B_{21}w_s^\varepsilon-Cy_s^\varepsilon.\tag{48}\]

Fix a compact \(K\subset\mathbb{R}^d\) and set \[W^\varepsilon:=\sup_{z\in K}\sup_{0\le s\le t}|w_s^\varepsilon(z)|,\qquad Y^\varepsilon:=\sup_{z\in K}\sup_{0\le s\le t}|y_s^\varepsilon(z)|.\] Variation of constants applied to the first equation in Eq. 48 , together with Eq. 47 , yields for all \(z\in K\) and \(s\in[0,t]\) \[|w_s^\varepsilon(z)| \le M e^{-\alpha s/\varepsilon}|w_0(z)| +M\!\int_0^s e^{-\alpha(s-r)/\varepsilon}\Big(\|B_{11}^{-1}B_{12}B_{21}\|\,|w_r^\varepsilon(z)| +\|B_{11}^{-1}B_{12}C\|\,|y_r^\varepsilon(z)|\Big)\,dr .\] Taking suprema and using \(\int_0^s e^{-\alpha(s-r)/\varepsilon}dr\le \varepsilon/\alpha\) gives \[\label{eq:Wineq-short} W^\varepsilon\le M \sup_{z\in K}|w_0(z)| + \varepsilon c_1 W^\varepsilon+ \varepsilon c_2 Y^\varepsilon,\tag{49}\] for constants \(c_1,c_2\) depending only on the matrices. Similarly, variation of constants for the second equation in Eq. 48 yields \[|y_s^\varepsilon(z)| \le \|e^{-Cs}\|\,|y|+\int_0^s \|e^{-C(s-r)}\|\,\|B_{21}\|\,|w_r^\varepsilon(z)|\,dr \le C_t\sup_{z\in K}|y|+ C_t\|B_{21}\|\,t\,W^\varepsilon,\] where \(C_t:=\sup_{0\le r\le t}\|e^{-Cr}\|<\infty\) since \(C\) is Hurwitz. Hence \[\label{eq:Yineq-short} Y^\varepsilon\le a_{t,K}+b_t W^\varepsilon,\tag{50}\] with \(a_{t,K}:=C_t\sup_{z\in K}|y|\) and \(b_t:=C_t\|B_{21}\|t\).

Combining Eq. 49 –Eq. 50 gives \[W^\varepsilon\le M \sup_{z\in K}|w_0(z)| + \varepsilon c_1 W^\varepsilon+ \varepsilon c_2(a_{t,K}+b_t W^\varepsilon) \le A_{t,K} + \varepsilon\tilde{c}_t W^\varepsilon,\] for suitable constants \(A_{t,K},\tilde{c}_t<\infty\). Choosing \(\varepsilon_0\in(0,1]\) such that \(\varepsilon_0\tilde{c}_t\le \tfrac12\) yields \(W^\varepsilon\le 2A_{t,K}\) for all \(\varepsilon\le \varepsilon_0\), and inserting this back into Eq. 49 implies \(W^\varepsilon\le C_{t,K}\varepsilon\) for all \(\varepsilon\le\varepsilon_0\), proving Eq. ?? (and trivially extending to \(\varepsilon\in[\varepsilon_0,1]\) by enlarging \(C_{t,K}\)).

Finally, using the variation-of-constants formula for \(y_s^\varepsilon\) and the averaged solution \(\bar y_s:=e^{-Cs}\Phi(z)\), \[y_s^\varepsilon(z)-\bar y_s =-\int_0^s e^{-C(s-r)}B_{21}w_r^\varepsilon(z)\,dr,\] so Eq. ?? implies \[\sup_{z\in K}\sup_{0\le s\le t}|y_s^\varepsilon(z)-e^{-Cs}\Phi(z)| \le C_t\|B_{21}\|\int_0^t \sup_{z\in K}|w_r^\varepsilon(z)|\,dr \le C_{t,K}\varepsilon,\] which is Eq. ?? . Since \(x_s^\varepsilon=w_s^\varepsilon-B_{11}^{-1}B_{12}y_s^\varepsilon\) and \(x_s^0=-B_{11}^{-1}B_{12}e^{-Cs}\Phi(z)\), the bounds of Eq. ?? –Eq. ?? imply \(\sup_{z\in K}\sup_{0\le s\le t}|m_s^\varepsilon(z)-m_s^0(z)|\to0\) as \(\varepsilon\to0\). ◻

Proof. Proof of 1

(i) Let \(M^\varepsilon:=I^\varepsilon B\). Since \(B_{11}\) and \(C:=B_{22}-B_{21}B_{11}^{-1}B_{12}\) are Hurwitz, 10 implies exponential stability of the mean dynamics when \(\varepsilon\) is small; in particular, there exists \(\varepsilon_0>0\) such that \(M^\varepsilon\) is Hurwitz for all \(\varepsilon\in(0,\varepsilon_0]\). Hence, for each \(\varepsilon\in(0,\varepsilon_0]\), the OU process with generator \(\mathcal{L}^\varepsilon\) admits a unique invariant Gaussian measure \(\pi^\varepsilon=\mathcal{N}(0,\Sigma^\varepsilon)\), where \(\Sigma^\varepsilon\succ0\) is the unique solution of the Lyapunov equation \[\label{eq:Lyap-eps-proof} I^{\varepsilon}B\,\Sigma^\varepsilon+\Sigma^\varepsilon B^\top I^{\varepsilon}=2I^{\varepsilon}.\tag{51}\]

Write \[\Sigma^\varepsilon= \begin{pmatrix} \Sigma_{xx}^\varepsilon & \Sigma_{xy}^\varepsilon\\ \Sigma_{yx}^\varepsilon & \Sigma_{yy}^\varepsilon \end{pmatrix}, \qquad \Sigma_{yx}^\varepsilon=(\Sigma_{xy}^\varepsilon)^\top.\] Expanding Eq. 51 in blocks gives \[\begin{align} &\mathrm{(TL)}\quad B_{11}\Sigma_{xx}^\varepsilon+\Sigma_{xx}^\varepsilon B_{11}^\top +B_{12}\Sigma_{yx}^\varepsilon+\Sigma_{xy}^\varepsilon B_{12}^\top =2I_{d_x},\tag{52}\\ &\mathrm{(TR)}\quad \varepsilon^{-1}\bigl(B_{11}\Sigma_{xy}^\varepsilon+B_{12}\Sigma_{yy}^\varepsilon\bigr) +\Sigma_{xx}^\varepsilon B_{21}^\top+\Sigma_{xy}^\varepsilon B_{22}^\top =0,\tag{53}\\ &\mathrm{(BL)}\quad B_{21}\Sigma_{xx}^\varepsilon+B_{22}\Sigma_{yx}^\varepsilon +\varepsilon^{-1}\bigl(\Sigma_{yx}^\varepsilon B_{11}^\top+\Sigma_{yy}^\varepsilon B_{12}^\top\bigr) =0,\tag{54}\\ &\mathrm{(BR)}\quad B_{21}\Sigma_{xy}^\varepsilon+B_{22}\Sigma_{yy}^\varepsilon +\Sigma_{yx}^\varepsilon B_{21}^\top+\Sigma_{yy}^\varepsilon B_{22}^\top =2I_{d_y}.\tag{55} \end{align}\]

Boundedness (by contradiction). Assume \(\|\Sigma^\varepsilon\|\to\infty\) along \(\varepsilon_n\downarrow0\) and set \(\widehat\Sigma_n:=\Sigma^{\varepsilon_n}/\|\Sigma^{\varepsilon_n}\|\). Extract \(\widehat\Sigma_n\to\widehat\Sigma_*\). Multiplying Eq. 53 by \(\varepsilon_n\) and letting \(n\to\infty\) gives \(B_{11}\widehat\Sigma_{xy,*}+B_{12}\widehat\Sigma_{yy,*}=0\), hence \(\widehat\Sigma_{xy,*}=-B_{11}^{-1}B_{12}\widehat\Sigma_{yy,*}\). Passing to the limit in the rescaled Eq. 55 yields \(C\,\widehat\Sigma_{yy,*}+\widehat\Sigma_{yy,*}C^\top=0\), so \(\widehat\Sigma_{yy,*}=0\) since \(C\) is Hurwitz, and thus \(\widehat\Sigma_{xy,*}=0\). Passing to the limit in the rescaled Eq. 52 gives \(B_{11}\widehat\Sigma_{xx,*}+\widehat\Sigma_{xx,*}B_{11}^\top=0\), hence \(\widehat\Sigma_{xx,*}=0\) since \(B_{11}\) is Hurwitz. This contradicts \(\|\widehat\Sigma_*\|=1\), so \((\Sigma^\varepsilon)\) is bounded.

Identification of the limit. Let \(\varepsilon_n\downarrow0\) and extract a subsequence \(\Sigma^{\varepsilon_n}\to\Sigma^0\). Multiplying Eq. 53 by \(\varepsilon_n\) and letting \(n\to\infty\) yields \(\Sigma^0_{xy}=-B_{11}^{-1}B_{12}\Sigma^0_{yy}\). Passing to the limit in Eq. 55 gives \(C\Sigma^0_{yy}+\Sigma^0_{yy}C^\top=2I_{d_y}\), which has a unique solution since \(C\) is Hurwitz. Hence \(\Sigma^0_{yy}\) and \(\Sigma^0_{xy},\Sigma^0_{yx}\) are uniquely determined. The remaining block \(\Sigma^0_{xx}\) is uniquely determined from the limit of Eq. 52 (Lyapunov equation with drift \(B_{11}\)). Therefore \(\Sigma^\varepsilon\to\Sigma^0\) as \(\varepsilon\to0\), and thus \(\pi^\varepsilon\Rightarrow \Pi:=\mathcal{N}(0,\Sigma^0)\) and \(\Phi_\#\Pi=\mathcal{N}(0,\Sigma^0_{yy})=:\bar\pi\).

(ii) Conditional kernel and projection. Let \((X,Y)\sim\Pi\). Since \(\Sigma^0_{yy}\succ0\), Gaussian conditioning yields \[\mathcal{L}(X\mid Y=y)=\mathcal{N}\!\Bigl(\Sigma^0_{xy}(\Sigma^0_{yy})^{-1}y,\, \Sigma^0_{xx}-\Sigma^0_{xy}(\Sigma^0_{yy})^{-1}\Sigma^0_{yx}\Bigr).\] Using \(\Sigma^0_{xy}=-B_{11}^{-1}B_{12}\Sigma^0_{yy}\) gives \(\Sigma^0_{xy}(\Sigma^0_{yy})^{-1}=-B_{11}^{-1}B_{12}\). Define \[\Sigma^x:=\Sigma^0_{xx}-\Sigma^0_{xy}(\Sigma^0_{yy})^{-1}\Sigma^0_{yx}, \qquad \mu_y:=\mathcal{N}(-B_{11}^{-1}B_{12}y,\Sigma^x),\] and \((\mathcal{P}f)(y):=\int_{\mathbb{R}^{d_x}} f(x,y)\,\mu_y(dx)\) for \(f\in\mathcal{M}\). Then \(\mathcal{P}f\in C_b(\bar E)\) by dominated convergence, and the tower property gives \[\int_E f(z)\,\varphi(\Phi(z))\,d\Pi(z) =\int_{\bar E}(\mathcal{P}f)(y)\,\varphi(y)\,d\bar\pi(y), \qquad \forall\,\varphi\in C_b(\bar E),\] which is 1(ii).

(iii) Dynamic convergence. Fix \(t>0\). For each \(\varepsilon>0\) and \(z\in\mathbb{R}^d\), \[Z_t^\varepsilon\mid(Z_0^\varepsilon=z)\sim\mathcal{N}(m_t^\varepsilon(z),Q_t^\varepsilon), \qquad m_t^\varepsilon(z)=e^{-M^\varepsilon t}z, \qquad Q_t^\varepsilon=\Sigma^\varepsilon-e^{-M^\varepsilon t}\Sigma^\varepsilon e^{-(M^\varepsilon)^\top t}.\] Let \[m_t^0(z):=\bigl(-B_{11}^{-1}B_{12}e^{-Ct}\Phi(z),\;e^{-Ct}\Phi(z)\bigr), \qquad L_t z:=m_t^0(z).\] By 10, for every compact \(K\subset\mathbb{R}^d\) we have \(\sup_{z\in K}|m_t^\varepsilon(z)-m_t^0(z)|\to0\) as \(\varepsilon\to0\), and in particular \(e^{-M^\varepsilon t}e_i\to L_t e_i\) for the standard basis \(\{e_i\}_{i=1}^d\), hence \[\label{eq:opnorm-semigroup} \|e^{-M^\varepsilon t}-L_t\|\to0.\tag{56}\] Together with \(\Sigma^\varepsilon\to\Sigma^0\), this implies \(Q_t^\varepsilon\to Q_t^0:=\Sigma^0-L_t\Sigma^0 L_t^\top\) in the matrix norm.

Let \(f\in\mathcal{M}\). Writing \[u^\varepsilon(t,z)=\mathbb{E}[f(Z_t^\varepsilon)\mid Z_0^\varepsilon=z]=\int_E f(\xi)\,\mathcal{N}(m_t^\varepsilon(z),Q_t^\varepsilon)(d\xi),\] the convergence of \((m_t^\varepsilon,Q_t^\varepsilon)\) on compacts and uniform continuity of the Gaussian integral map on compact parameter sets yield local uniform convergence \(u^\varepsilon(t,\cdot)\to u^0(t,\cdot)\), where \[u^0(t,z):=\int_E f(\xi)\,\mathcal{N}(m_t^0(z),Q_t^0)(d\xi).\] Finally, under \(\mathcal{N}(m_t^0(z),Q_t^0)\) the marginal law of \(Y\) coincides with that of the averaged OU \(\bar Y_t\) started at \(\Phi(z)\), and the conditional law of \(X\) given \(Y=y'\) equals \(\mu_{y'}\) from (ii) (Gaussian conditioning using \(\Sigma^0_{xy}=-B_{11}^{-1}B_{12}\Sigma^0_{yy}\) and the definition of \(L_t\)). Therefore, \[u^0(t,z)=\mathbb{E}\big[\mathbb{E}[f(X,Y)\mid Y]\big]=\mathbb{E}\big[(\mathcal{P}f)(Y)\big] =\mathbb{E}_{\Phi(z)}\big[(\mathcal{P}f)(\bar Y_t)\big]=\bar u(t,\Phi(z)),\] which is 1(iii).

Proof of 3.

(ii) Projected diffusivity. Here \(A^\varepsilon\equiv I^\varepsilon\), hence \(D\Phi A^\varepsilon D\Phi^\top\equiv I_{d_y}\) and \(Q^\varepsilon=(D\Phi A^\varepsilon D\Phi^\top)\pi^\varepsilon=I_{d_y}\pi^\varepsilon\). Thus \(Q^\varepsilon\rightharpoonup I_{d_y}\Pi\) and \(\Phi_\#(I_{d_y}\Pi)=I_{d_y}\bar\pi=:\bar Q\).

(iii) Uniform projected housekeeping dissipation. Since \(\gamma^\varepsilon(z)=I^\varepsilon((\Sigma^\varepsilon)^{-1}-B)z\), \(D\Phi\,\gamma^\varepsilon\) is linear in \(z\) and, because \(\Sigma^\varepsilon\to\Sigma^0\succ0\), we have \(\sup_{\varepsilon\le\varepsilon_0}\|(\Sigma^\varepsilon)^{-1}\|<\infty\). Hence \(|D\Phi\,\gamma^\varepsilon(z)|\le C|z|\) uniformly for \(\varepsilon\le\varepsilon_0\), and therefore \[\sup_{0<\varepsilon\le\varepsilon_0}\int_E |D\Phi\,\gamma^\varepsilon(z)|^2\,d\pi^\varepsilon(z) \le C\sup_{0<\varepsilon\le\varepsilon_0}\int_E |z|^2\,d\pi^\varepsilon(z) = C\sup_{0<\varepsilon\le\varepsilon_0}\mathrm{tr}(\Sigma^\varepsilon)<\infty.\]

(i) Weak convergence of projected current. We apply [prop:drift-implies-current]. Here \(\bar E=\mathbb{R}^{d_y}\) is open and \(\Phi(x,y)=y\) is affine. Write \(\pi^\varepsilon(dz)=e^{-V^\varepsilon(z)}dz\) with \(V^\varepsilon(z)=\tfrac12 z^\top(\Sigma^\varepsilon)^{-1}z+\mathrm{const}\). Since \(A^\varepsilon\) is constant, \(\nabla\!\cdot A^\varepsilon\equiv0\), and \[b^\varepsilon=\gamma^\varepsilon-A^\varepsilon\nabla V^\varepsilon =I^\varepsilon\big((\Sigma^\varepsilon)^{-1}-B\big)z-I^\varepsilon(\Sigma^\varepsilon)^{-1}z =-I^\varepsilon Bz,\] so \(D\Phi\,b^\varepsilon(x,y)=-(B_{21}x+B_{22}y)\) is independent of \(\varepsilon\).

Let \(\xi\in C_b(\bar E;\mathbb{R}^{d_y})\) and set \(\Xi:=\xi\circ\Phi\). Using \(\pi^\varepsilon\Rightarrow\Pi\) and \(\sup_{\varepsilon\le\varepsilon_0}\int|z|^2\,d\pi^\varepsilon<\infty\), we may pass to the limit: \[\int_E \langle \xi(\Phi(z)),D\Phi b^\varepsilon(z)\rangle\,d\pi^\varepsilon(z) \to \int_E \langle \xi(\Phi(z)),D\Phi b^0(z)\rangle\,d\Pi(z).\] Under \(\Pi\), \(\mathbb{E}[X\mid Y=y]=-B_{11}^{-1}B_{12}y\), hence the right-hand side equals \[\int_{\bar E}\big\langle \xi(y),-\bigl(B_{22}-B_{21}B_{11}^{-1}B_{12}\bigr)y\big\rangle\,d\bar\pi(y) = \int_{\bar E}\langle \xi(y),-Cy\rangle\,d\bar\pi(y).\] Thus \(B^\varepsilon=\Phi_\#(D\Phi\,b^\varepsilon\,\pi^\varepsilon)\rightharpoonup \bar B\) with \(\bar B(dy)=(-Cy)\bar\pi(dy)\), and [prop:drift-implies-current] yields 3(i).

This completes the proof. ◻

Proof. For the curvature–dimension estimate, we write \[\mathrm{Sym}(B) := \tfrac12\bigl(B+B^\top\bigr) = \begin{pmatrix} S_{11} & S_{12}\\[2pt] S_{21} & S_{22} \end{pmatrix}, \qquad S_{11} = \mathrm{Sym}(B_{11}).\] By assumption, \(S_{11}\) is positive definite and hence invertible. Denote by \[S := S_{22} - S_{21}S_{11}^{-1}S_{12}\] the Schur complement of \(\mathrm{Sym}(B)\) with respect to \(S_{11}\), and set \[\rho := \min\bigl\{0,\lambda_{\min}(S)\bigr\} \le 0.\] For the generator \[\mathcal{L}^\varepsilon f(z) = \mathrm{tr}\bigl(I^{\varepsilon} D^2 f(z)\bigr) - (I^{\varepsilon} Bz)\cdot\nabla f(z),\] one computes (see e.g.the Bakry–Émery calculus for linear diffusions) that \[\Gamma_1^\varepsilon(f) = \langle\nabla f, I^{\varepsilon}\nabla f\rangle, \qquad \Gamma_2^\varepsilon(f) = \|D^2 f\|_{I^{\varepsilon}}^2 + \bigl\langle\nabla f,\,I^{\varepsilon} \mathrm{Sym}(B) I^{\varepsilon}\,\nabla f\bigr\rangle,\] where \(\|D^2 f\|_{I^{\varepsilon}}^2\ge0\) is the Hessian term. A block decomposition of the quadratic form \(v\mapsto \langle v, I^{\varepsilon} \mathrm{Sym}(B) I^{\varepsilon} v\rangle\) using the Schur complement shows that, for every \(v\in\mathbb{R}^d\) and every \(\varepsilon>0\), \[\bigl\langle v,\,I^{\varepsilon} \mathrm{Sym}(B) I^{\varepsilon} v\bigr\rangle \;\ge\; \rho\,\langle v, I^{\varepsilon} v\rangle.\] In particular, \[\Gamma_2^\varepsilon(f) \;\ge\; \bigl\langle\nabla f,\,I^{\varepsilon} \mathrm{Sym}(B) I^{\varepsilon} \nabla f\bigr\rangle \;\ge\; \rho\,\Gamma_1^\varepsilon(f), \qquad \forall f\in C_c^\infty(\mathbb{R}^d).\] Hence \(\mathcal{L}^\varepsilon\) satisfies a uniform curvature–dimension bound \(\mathrm{CD}(\rho,\infty)\) for all \(0<\varepsilon<\varepsilon_0\). ◻

Proof. Fix \(t>0\) and suppose 1 holds. Then for some \(\varepsilon_0>0\), \(\Sigma^\varepsilon\to\Sigma^0\) as \(\varepsilon\to0\) along \((0,\varepsilon_0]\) and \[\sup_{0<\varepsilon\le\varepsilon_0}\mathrm{tr}(\Sigma^\varepsilon)<\infty,\qquad \sup_{0<\varepsilon\le\varepsilon_0}\|(\Sigma^\varepsilon)^{-1}\|<\infty.\] Since \(B_{11}\) is Hurwitz and symmetric, \(B_{11}\succ0\). Throughout \(0<\varepsilon\le\varepsilon_0\).

In the OU setting \(A^\varepsilon\equiv I^\varepsilon\), \(\Phi(x,y)=y\), and \[(A^\varepsilon)^{-1}\gamma^\varepsilon(z)=\big((\Sigma^\varepsilon)^{-1}-B\big)z=:K^\varepsilon z,\qquad D\Phi^\top\psi(\Phi(z))=(0,\psi(y)),\qquad \|v\|_{I^\varepsilon}^2=\varepsilon^{-1}|v_x|^2+|v_y|^2.\] Moreover \(\bar A\equiv I_{d_y}\) and \[\bar F(y):=\bar A^{-1}\bar\gamma(y)=\big((\Sigma^y)^{-1}-C\big)y,\qquad C:=B_{22}-B_{21}B_{11}^{-1}B_{12},\qquad \Sigma^y=\Sigma^0_{yy}.\]

Step 1 (CI). Here \(D\Phi A^\varepsilon D\Phi^\top\equiv I_{d_y}\) and \(\bar A\equiv I_{d_y}\), so Eq. 33 reduces to \[\int_E u^\varepsilon(t,z)\,|\bar F(\Phi(z))|^2\,d\pi^\varepsilon(z)\;\to\; \int_{\bar E}\bar u(t,y)\,|\bar F(y)|^2\,d\bar\pi(y).\] Since \(\bar F\) is linear, \(|\bar F(y)|^2\le c_F^2|y|^2\). Let \(\chi_k\in C_b^\infty(\bar E;[0,1])\) be a cutoff and set \(\bar F_k:=\chi_k\bar F\in C_b(\bar E;\mathbb{R}^{d_y})\). By Lemma A.2, Eq. ?? applied to \(\psi=\bar F_k\) we have convergence for \(\bar F_k\): \[\int_E u^\varepsilon|\bar F_k\circ\Phi|^2\,d\pi^\varepsilon\to \int_{\bar E}\bar u|\bar F_k|^2\,d\bar\pi.\] To pass \(k\to\infty\), note that \(u^\varepsilon(t,\cdot)\) is bounded uniformly in \(\varepsilon\) by the maximum principle (for bounded terminal data), and write \(\nu^\varepsilon:=\Phi_\#\pi^\varepsilon=\mathcal{N}(0,\Sigma_{yy}^\varepsilon)\). Then \[\int_E u^\varepsilon|\bar F\circ\Phi-\bar F_k\circ\Phi|^2\,d\pi^\varepsilon \le \|u^\varepsilon(t,\cdot)\|_\infty\,c_F^2\int_{\bar E}|y|^2\mathbf{1}_{\{|y|>k\}}\,d\nu^\varepsilon(y).\] For centered Gaussians \(Y\sim\mathcal{N}(0,\Sigma)\) one has the moment bound \(\mathbb{E}|Y|^4\le C_{d_y}(\mathrm{tr}\Sigma)^2\), hence by Markov inequality \[\int |y|^2\mathbf{1}_{\{|y|>k\}}\,d\nu^\varepsilon(y) \le k^{-2}\int |y|^4\,d\nu^\varepsilon(y) \le \frac{C}{k^2}, \qquad \text{uniformly in }0<\varepsilon\le\varepsilon_0,\] because \(\sup_{\varepsilon\le\varepsilon_0}\mathrm{tr}(\Sigma_{yy}^\varepsilon)\le \sup_{\varepsilon\le\varepsilon_0}\mathrm{tr}(\Sigma^\varepsilon)<\infty\). The same estimate holds for \(\bar\pi\). Therefore the tail errors vanish uniformly as \(k\to\infty\), and Eq. 33 follows.

Step 2 (canonical reduction). By [prop:canonical-locking-CI], it suffices to show \[\limsup_{\varepsilon\to0}R^\varepsilon(t;\bar F)=0, \qquad R^\varepsilon(t;\bar F)=\int_E u^\varepsilon(t,z)\,\|K^\varepsilon z-(0,\bar F(y))\|_{I^\varepsilon}^2\,d\pi^\varepsilon(z).\] Equivalently, \[\label{eq:Rsplit-OU} R^\varepsilon(t;\bar F) =\int_E u^\varepsilon(t,z)\Big(\varepsilon^{-1}|(K^\varepsilon z)_x|^2+|(K^\varepsilon z)_y-\bar F(y)|^2\Big)\,d\pi^\varepsilon(z).\tag{57}\]

Step 3 (fast row estimate). Write \(\Sigma^\varepsilon\) in blocks and use \(B_{11}=B_{11}^\top\), \(B_{21}=B_{12}^\top\). From the \((\mathrm{TR})\) Lyapunov block in Eq. 53 , we have \[\label{eq:TR-lock-short} B_{11}\Sigma_{xy}^\varepsilon+B_{12}\Sigma_{yy}^\varepsilon =-\varepsilon\big(\Sigma_{xx}^\varepsilon B_{12}+\Sigma_{xy}^\varepsilon B_{22}^\top\big).\tag{58}\] Since \(\Sigma^\varepsilon\to\Sigma^0\) and \(\Sigma_{yy}^0\succ0\), there is \(M<\infty\) with \[\label{eq:blocks-bdd-short} \sup_{\varepsilon\le\varepsilon_0}\Big(\|\Sigma_{xx}^\varepsilon\|+\|\Sigma_{xy}^\varepsilon\|+\|\Sigma_{yy}^\varepsilon\| +\|(\Sigma_{yy}^\varepsilon)^{-1}\|\Big)\le M.\tag{59}\] Right-multiplying Eq. 58 by \((\Sigma_{yy}^\varepsilon)^{-1}\) gives \[\label{eq:xyyy-short} \Big\|\Sigma_{xy}^\varepsilon(\Sigma_{yy}^\varepsilon)^{-1}+B_{11}^{-1}B_{12}\Big\|\le C\varepsilon.\tag{60}\]

Let \(S^\varepsilon:=\Sigma_{xx}^\varepsilon-\Sigma_{xy}^\varepsilon(\Sigma_{yy}^\varepsilon)^{-1}\Sigma_{yx}^\varepsilon\succ0\). Using the definition of \(S^\varepsilon\) and substituting Eq. 58 and its transpose into \(B_{11}\Sigma_{xy}^\varepsilon(\Sigma_{yy}^\varepsilon)^{-1}\) and \((\Sigma_{yy}^\varepsilon)^{-1}\Sigma_{yx}^\varepsilon B_{11}\) yields \[\begin{align} B_{11}S^\varepsilon+S^\varepsilon B_{11} &=(B_{11}\Sigma_{xx}^\varepsilon+\Sigma_{xx}^\varepsilon B_{11}) +(B_{12}\Sigma_{yx}^\varepsilon+\Sigma_{xy}^\varepsilon B_{12}^\top) +\varepsilon\,R_S^\varepsilon,\label{eq:Schur-3}\\ &=2I_{d_x}+\varepsilon\,R_S^\varepsilon,\nonumber \end{align}\tag{61}\] where the last line uses the \((\mathrm{TL})\) block in Eq. 52 . Moreover, Eq. 59 implies \(\sup_{\varepsilon\le\varepsilon_0}\|R_S^\varepsilon\|\le C\).

Since \(B_{11}\succ0\), the Lyapunov operator \(X\mapsto B_{11}X+XB_{11}\) is invertible on symmetric matrices, hence \(\|S^\varepsilon-B_{11}^{-1}\|\le C\varepsilon\). In particular, \(S^\varepsilon\to B_{11}^{-1}\succ0\), so shrinking \(\varepsilon_0\) if needed we may assume \(\sup_{\varepsilon\le\varepsilon_0}\|(S^\varepsilon)^{-1}\|<\infty\). Using \[(S^\varepsilon)^{-1}-B_{11}=(S^\varepsilon)^{-1}(B_{11}^{-1}-S^\varepsilon)B_{11},\] we obtain \(\|(S^\varepsilon)^{-1}-B_{11}\|\le C\varepsilon\) for \(0<\varepsilon\le\varepsilon_0\). By the block inverse formula, \[(\Sigma^\varepsilon)^{-1}_{xx}=(S^\varepsilon)^{-1},\qquad (\Sigma^\varepsilon)^{-1}_{xy}=-(S^\varepsilon)^{-1}\Sigma_{xy}^\varepsilon(\Sigma_{yy}^\varepsilon)^{-1},\] so combining with Eq. 60 gives \[\label{eq:fastrow} \|(\Sigma^\varepsilon)^{-1}_{xx}-B_{11}\|+\|(\Sigma^\varepsilon)^{-1}_{xy}-B_{12}\|\le C\varepsilon.\tag{62}\] Therefore for \(K^\varepsilon=(\Sigma^\varepsilon)^{-1}-B\), \[\label{eq:Kxstar} \|K_{x*}^\varepsilon\|\le C\varepsilon,\qquad \|K_{yx}^\varepsilon\|\le C\varepsilon,\tag{63}\] where the second bound uses symmetry of \((\Sigma^\varepsilon)^{-1}\) and \(B_{21}=B_{12}^\top\).

Step 4 (estimate \(R^\varepsilon\)). Let \(M_u:=\sup_\varepsilon\|u^\varepsilon(t,\cdot)\|_\infty\). Since \(\pi^\varepsilon=\mathcal{N}(0,\Sigma^\varepsilon)\), \(\int|z|^2\,d\pi^\varepsilon=\mathrm{tr}(\Sigma^\varepsilon)\le C\) uniformly.

By Eq. 63 , \(|(K^\varepsilon z)_x|\le \|K_{x*}^\varepsilon\||z|\le C\varepsilon|z|\), hence \[\int u^\varepsilon\,\varepsilon^{-1}|(K^\varepsilon z)_x|^2\,d\pi^\varepsilon\le M_u\,\varepsilon^{-1}(C\varepsilon)^2\int|z|^2\,d\pi^\varepsilon\le C\varepsilon\to0.\]

Since \((\Sigma^\varepsilon)^{-1}\to(\Sigma^0)^{-1}\), we have \(K_{yy}^\varepsilon\to K_{yy}^0\). Letting \(\varepsilon\to0\) in Eq. 62 gives \((\Sigma^0)^{-1}_{xx}=B_{11}\) and \((\Sigma^0)^{-1}_{xy}=B_{12}\), hence \((\Sigma^0)^{-1}_{yx}=B_{21}\). The Schur identity yields \[(\Sigma^0)^{-1}_{yy}=(\Sigma^0_{yy})^{-1}+B_{21}B_{11}^{-1}B_{12}=(\Sigma^y)^{-1}+B_{21}B_{11}^{-1}B_{12},\] so \(K_{yy}^0=(\Sigma^y)^{-1}-C\) and therefore \(\bar F(y)=K_{yy}^0y\). Thus \[(K^\varepsilon z)_y-\bar F(y)=K_{yx}^\varepsilon x+(K_{yy}^\varepsilon-K_{yy}^0)y,\] and using Eq. 63 and \(\|K_{yy}^\varepsilon-K_{yy}^0\|\to0\) gives \[\int u^\varepsilon\,|(K^\varepsilon z)_y-\bar F(y)|^2\,d\pi^\varepsilon \le C\|K_{yx}^\varepsilon\|^2\int|x|^2\,d\pi^\varepsilon+ C\|K_{yy}^\varepsilon-K_{yy}^0\|^2\int|y|^2\,d\pi^\varepsilon\to0.\]

Together with Eq. 57 , it yields \(R^\varepsilon(t;\bar F)\to0\), hence Eq. 34 . By [prop:canonical-locking-CI] we conclude 4. ◻

Proof. of 7.] We use the operator conventions stated above; in particular \(a_1:\nabla_x^2\phi=\mathrm{Tr}(a_1\,\nabla_x^2\phi)\) and \(a_2:\nabla_y^2\phi=\mathrm{Tr}(a_2\,\nabla_y^2\phi)\).

Step 1: Synchronous coupling and weighted energies. Let \((X_t^1,Y_t^1)\) and \((X_t^2,Y_t^2)\) be synchronously coupled solutions of Eq. 13 , and set \(\Delta X_t=X_t^1-X_t^2\), \(\Delta Y_t=Y_t^1-Y_t^2\), \(Z_t^i=(X_t^i,Y_t^i)\). Define \[V_1(t):=\Delta X_t^\top B_1(Z_t^1)\Delta X_t,\qquad V_2(t):=\Delta Y_t^\top B_2(Y_t^1)\Delta Y_t,\] and \(h(t):=\mathbb{E} V_1(t)\), \(j(t):=\mathbb{E} V_2(t)\), \(g_\varepsilon(t):=\varepsilon h(t)+j(t)\). Uniform ellipticity yields \[\frac{\varepsilon}{\Lambda_1}|\Delta X_t|^2\le \varepsilon V_1(t)\le \frac{\varepsilon}{\lambda_1}|\Delta X_t|^2,\qquad \frac{1}{\Lambda_2}|\Delta Y_t|^2\le V_2(t)\le \frac{1}{\lambda_2}|\Delta Y_t|^2.\]

Step 2: Itô expansions for \(B_1\) and \(B_2\). Write \(L_\varepsilon=\varepsilon^{-1}\mathcal{L}_f+\mathcal{L}_s\) with \[\mathcal{L}_f\phi=b_1\cdot\nabla_x\phi+a_1:\nabla_x^2\phi,\qquad \mathcal{L}_s\phi=b_2\cdot\nabla_y\phi+a_2:\nabla_y^2\phi.\] Along the first path \(Z_t^1\), \[\begin{align} dB_1(Z_t^1) &=\Big(\tfrac{1}{\varepsilon}(\mathcal{L}_f B_1)+\mathcal{L}_s B_1\Big)(Z_t^1)\,dt +\tfrac{1}{\sqrt\varepsilon}\underbrace{(\nabla_x B_1\,\eta_1)(Z_t^1)}_{=:N_1^{(1)}(t)}\,dW_t^{(1)} +\underbrace{(\nabla_y B_1\,\eta_2)(Z_t^1)}_{=:N_1^{(2)}(t)}\,dW_t^{(2)},\\ dB_2(Y_t^1) &=\underbrace{\big((\nabla_y B_2)b_2+a_2:\nabla_y^2 B_2\big)(Z_t^1)}_{=:M_2(Z_t^1)}\,dt +\underbrace{(\nabla_y B_2\,\eta_2)(Y_t^1)}_{=:N_2(Y_t^1)}\,dW_t^{(2)}. \end{align}\]

Step 3: Differential inequality for \(h'(t)\). Applying Itô’s lemma to \(V_1=\Delta X^\top B_1\Delta X\), \[dV_1=D_1\,dt+D_2\,dt+Q_1\,dt+Q_2\,dt+dM_t,\] where \(D_1\) collects drift terms from \(d\Delta X_t\), \(D_2\) drift terms from \(dB_1\), \(Q_1=(d\Delta X_t)^\top B_1\,d\Delta X_t\) is the quadratic variation, \(Q_2\) is the cross-variation between \(d\Delta X_t\) and \(dB_1\), and \(M_t\) is a local martingale. By a standard localisation argument (see the “Remark (local martingales)” at the end of this proof), we can take expectations and use \(\mathbb{E}[dM_t]=0\) at the level of differentials.

(i) Drift \(D_1\). Linearise \[b_1(Z_t^1)-b_1(Z_t^2)=\int_0^1\big((\nabla_x b_1)\Delta X_t+(\nabla_y b_1)\Delta Y_t\big)\big(Z_t^\theta\big)\,d\theta.\] Using \(K_x^{(W)}\), \(B_{xy}^{(W)}\) and a one-line Young inequality, \[\mathbb{E}[D_1]\le -\frac{2K_x^{(W)}}{\varepsilon}h(t)+\frac{\Lambda_1 B_{xy}^{(W)}}{\varepsilon}h(t)+\frac{\Lambda_2 B_{xy}^{(W)}}{\varepsilon}j(t).\]

(ii) Quadratic variation \(Q_1\). With the Lipschitz bounds for \(\eta_1\) and ellipticity, \[\mathbb{E}[Q_1] =\frac{1}{\varepsilon}\,\mathbb{E}\big[\mathrm{Tr}(\Delta\eta_1^\top B_1\Delta\eta_1)\big] \le \frac{C_{1h}}{\varepsilon}\,h(t)+\frac{C_{1j}}{\varepsilon}\,j(t).\]

(iii) Drift \(D_2\) from \(dB_1\). Using \(\|(\mathcal{L}_f B_1)\|\) and \(\|(\mathcal{L}_s B_1)\|\), \[\mathbb{E}[D_2] =\mathbb{E}\big[\Delta X^\top(\tfrac1\varepsilon\mathcal{L}_f B_1+\mathcal{L}_s B_1)\Delta X\big] \le \frac{\tilde{C}_{LfB_1}}{\varepsilon}\,h(t)+C_h\,h(t).\]

(iv) Cross-variation \(Q_2\). Only the \(W^{(1)}\)-channel contributes. By Lemma 14, \[\mathbb{E}[Q_2(t)]\le \frac{C_{X1}}{\varepsilon}\,h(t)+\frac{C_{Y1}}{\varepsilon}\,j(t).\]

Combining (i)–(iv) yields \[\label{eq:H} h'(t)\le -\frac{\alpha_0}{\varepsilon}\,h(t)+\frac{\beta_0}{\varepsilon}\,j(t)+C_h\,h(t).\tag{64}\]

Step 4: Differential inequality for \(j'(t)\). A similar computation for \(V_2=\Delta Y^\top B_2\Delta Y\) (using \(a_2=a_2(y)\)) gives \[\label{eq:J} j'(t)\le c\,h(t)+d\,j(t).\tag{65}\]

Step 5: Total energy and Gronwall inequality. For \(g_\varepsilon=\varepsilon h+j\), from Eqs. 64 ,65 , \[g_\varepsilon'(t)\le \big(-\alpha_0+c+\varepsilon C_h\big)h(t)+(\beta_0+d)\,j(t).\] By \(\alpha_0>c\), choose \(\varepsilon_0\in(0,1]\) with \(-\alpha_0+c+\varepsilon C_h\le -(\alpha_0-c)/2<0\) for all \(\varepsilon\le\varepsilon_0\). Since \(j\le g_\varepsilon\), we deduce \[g_\varepsilon(t)\le e^{(\beta_0+d)t}\,g_\varepsilon(0).\] Equivalently, for any \(z^1,z^2\), \[\label{eq:sync-energy} \mathbb{E}\big[(Z_t^1-Z_t^2)^\top G_\varepsilon(Z_t^1)(Z_t^1-Z_t^2)\big] \le e^{(\beta_0+d)t}\,(z^1-z^2)^\top G_\varepsilon(z^1)(z^1-z^2).\tag{66}\]  ◻

Proof. Keep only the noise parts that covary: \[d\Delta X_t^{\mathrm{noise}}=\tfrac1{\sqrt\varepsilon}\,\Delta\eta_1(t)\,dW_t^{(1)},\qquad dB_1^{\mathrm{noise}}(Z_t^1)=\tfrac1{\sqrt\varepsilon}(\nabla_x B_1\,\eta_1)(Z_t^1)\,dW_t^{(1)} +(\nabla_y B_1\,\eta_2)(Z_t^1)\,dW_t^{(2)}.\] Independence of \(W^{(1)}\) and \(W^{(2)}\) implies only \(W^{(1)}\) contributes: \[\mathbb{E}[Q_2(t)] \le \frac{2}{\varepsilon}\,\mathbb{E}\Big[\sum_{i=1}^{r_1}\|\Delta\eta_1^{(i)}\|\,\|(\nabla_x B_1\,\eta_1)^{(i)}\|\,|\Delta X_t|\Big].\] Using the elementary bound \[\sum_{i=1}^{r_1} A_i B_i \;\le\;r_1\,\max_i A_i\,\max_i B_i\;\le\;r_1\,\|A\|_{\mathrm{op}}\,\|B\|_{\mathrm{op}},\] we obtain, with \(c_{r_1}=2r_1\), \[\mathbb{E}[Q_2(t)]\le \frac{c_{r_1}}{\varepsilon}\,\mathbb{E}\big[\;\|\Delta\eta_1(t)\|_{\mathrm{op}}\cdot \|(\nabla_x B_1\,\eta_1)(Z_t^1)\|_{\mathrm{op}}\cdot |\Delta X_t|\;\big].\] Use \(\|(\nabla_x B_1\,\eta_1)\|_{\mathrm{op}}\le L_{B_1,x}H_{1,\infty}\) and \(\|\Delta\eta_1\|_{\mathrm{op}}\le L_{\eta_1,x}|\Delta X_t|+L_{\eta_1,y}|\Delta Y_t|\) to get \[\mathbb{E}[Q_2(t)]\le \frac{c_{r_1}L_{B_1,x}H_{1,\infty}}{\varepsilon} \,\mathbb{E}\big[L_{\eta_1,x}|\Delta X_t|^2+L_{\eta_1,y}|\Delta X_t|\,|\Delta Y_t|\big].\] Apply \(2ab\le a^2+b^2\) to the cross term and the ellipticity bounds \(\mathbb{E}|\Delta X_t|^2\le \Lambda_1 h(t)\), \(\mathbb{E}|\Delta Y_t|^2\le \Lambda_2 j(t)\); this yields the stated bound with the displayed \(C_{X1},C_{Y1}\). ◻

Proof. We use the standard closability criterion. Let \(v_n\in C_c^\infty(E)\) satisfy \(v_n\to0\) in \(L^2(\Pi)\) and \(\mathcal{E}_{y,0}(v_n-v_k,v_n-v_k)\to0\) as \(n,k\to\infty\). Set \(w_n:=a_2^{1/2}\nabla_y v_n\). Then \[\|w_n-w_k\|_{L^2(\Pi)}^2=\mathcal{E}_{y,0}(v_n-v_k,v_n-v_k)\to0,\] so \(w_n\to w\) in \(L^2(\Pi)\) for some \(w\).

Fix \(\ell\in\mathbb{N}\) and \(K_\ell=\{|z|\le \ell\}\). By uniform ellipticity on \(K_\ell\), \[\int_{K_\ell}|\nabla_y(v_n-v_k)|^2\,d\Pi \le \lambda_{K_\ell}^{-1}\mathcal{E}_{y,0}(v_n-v_k,v_n-v_k)\to0,\] hence \(\nabla_y v_n\to g^{(\ell)}\) in \(L^2(K_\ell,\Pi)\) for some \(g^{(\ell)}\). For any \(\varphi\in C_c^\infty(E)\) with \(\mathrm{supp}\varphi\subset K_\ell^\circ\) and any \(i\), integration by parts gives \[\int \partial_{y_i}v_n\,\varphi\,d\Pi =-\int v_n(\partial_{y_i}\varphi-\varphi\,\partial_{y_i}V)\,d\Pi\to0,\] since \(v_n\to0\) in \(L^2(\Pi)\) and the coefficient is bounded on \(\mathrm{supp}\varphi\). Passing to the limit also yields \(\int g^{(\ell)}_i\,\varphi\,d\Pi=0\) for all such \(\varphi\), hence \(g^{(\ell)}=0\) \(\Pi\)–a.e.on \(K_\ell^\circ\). Therefore \(\nabla_y v_n\to0\) in \(L^2(K_\ell^\circ,\Pi)\).

By local boundedness of \(a_2\), \(\|a_2^{1/2}\|_{L^\infty(K_\ell)}<\infty\), so \[\|w_n\|_{L^2(K_\ell^\circ,\Pi)}\le \|a_2^{1/2}\|_{L^\infty(K_\ell)}\|\nabla_y v_n\|_{L^2(K_\ell^\circ,\Pi)}\to0.\] Since also \(w_n\to w\) in \(L^2(\Pi)\), we get \(w=0\) on each \(K_\ell^\circ\), hence \(w=0\) \(\Pi\)–a.e.on \(E\). Thus \(w_n\to0\) in \(L^2(\Pi)\) and \[\mathcal{E}_{y,0}(v_n,v_n)=\|w_n\|_{L^2(\Pi)}^2\to0,\] which proves closability. ◻

Proof of 8.. Set \(v^\varepsilon:=\sqrt{u^\varepsilon(t)}\) and \(v:=\sqrt{\bar u(t)\circ\Phi}\). By the Markov property, \(0\le u^\varepsilon(t,\cdot),\bar u(t,\cdot)\le \|f\|_\infty\), hence \(0\le v^\varepsilon,v\le \|f\|_\infty^{1/2}\).

By 1(iii) and the standard compact/tail argument on the probability space \((E,\pi)\), \(u^\varepsilon(t)\to \bar u(t)\circ\Phi\) in \(L^1(\pi)\); thus \[\|v^\varepsilon-v\|_{L^2(\pi)}^2\le \|u^\varepsilon-\bar u(t)\circ\Phi\|_{L^1(\pi)}\to0 \qquad(|\sqrt a-\sqrt b|^2\le |a-b|).\]

By Eq. ?? and the block form of \(A^\varepsilon\), \[\mathcal{I}^\varepsilon(t)=4\Big(\tfrac1\varepsilon\|\nabla_x v^\varepsilon\|_{L^2(\pi;a_1)}^2 +\|\nabla_y v^\varepsilon\|_{L^2(\pi;a_2)}^2\Big)=4(\alpha^\varepsilon+\beta^\varepsilon).\] Under [ass:standing,ass:CDkappa], 3 yields \(\mathcal{I}^\varepsilon(t)\to\bar\mathcal{I}(t)\), hence \(\sup_\varepsilon\beta^\varepsilon<\infty\). By 15, \[a_2^{1/2}\nabla_y v^\varepsilon \rightharpoonup a_2^{1/2}\nabla_y v \text{ in }L^2(\pi), \qquad \liminf_{\varepsilon\to0}\beta^\varepsilon\ge \|a_2^{1/2}\nabla_y v\|_{L^2(\pi)}^2.\] By 5, \(\|a_2^{1/2}\nabla_y v\|_{L^2(\pi)}^2=\bar\mathcal{I}(t)/4\). Since \(\alpha^\varepsilon\ge0\) and \(\alpha^\varepsilon+\beta^\varepsilon=\mathcal{I}^\varepsilon(t)/4\to\bar\mathcal{I}(t)/4\), we get \(\beta^\varepsilon\to\bar\mathcal{I}(t)/4\) and \(\alpha^\varepsilon\to0\), proving (i); moreover weak convergence plus norm convergence give \(a_2^{1/2}\nabla_y v^\varepsilon\to a_2^{1/2}\nabla_y v\) strongly in \(L^2(\pi)\), proving (ii).

Using \(u^\varepsilon=(v^\varepsilon)^2\) and the chain rule, \(\nabla_x u^\varepsilon=2v^\varepsilon\nabla_x v^\varepsilon\) and \(\nabla_y u^\varepsilon=2v^\varepsilon\nabla_y v^\varepsilon\). Then (iii) follows from \(\|v^\varepsilon\|_\infty\le \|f\|_\infty^{1/2}\) and \(\alpha^\varepsilon\to0\). For (iv), with \(h^\varepsilon:=a_2^{1/2}\nabla_y v^\varepsilon\) and \(h:=a_2^{1/2}\nabla_y v\), \[a_2^{1/2}\nabla_y u^\varepsilon-a_2^{1/2}\nabla_y(\bar u(t)\circ\Phi) =2(v^\varepsilon-v)h^\varepsilon+2v(h^\varepsilon-h),\] and the RHS \(\to0\) in \(L^2(\pi)\) by \(v^\varepsilon\to v\) in \(L^2(\pi)\), \(h^\varepsilon\to h\) in \(L^2(\pi)\), the \(L^\infty\) bound on \(v^\varepsilon\), and the standard Vitali theorem. ◻