Proof. The convergence of \({Q^{(\epsilon)}}\) to \(Q\) will follow from the convergence of the associated semigroups. Because of theorem 6.1 in the first chapter of [9], since \(D\) is a core for \(A\), in order to prove uniform convergence of the semigroups on bounded intervals, it suffices to show that for any \(f \in D\), there exist \({f^{(\epsilon)}} \in \mathcal{D}\left({A^{(\epsilon)}}\right)\) for all \(\epsilon > 0\) such that \({f^{(\epsilon)}} \to f\) and \({A^{(\epsilon)}} {f^{(\epsilon)}} \to A f\) as \(\epsilon \downarrow 0\). Given \(f \in D\), we will construct \({f^{(\epsilon)}} \in \mathcal{D}\left({A^{(\epsilon)}}\right)\) as \({f^{(\epsilon)}} := f + \epsilon f_1 + \epsilon^2 f_2\) with \(f_1, f_2 \in C_0(R; \mathcal{B})\).

Because \(f \in D\), for every \(\alpha\), we have \(\nabla_{v_\alpha} f \in C_0(R)\). Also, \(K\) is right-invertible on \(\ker(\omega) \cap \Omega\) and \(\phi_\alpha \in \ker(\omega) \cap \Omega\) so \(K^{-1} \phi_\alpha \in \ker(\omega) \cap \Omega \subseteq \Omega\). Then, \(\eta^{-1} \nabla_{v_\alpha} f \in C_0(R)\) so \(\eta^{-1} (K^{-1} \phi_\alpha) \nabla_{v_\alpha} f \in C_0(R) \otimes \Omega\) and we define \[\begin{align} \tag{18} &f_1 := -\eta^{-1} K^{-1} J f = - \eta^{-1} \sum_{\alpha=1}^n (K^{-1} \phi_\alpha) \nabla_{v_\alpha} f \in C_0(R) \otimes \Omega \\ \tag{19} \text{and, sincef \in C_0(R) \subseteq \ker(E) ,} \quad &\eta K f_1 + (E + J) f = \eta K (- \eta^{-1} K^{-1} J f) + J f = - J f + J f = 0. \end{align}\] For all \(\alpha\), since \(K^{-1} \phi_\alpha \in \ker(\omega) \cap \Omega\) and \(E\) preserves \(\ker(\omega) \cap \Omega\), we know \(E K^{-1} \phi_\alpha \in \ker(\omega) \cap \Omega\). Also, for all \(\alpha\) and \(\beta\), by assumption, \(\phi_\alpha \cdot K^{-1} \phi_\beta \in \Omega\) The second derivatives of \(f\) are continuous and vanish at infinity and \(\{v_\alpha\}\) have bounded continuous derivatives so \(f \in \mathcal{D}\left(\nabla_{v_\alpha} \nabla_{v_\beta}\right)\). Note that since \(\nabla_{v_\alpha}\) is a derivation, \(\nabla_{v_\alpha} (\eta^{-1/2}) = 0\) implies that \(\nabla_{v_\alpha}\) commutes with multiplication by \(\eta^{-1/2}\). Firt, assume that \(L f \in L(D) \subseteq C_0(R) \otimes \Omega\), since \(E\) is \(C_0(R)\)-linear, we can define \[g := L f + (E + J) f_1 = L f - \eta^{-1} \sum_{\alpha=1}^n (E K^{-1} \phi_\alpha) \nabla_{v_\alpha} f - \eta^{-1} \sum_{\alpha,\beta=1}^n (\phi_\alpha \cdot K^{-1} \phi_\beta) \;\; \nabla_{v_\alpha} \nabla_{v_\beta} f \in C_0(R) \otimes \Omega.\] We naturally extend \(\omega\) to \(C_0(R; \mathcal{B})\) by applying \(\omega\) pointwise. Since \(E K^{-1} \phi_\alpha \in \ker(\omega)\), recalling the expression for \(A f\) from 15 , we get \[\begin{align} \omega(g) &= \omega(L f) - \eta^{-1} \sum_{\alpha=1}^n \omega(E K^{-1} \phi_\alpha) \nabla_{v_\alpha} f - \eta^{-1} \sum_{\alpha,\beta=1}^n \omega(\phi_\alpha \cdot K^{-1} \phi_\beta) \;\; (\nabla_{v_\alpha} \nabla_{v_\beta} f) \\ &= \omega(L f) + \eta^{-1} \sum_{\alpha,\beta=1}^n \langle \phi_\alpha, \phi_\beta \rangle_K \;\; (\nabla_{v_\alpha} \nabla_{v_\beta} f) = A f \in C_0(R). \end{align}\] Since \(1 \in \Omega\), we have \(\omega(g) \in C_0(R) \subseteq C_0(R) \otimes \Omega\) and so \(\omega(g) - g \in C_0(R) \otimes \Omega\). Because \(\omega\) is a projection, we have \(\omega(\omega(g) - g) = \omega(g) - \omega(g) = 0\) so \(\omega(g) - g \in \ker(\omega) \cap (C_0(R) \otimes \Omega)\). We extend \(K^{-1}\) to \(\ker(\omega) \cap (C_0(R) \otimes \Omega)\) by linearity and define \(f_2 := \eta^{-1} K^{-1} (\omega(g) - g) \in \ker(\omega) \cap (C_0(R) \otimes \Omega)\).

Now, assume that \(K^{-1}\) is bounded on \(\ker{\omega} \cap \Omega\) and \(L(D) \subseteq C_0(R; \Omega)\). Then, \(K^{-1}\) can be extended to all of \(\ker(\omega) \cap C_0(R; \Omega)\) and we allow \(g \in C_0(R; \Omega)\). Then, \[\label{eq:zeroth-ord-val} g + \eta K f_2 = g + \eta K \eta^{-1} K^{-1} (\omega(g) - g) = g + \omega(g) - g = \omega(g) = A f.\tag{20}\]

Using \(K f = f K(1) = 0\), 19 , and 20 , it follows that \[\begin{align} {A^{(\epsilon)}} {f^{(\epsilon)}} &= \left(L + \frac{1}{\epsilon} (E + J) + \frac{1}{\epsilon^2} \eta K\right) (f + \epsilon f_1 + \epsilon^2 f_2) \\ &= \frac{1}{\epsilon^2} \eta K f + \frac{1}{\epsilon} \left(\big. \eta K f_1 + (E + J) f \right) \\ &\quad + \left(\big. L f + (E + J) f_1 + \eta K f_2 \right) + \epsilon \left(\big. J f_2 + L f_1 + \epsilon L f_2 \right) \\ &= 0 + 0 + (g + \eta K f_2) + O(\epsilon) = A f + O(\epsilon) \to A f \end{align}\] as \(\epsilon \downarrow 0\) proving the condition necessary to apply theorem 6.1 of [9]. Thus, for every \(f \in C_0(R)\), \({T_t^{(\epsilon)}} f \to T_t f\) uniformly on bounded intervals of \(t \in [0, \infty)\). Then, by Corollary 8.7 in chapter 4 of [9], since \(C_c^\infty(R) \subseteq C_0(R)\) strongly separates points in \(\mathbb{R}^{d_X}\) (using bump functions), we conclude that \({Q^{(\epsilon)}} \to Q\) in law. ◻

Proof. Because \(L = b \cdot \nabla_R\) and \(\omega\) is \(C_b(R)\)-linear, we know \(\omega(L f) = \omega(b \cdot \nabla_R f) = \overline{b} \cdot \nabla_R f\). We define the matrix \(C = \{c_{\alpha \beta}\}_{\alpha,\beta=1}^n := \{\langle \phi_\alpha, \phi_\beta \rangle_K\}_{\alpha,\beta=1}^n\). By assumption, \(C\) is symmetric and positive semi-definite so there exists a symmetric positive semi-definite matrix \(H = \{h_{\alpha \gamma}\}_{\alpha,\gamma=1}^n := \sqrt{C}\). Then, since \(C = H^2\), for all \(\alpha\) and \(\beta\), we have \(c_{\alpha \beta} = \sum_{\gamma=1}^n h_{\alpha \gamma} h_{\gamma \beta}\). Using the definition of \(\{\tilde{v}_\gamma\}\), for all \(f \in D\), \[\begin{align} \sum_{\alpha,\beta=1}^n \langle \phi_\alpha, \phi_\beta \rangle_K \nabla_{v_\alpha} \nabla_{v_\beta} f &= \sum_{\alpha,\beta=1}^n c_{\alpha \beta} \nabla_{v_\alpha} \nabla_{v_\beta} f = \sum_{\alpha,\beta=1}^n \sum_{\gamma=1}^n h_{\alpha \gamma} h_{\beta \gamma} \nabla_{v_\alpha} \nabla_{v_\beta} f \\ &= \sum_{\gamma=1}^n \left(\sum_{\alpha=1}^n \nabla_{h_{\alpha \gamma} v_\alpha} \right) \left(\sum_{\beta=1}^n \nabla_{h_{\beta \gamma} v_\beta}\right) f = \sum_{\gamma=1}^n \nabla_{\tilde{v}_\gamma} \nabla_{\tilde{v}_\gamma} f \\ &= \sum_{\gamma=1}^n \left[\Big. (\nabla_{\tilde{v}_\gamma} \tilde{v}_\gamma) \cdot \nabla_R f + (\nabla \nabla_R f)(\tilde{v}_\gamma, \tilde{v}_\gamma) \right] \end{align}\] which, along with the commutation of \(\{\nabla_{v_\alpha}\}\) with \(\eta^{-1/2}\), demonstrates the equation for \(A f\) in 16 . Consequently, \[A f = \overline{b} \cdot \nabla_R f + \sum_{\gamma=1}^n \left[ \frac{1}{2} \left(\nabla_{\sqrt{2 \eta^{-1}} \tilde{v}_\gamma} \sqrt{2 \eta^{-1}} \tilde{v}_\gamma\right) \cdot \nabla_R f + \frac{1}{2} (\nabla \nabla_R f)\left(\sqrt{2 \eta^{-1}} \tilde{v}_\gamma, \sqrt{2 \eta^{-1}} \tilde{v}_\gamma\right) \right]\] which, by Itô’s formula, is the infinitesimal generator for the Itô SDE: \[\label{eq:ito-limit-sde} \; \mathrm{d}Q(t) = \overline{b}(Q(t)) \; \mathrm{d}t + \sum_{\gamma=1}^n \left[ \frac{1}{2} \left(\nabla_{\sqrt{2 \eta^{-1}} \tilde{v}_\gamma} \sqrt{2 \eta^{-1}} \tilde{v}_\gamma\right)(Q(t)) \; \mathrm{d}t + \sqrt{2 \eta^{-1}} \tilde{v}_\gamma(Q(t)) \; \mathrm{d}B_\gamma(t) \right]\tag{21}\] where \(\{B_\gamma\}_{\gamma=1}^n\) is a collection of independent standard Wiener processes so that \(\mathbb{E}\left[B_\gamma(s) B_\gamma(t)\right] = \min(s, t)\) for all \(\gamma\). The \(\nabla_{\sqrt{2} \tilde{v}_\gamma} \sqrt{2} \tilde{v}_\gamma\) term is a Stratonovich correction (see section 4.9 of [10]). Additionally, for each \(\alpha\), we define \(\tilde{B}_\alpha := \sum_{\gamma=1}^n h_{\alpha \gamma} B_\gamma\) so that we can rewrite 21 as the Stratonovich SDE: \[\label{eq:linear-noise-trf} \begin{align} \; \mathrm{d}Q(t) = \overline{b}(Q(t)) \; \mathrm{d}t + \sum_{\gamma=1}^n \sqrt{2 \eta^{-1}} \tilde{v}_\gamma(Q(t)) \circ \; \mathrm{d}B_\gamma(t) &= \overline{b}(Q(t)) \; \mathrm{d}t + \frac{1}{\sqrt{\eta(Q(t))}} \sum_{\gamma=1}^n \sum_{\alpha=1}^n \sqrt{2} h_{\alpha \gamma} v_\alpha(Q(t)) \circ \; \mathrm{d}B_\gamma(t) \\ &= \overline{b}(Q(t)) \; \mathrm{d}t + \frac{1}{\sqrt{\eta(Q(t))}} \sum_{\alpha=1}^n \sqrt{2} v_\alpha(Q(t)) \circ \; \mathrm{d}\tilde{B}_\alpha(t) \end{align}\tag{22}\] where \(\{\tilde{B}_\alpha\}_{\alpha=1}^n\) is a multivariate Wiener process with covariance \[\mathbb{E}\left[\tilde{B}_\alpha(s) \tilde{B}_\beta(t)\right] = \sum_{\gamma=1}^n h_{\alpha \gamma} h_{\gamma \beta} \mathbb{E}\left[B_\gamma(s) B_\gamma(t)\right] = \min(s, t) \sum_{\gamma=1}^n h_{\alpha \gamma} h_{\gamma \beta} = c_{\alpha \beta} \cdot \min(s, t) = \langle \phi_\alpha, \phi_\beta \rangle_K \cdot \min(s, t)\] for all \(1 \leq \alpha, \beta \leq n\) and \(s, t \geq 0\) proving ?? . Lastly, we show that \(D\) is a core for \(A\). For any \(f \in C_c^\infty(R)\) (i.e. smooth compactly supported functions), let \((r, t) \mapsto u(r, t)\) be a solution to the parabolic PDE \(\partial_t u = T_t u\) with initial condition \(u(r, 0) = f(r)\). Then, by parabolic regularity, since \(\overline{b}\) and \(\eta\) are second-differentiable and \(\{v_\alpha\}\) are third-differentiable, for any \(t \geq 0\), we know \(r \mapsto u(r, t)\) is second-differentiable and vanishes at infinity. Thus, from proposition 3.3 of [9] (with \(D_0 = C_c^\infty(R)\)), we see that \(D\) is a core for the infinitesimal generator \(A\). ◻

Proof. Note that \(C_c(R)\) (i.e. compactly supported continuous functions) is dense in \(C_0(R)\) and \(C_c(S)\) is dense in \(L^p(S, \mu)\). Then, \(C_c(R) \times C_c(S) \subseteq C_c(R \times S)\) is dense in \(C_0(R) \otimes L^p(S, \mu)\) which is, in turn, dense in \(C_0(R; L^p(S, \mu))\). Thus, \(C_c(R \times S)\) is dense in \(C_0(R; L^p(S, \mu))\). We use \(\|\cdot\|_\infty\) to denote the norm on \(C_0(R \times S)\) and \(\|\cdot\|_p\) for the norm on \(C_0(R; L^p(S, \mu))\).

Because \(\{T_t\}\) is a semigroup, it suffices to prove strong continuity at \(t = 0\). Let \(f \in C_0(R; L^p(S, \mu))\) and \(\epsilon > 0\). By density, there exists \(g \in C_c(R \times S)\) such that \(\|f - g\|_p \leq \epsilon / 3\). Then, since \(\{T_t\}\) is strongly continuous on \(C_0(R \times S)\), there exists \(\delta > 0\) such that \(\|T_t g - g\|_\infty < \epsilon / 3\) for all \(0 \leq t < \delta\). For any \(f \in L^p(S, \mu)\), since \(\mu\) is an invariant measure, using Jensen’s inequality, \[\|T_t f\|_p^p = \int_S \left|\int_S f(y) \cdot p_t(x, \!\; \mathrm{d}y)\right|^p \mu(\!\; \mathrm{d}x) \leq \int_S |f(y)|^p \int_S p_t(x, \!\; \mathrm{d}y) \mu(\!\; \mathrm{d}x) = \int_S |f(y)|^p \cdot \mu(\!\; \mathrm{d}y) = \|f\|_p^p\] so \(\{T_t\}\) is contractive on \(L^p(S, \mu)\). Thus, for any \(t \in [0, \delta)\), \[\|T_t f - f\|_p \leq \|T_t f - T_t g\|_p + \|T_t g - g\|_p + \|g - f\|_p \leq \|T_t(f - g)\|_p + \|T_t g - g\|_\infty + \|g - f\|_p \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.\] Therefore, \(t \mapsto T_t f\) is continuous and so \(\{T_t\}\) is strongly continuous with respect to \(\|\cdot\|_p\). ◻

Proof. For \(\theta \in [0,2\pi]\), we define \[\Phi(\theta) := \int_0^{2\pi} k(\theta, \alpha) \phi(\alpha) \; \mathrm{d}\alpha \quad \text{where} \quad k(\theta, \alpha) := \frac{1}{2\pi} \begin{cases} \alpha & \text{for } \alpha \leq \theta \\ \alpha - 2\pi & \text{for } \alpha > \theta \end{cases}.\] Since \(\Phi(0) = \Phi(2\pi)\), \(\Phi\) can be extended to a continuous \(2\pi\)-periodic function on \(\mathbb{R}\). Morally, \((\partial_\theta k)(\theta, \alpha) = \delta(\theta - \alpha)\) so \(\Phi' = \phi\). Since \(|k(\theta, \alpha)| \leq 1\), we have \(\|\Phi\| \leq 2\pi \|\phi\|\). Because \(\int_0^{2\pi} k(\theta, \alpha) \; \mathrm{d}\theta = 0\), we see that \(\Phi\) is mean zero. Applying the same construction to \(\Phi\) gives \(\Psi\). ◻

Proof of 1. We prove the convergence in law of the processes using 9. First, we introduce the random process \({\Theta^{(\epsilon)}}(t) = \frac{1}{\epsilon} \vartheta(M(t))\) so that 8 becomes \[\label{eq:ampl-scaling-extended} \begin{align} \; \mathrm{d}{X^{(\epsilon)}}(t) &= b\left( {X^{(\epsilon)}}(t), {\Theta^{(\epsilon)}}(t) \right) \; \mathrm{d}t + \frac{1}{\epsilon} \sum_{\alpha=1}^{d_B} v_\alpha\left( {X^{(\epsilon)}}(t) \right) \phi_\alpha\left( {\Theta^{(\epsilon)}}(t) \right) \; \mathrm{d}t \\ \; \mathrm{d}M(t) &= \nu(M(t)) \; \mathrm{d}t + \sum_{k=1}^{d_W} \sigma_k(M(t)) \; \mathrm{d}W_k(t) \\ \; \mathrm{d}{\Theta^{(\epsilon)}}(t) &= \frac{1}{\epsilon} \rho(M(t)) \; \mathrm{d}t + \frac{1}{\epsilon} \sum_{k=1}^{d_W} \varsigma_k(M(t)) \; \mathrm{d}W_k(t). \end{align}\tag{23}\] where the functions \(\rho : \mathbb{R}^{d_M} \to \mathbb{R}\) and \(\{\varsigma_k : \mathbb{R}^{d_M} \to \mathbb{R}\}_{k=1}^{d_W}\) are defined as \[\begin{align} \rho(m) &:= \nu(m) \cdot \nabla \vartheta(m) + \frac{1}{2} \sum_{k=1}^{d_W} \sum_{i,j=1}^{d_M} \sigma_{k i}(m) \sigma_{k j}(m) \;\; (\partial_{m_i} \partial_{m_j} \vartheta)(m) \\ \varsigma_k(m) &:= \sigma_k(m) \cdot \nabla \vartheta(m) \quad \text{ for all } k \in \{1,\dots, d_W\}. \end{align}\]

Note that \(M\) is a Markov process. While neither \({X^{(\epsilon)}}(t)\) nor \({\Theta^{(\epsilon)}}(t)\) are Markovian on their own, the whole system \({Y^{(\epsilon)}}(t) := \left({X^{(\epsilon)}}(t), M(t), {\Theta^{(\epsilon)}}(t)\right)\) is Markovian. Since only \(2\pi\)-periodic functions of \({\Theta^{(\epsilon)}}(t)\) are considered, we can take the state space of \({Y^{(\epsilon)}}(t)\) to be \(\mathbb{R}^{d_X} \times \mathbb{R}^{d_M} \times \mathbf{S}^1\) where we identify \(\mathbf{S}^1\) with \(\mathbb{R} / 2\pi \mathbb{Z}\). We represent points of this space using the variables \((x, m, \theta)\). \({Y^{(\epsilon)}}(t)\) is a Feller-Dynkin process in the sense of [5]. Let \(T^{(\epsilon)}\) be the associated semigroup on the Banach space \(C_0(\mathbb{R}^{d_X} \times \mathbb{R}^{d_M}; C_b(\mathbf{S}^1)) = C_0(\mathbb{R}^{d_X} \times \mathbb{R}^{d_M} \times \mathbf{S}^1)\). We denote its infinitesimal generator by \({A^{(\epsilon)}}\). For differentiable functions \(f\), we write \(\nabla_x f = (\partial_{x_1}, \ldots, \partial_{x_{d_X}})\) and similarly for \(\nabla_m f\). It follows from the Itô formula that for sufficiently differentiable \(f\), \[\begin{align} \left({A^{(\epsilon)}} f \right)(x, m, \theta) &= b(x, \theta) \cdot \nabla_x f + \frac{1}{\epsilon} \sum_{\alpha=1}^{d_B} \phi_\alpha(\theta) v_\alpha(x) \cdot \nabla_x f + \frac{1}{\epsilon} \rho(m) \partial_\theta f + \nu(m) \cdot \nabla_m f \\ &\qquad + \frac{1}{2} \sum_{k=1}^{d_W} \left[ \frac{1}{\epsilon^2} \varsigma_k(m)^2 \partial_\theta^2 f + \frac{2}{\epsilon} \varsigma_k(m) (\sigma_k(m) \cdot \nabla_m (\partial_\theta f)) + \sum_{i,j=1}^{d_M} \sigma_{k i}(m) \sigma_{k j}(m) \;\; \partial_{m_i} \partial_{m_j} f \right]. \end{align}\] The process \(M\) is a Feller-Dynkin process in its own right and so has an infinitesimal generator \(A_M\). For sufficiently differentiable \(f\), we have \[(A_M f)(x, m, \theta) = \nu(m) \cdot \partial_m f + \frac{1}{2} \sum_{k=1}^{d_W} \sum_{i,j=1}^{d_M} \sigma_{k i}(m) \sigma_{k j}(m) \;\;\; \partial_{m_i} \partial_{m_j} f.\] We wish to apply 9 so we decompose \({A^{(\epsilon)}}\) as \[\begin{align} {A^{(\epsilon)}} f = L f + \frac{1}{\epsilon} (J &+ E) f + \frac{1}{\epsilon^2} \eta K f \quad \text{for sufficiently differentiablefwhere} \\ L := b(x,\theta) \cdot \nabla_x + A_M &\qquad J := \sum_{\alpha=1}^{d_B} \phi_\alpha(\theta) v_\alpha(x) \cdot \nabla_x \\ E := \rho(m) \partial_\theta + \sum_{k=1}^{d_W} \varsigma_k(m) (\sigma_k(m) \cdot \nabla_m) \partial_\theta &\qquad \eta(m) := \frac{1}{2} \kappa(m)^2 = \frac{1}{2} \sum_{k=1}^{d_W} \varsigma_k(m)^2 \qquad K := \partial_\theta^2 \end{align}\] Now, we take \(R = \mathbb{R}^{d_X} \times \mathbb{R}^{d_M}\) and \(S = \mathbf{S}^1\). Also, we take \(\mu\) to be the uniform probability measure on \(\mathbf{S}^1\) and \(\mathcal{B} := \Omega := C_b(S)\).

Next, we verify that the operators fulfill the necessary conditions.

1. \(\partial_\theta^2 (1) = 0\) and, by 12, \(\partial_\theta^2\) is invertible on \(\ker(\omega) \cap \Omega = \ker(\omega) \cap C_b(S)\).

2. Since \(\eta^{-1/2} = \sqrt{2} / \kappa\) is a function of only \(m\), \(\nabla_x \eta^{-1/2} = 0\). Since \(\phi_\alpha, K^{-1} \phi_\beta \in C_b(S)\), we have \(\phi_\alpha \cdot K^{-1} \phi_\beta \in C_b(S) = \Omega\).

3. \(\partial_\theta\) is \(C_0(R)\)-linear so \(E\) is as well and, further, \(C_0(R) \subseteq \ker(\partial_\theta) \subseteq \ker(E)\). If \(\int \phi \; \mathrm{d}\theta = 0\) and \(\phi\) is differentiable then \(\int \partial_\theta \phi \; \mathrm{d}\theta = 0\) too so \(E\) preserves \(\ker(\omega) \cap \Omega\).

4. The elements of \(D\) have first derivatives in \(C_0(R)\) so \(D \subseteq \mathcal{D}\left(L\right)\). Again, by 12, \(K^{-1}\) is bounded on \(\ker(\omega) \cap \Omega\) and, by the assumptions on \(b\), \(L f = b \cdot \nabla_x f \in C_0(R \times S) = C_0(R;\Omega)\). Since \(\nabla_x\) is \(\mathcal{H}\)-linear, so is \(L\).

For each \(\alpha\), let \(\Psi_\alpha, \Phi_\alpha \in C_b(S)\) such that \(\partial_\theta^2 \Psi_\alpha = \partial_\theta \Phi_\alpha = \phi_\alpha\). Then, the bilinear form \(\langle \cdot, \cdot \rangle_K\) is \[\begin{align} c_{\alpha \beta} = \langle \phi_\alpha, \phi_\beta \rangle_K &= -\mu(\phi_\alpha \cdot K^{-1} \phi_\beta) = - \frac{1}{2\pi} \int_0^{2\pi} \phi_\alpha(\theta) \cdot \Psi_\beta(\theta) \; \mathrm{d}\theta \\ &= - \frac{1}{2\pi} \phi_\alpha(\theta) \Phi_\beta(\theta) \Big|_{\theta=0}^{2\pi} + \frac{1}{2\pi} \int_0^{2\pi} \Phi_\alpha(\theta) \cdot \Phi_\beta(\theta) \; \mathrm{d}\theta = \frac{1}{2\pi} \langle \Phi_\alpha, \Phi_\beta \rangle_{L^2([0, 2\pi))} \end{align}\] By 10, we know the solution \(X\) to the SDE (?? ) has infinitesimal generator \(A\) fulfilling 15 . Then, by 9, the solutions \({X^{(\epsilon)}}\) to the SDE (8 ) converge in law to \(X\). ◻

Proof of 3. Let \(T_M\) be the semigroup for \(M\) on \(\mathcal{B}_M\). We define the operator \(A_M^{-1} : D_0 \to D_0\) as \[A_M^{-1} \phi := -\int_0^\infty T_M(t) \phi \; \mathrm{d}t \in D_0.\] for all \(\phi \in D_0 \subseteq \mathcal{B}_M\). It remains to show that \(A_M^{-1}\) is well-defined while its linearity is immediate. For any \(t \geq 0\) and \(m \in S\), let \(\mu_{t,m}\) be the distribution of \(M(t)\) given that \(M(0) = m\). Thus, \(\mu_{t,m}(E) = \mathbb{P}\left(M(t) \in E\,\middle|\,M(0) = m\right)\) for every \(E \in \mathcal{M}\). From the strong ergodicity of \(M\), we get \(\|\mu_{t,m} - \mu\| \leq C e^{-\lambda t}\) for any \(m \in S\) and \(t \geq t_0\). For any \(t \geq t_0\), since \(\int \phi \; \mathrm{d}\mu = 0\), \[\label{eq:semigrp-bnd} \begin{align} \left|\Big. (T_M(t) \phi)(m) \right| &= \left|\Big.\mathbb{E}\left[\Big. \phi(M(t))\middle|M(0) = m\right] \right| = \left| \int \phi \; \mathrm{d}\mu_{t,m} \right| \\ &= \left| \int \phi \; \mathrm{d}\mu_{t,m} - \int \phi \; \mathrm{d}\mu \right| \\ &\leq \int |\phi| \cdot \; \mathrm{d}|\mu_{t,m} - \mu| \leq \|\phi\|_\infty \cdot C e^{-\lambda t}. \end{align}\tag{24}\] Thus, we can bound the integral with respect to \(t\) by \[\begin{align} \int_0^\infty \left\|\Big. T_M(t) \phi \right\|_\infty \; \mathrm{d}t &= \int_0^{t_0} \left\|\Big. T_M(t) \phi \right\|_\infty \; \mathrm{d}t + \int_{t_0}^\infty \left\|\Big. T_M(t) \phi \right\|_\infty \; \mathrm{d}t \\ &\leq t_0 \|\phi\|_\infty + \|\phi\|_\infty \int_0^\infty C e^{-\lambda t} \; \mathrm{d}t < \infty \end{align}\] implying that \(A_M^{-1} \phi\) is well-defined. From the above inequality, for any \(\phi \in D_0\), we also have \[\|A_M^{-1} \phi\|_\infty \leq t_0 \|\phi\|_\infty + \|\phi\|_\infty \int_0^\infty C e^{-\lambda t} \; \mathrm{d}t = \left(t_0 + \frac{C}{\lambda}\right) \|\phi\|_\infty\] so \(\|A_M^{-1}\| \leq t_0 + \frac{C}{\lambda} < \infty\) and \(A_M^{-1}\) is bounded. Also, using the invariance of \(\mu\), we get \[\int_S (A_M^{-1} \phi)(m) \; \mathrm{d}\mu(m) = -\int_0^\infty \int_S T_M(t) \phi \; \mathrm{d}\mu \; \mathrm{d}t = \int_0^\infty \left(\int_S \phi \; \mathrm{d}\mu\right) \; \mathrm{d}t = 0\] so the range of \(A_M^{-1}\) lies in \(D_0\). Lastly, we know \(t \mapsto T_M(t) \phi\) is continuous so \[\begin{align} A_M A_M^{-1} \phi &= -\lim_{h \downarrow 0} \frac{T_M(h) - I}{h} \int_0^\infty T_M(t) \phi \; \mathrm{d}t = \lim_{h \downarrow 0} \frac{1}{h} \left[ \int_0^\infty T_M(t) \phi \; \mathrm{d}t - \int_0^\infty T_M(t+h) \phi \; \mathrm{d}t \right] \\ &= \lim_{h \downarrow 0} \frac{1}{h} \left[ \int_0^\infty T_M(t) \phi \; \mathrm{d}t - \int_h^\infty T_M(t) \phi \; \mathrm{d}t \right] = \lim_{h \downarrow 0} \frac{1}{h} \int_0^h T_M(t) \phi \; \mathrm{d}t = T_M(0) \phi = \phi \end{align}\] implying that \(A_M A_M^{-1} = I_{D_0}\). ◻

Proof of 5. We can rewrite 9 by defining \({M^{(\epsilon)}}(t) = M(t / \epsilon^2)\) to obtain \[\label{eq:time-scaling-mod} \; \mathrm{d}{X^{(\epsilon)}}(t) = b\left({X^{(\epsilon)}}(t), {M^{(\epsilon)}}(t)\right) \; \mathrm{d}t + \frac{1}{\epsilon} \sum_{\alpha=1}^{d_B} \phi_\alpha\left({M^{(\epsilon)}}(t)\right) \, v_\alpha\left( {X^{(\epsilon)}}(t) \right) \; \mathrm{d}t.\tag{25}\] By assumption, \(M\) is a Feller-Dynkin process so let \(\{T_M(t)\}_{t \geq 0}\) be its Feller semigroup on \(\mathcal{B}_M \subseteq C_b(S)\) and \(A_M\) be the associated infinitesimal generator. If we take \(T_{{M^{(\epsilon)}}}\) to be the semigroup of \({M^{(\epsilon)}}\) then for any \(m \in S\), \[(T_{{M^{(\epsilon)}}}(t) f)(m) = \mathbb{E}\left[f\left( {M^{(\epsilon)}}(t) \right)\middle|{M^{(\epsilon)}}(0) = m\right] = \mathbb{E}\left[f(M(t / \epsilon^2))\middle|M(0) = m\right] = (T_M(t / \epsilon^2) f)(m).\] Therefore, the infinitesimal generator \(A_{{M^{(\epsilon)}}}\) of \({M^{(\epsilon)}}\) is \[A_{{M^{(\epsilon)}}} f = \lim_{t \downarrow 0} \frac{T_{{M^{(\epsilon)}}}(t) f - f}{t} = \lim_{t \downarrow 0} \frac{T_M(t / \epsilon^2) f - f}{t} = \lim_{t \downarrow 0} \frac{T_M(t) f - f}{\epsilon^2 t} = \frac{1}{\epsilon^2} A_M f\] for all \(f \in \mathcal{D}\left(A_M\right)\). Now, \(\left( {X^{(\epsilon)}}, {M^{(\epsilon)}} \right)\) is a Feller-Dynkin process with infinitesimal generator \({A^{(\epsilon)}}\) which satisfies \[{A^{(\epsilon)}} f = b(x, m) \cdot \nabla_x f + \frac{1}{\epsilon} \sum_{\alpha=1}^{d_B} \phi_\alpha(m) v_\alpha(x) \cdot \nabla_x f + \frac{1}{\epsilon^2} A_M f\] for all sufficiently differentiable \(f\). We wish to apply 9. We take \(R = \mathbb{R}^{d_X}\) and \(S\) is given. We take \(\mu\) to be the invariant measure for \(M\) and \(\mathcal{B} := \Omega := \mathcal{B}_M\). Next, we decompose \({A^{(\epsilon)}}\) for sufficiently differentiable \(f\) as \[\begin{align} {A^{(\epsilon)}} f = L f + \frac{1}{\epsilon} &J + \frac{1}{\epsilon^2} K f \quad \text{where} \\ L := b(x, m) \cdot \nabla_x \qquad J := \sum_{\alpha=1}^{d_B} &\phi_\alpha(m) v_\alpha(x) \cdot \nabla_x \qquad K := A_M \end{align}\] while \(\eta = 1\) and \(E = 0\). We now verify that these operators fulfill the conditions of 9.

1. \(K(1) = A_M(1) = 0\) and, by 3, \(A_M\) is right-invertible on \(\ker(\omega) \cap \Omega\).

2. For all \(\alpha\) and \(\beta\), \(\nabla_{v_\alpha} (\eta^{-1/2}) = \nabla_{v_\alpha} (1) = 0\) while \(\phi_\alpha, K^{-1} \phi_\beta \in \mathcal{B}_M\) which, since \(\mathcal{B}_M\) is an algebra, means \(\phi_\alpha \cdot K^{-1} \phi_\beta \in \mathcal{B}_M = \Omega\).

3. Since \(E = 0\), it fulfills all of the necessary properties.

4. The elements of \(D\) have first derivatives in \(C_0(R)\) so \(D \subseteq \mathcal{D}\left(L\right)\). From 3, we know \(K^{-1} = A_M^{-1}\) is bounded on \(\ker(\omega) \cap \Omega\) and, by the assumptions on \(b\), \(L f = b \cdot \nabla_x f \in C_0(R; \mathcal{B}_M) = C_0(R; \Omega)\). Since \(\nabla_x\) is \(\mathcal{H}\)-linear so is \(L\).

By 10, we see that the diffusion process solving \(X\) of the SDE (?? ) has an infinitesimal generator \(A\) fulfilling 15 . Thus, by 9, the solutions \({X^{(\epsilon)}}\) to the SDE (25 ) converge in law to \(X\). ◻

Proof of 6. Let \(A_M\) be the infinitesimal generator for \(M\) so that \[A_M f = \frac{1}{2} \nabla_m^2 f - (\nabla_m U) \cdot (\nabla_m f)\] for all sufficiently differentiable \(f \in C_b(\mathbb{R}^{d_M})\). By 1.3B of [6], \(\mathcal{B}_M = \overline{\mathcal{D}\left(A_M\right)}\) so \(\{\phi_\alpha\} \subseteq \mathcal{D}\left(A_M\right) \subseteq \mathcal{B}_M\). Further, second differentiable functions on \(\mathbb{R}^{d_M}\) with appropriately bounded derivatives form a core for \(A_M\) and, since these functions form an algebra, \(\mathcal{B}_M\) is an algebra as well. Then, we can use \(C_0^2(\mathbb{R}^{d_M})\) as a core and for any \(f \in C_0^2(\mathbb{R}^{d_M})\), \[\int (A_M f)(m) \frac{e^{-2 U(m)}}{Z} \; \mathrm{d}m = \int \left(\frac{1}{2} \nabla^2 f - (\nabla U) \cdot (\nabla f) \right) \frac{e^{-2 U}}{Z} \; \mathrm{d}m = \int f \left(\frac{1}{2} \nabla^2 e^{-2 U} + \nabla \cdot ((\nabla U) e^{-2 U}) \right) \frac{1}{Z} \; \mathrm{d}m\] where \(Z > 0\) is an appropriate normalization to ensure \(\frac{1}{Z} e^{-2U} \; \mathrm{d}m\) is a probability measure. Now, we observe that \(\frac{1}{2}\nabla e^{-2U} + \left(\nabla U\right) e^{-2U} = 0\) which implies that \(\frac{1}{Z} e^{-2U} \; \mathrm{d}m\) is an invariant measure for \(M\). Conjugating \(A_M\) by the multiplication operator \(e^U\), we obtain \[\begin{align} e^{-U} A_M (e^U f) &= \frac{1}{2} e^{-U} \nabla^2 (e^U f) - e^{-U} (\nabla U) \cdot \nabla (e^U f) \\ &= \frac{1}{2} e^{-U} \left[e^U \nabla^2 f + 2 e^U (\nabla U) \cdot (\nabla f) + e^U (|\nabla U|^2 + \nabla^2 U) f \right] - e^{-U} (\nabla U) \cdot \left[ e^U \nabla f + e^U (\nabla U) f \right] \\ &= \frac{1}{2} \nabla^2 f + (\nabla U) \cdot (\nabla f) + \frac{1}{2} (|\nabla U|^2 + \nabla^2 U) f - (\nabla U) \cdot (\nabla f) - |\nabla U|^2 f \\ &= \frac{1}{2} \nabla^2 f + \frac{1}{2} (\nabla^2 U - |\nabla U|^2) f = \frac{1}{2} \nabla^2 f - V f = - H f \end{align}\] implying that \(A_M = - e^U H e^{-U}\) and \(A_M^{-1} = - e^U H^{-1} e^{-U}\) where \(H^{-1}\) is the inverse of \(H\) for \(\phi \in D_0 := \{\psi \in \mathcal{B}_M \,:\, 0 = \int_S \psi \; \mathrm{d}\mu = \int Z^{-1} e^{-2U} \psi \; \mathrm{d}m \}\). Then, we know that the bilinear form associated with \(M\) is \[\begin{align} \langle \phi, \psi \rangle_M &= - \int_{\mathbb{R}^{d_M}} \phi(m) (A_M^{-1} \psi)(m) \; \mathrm{d}\mu(m) = \int \phi(m) (e^U H^{-1} e^{-U} \psi)(m) \frac{1}{Z} e^{-2U} \; \mathrm{d}m \\ &= \frac{1}{Z} \int (e^{-U} \phi)(m) \cdot (H^{-1} e^{-U} \psi)(m) \; \mathrm{d}m. \end{align}\] Since \(H\) is self-adjoint with respect to\(\; \mathrm{d}m\), the form \(\langle \cdot, \cdot \rangle_M\) symmetric. It is known that \(\phi = e^{-U}\) is a non-degenerate eigenvector of \(H\) when the potential \(V = \frac{1}{2} (|\nabla U|^2 - \nabla^2 U) \to \infty\) as \(z \to \infty\) (theorem XIII.47 of [11]). Thus, \(H^{-1}\) is positive on \(D_0\) and so \(\langle \cdot, \cdot \rangle_M\) is positive semi-definite on \(\{\phi_\alpha\}\).

From the first part of ?? , for all \(x \in \mathbb{R}^{d_M}\), we get the bound \[\label{eq:dot-prod-bdd} x \cdot \nabla U(x) = x \cdot \int_0^1 |x| (\nabla \nabla U(tx)) x \; \mathrm{d}t \geq \int_0^1 |x|^3 h_1'(t |x|) \; \mathrm{d}t = |x|^2 (h_1(|x|) - h_1(0)) = |x|^2 h_1(|x|)\tag{26}\] and similarly \(x \cdot \nabla U(x) \leq |x|^2 h_2(|x|)\). By theorem 1.1(b) in Wang [7], from 26 , we know \(h_1 \leq g \leq h_2\) and so the conditions in ?? imply that \(M\) is strongly ergodic. Thus, from 5, we obtain the weak convergence. ◻

Proof. For any \(n\), we want to solve the equation \[\left[ \big. \frac{1}{2} \nabla_z^2 - (\nabla U) \cdot \nabla_z\right] g + in\rho g = f.\] The operator \[g \mapsto Lg = \left[ \big. \frac{1}{2} \nabla_z^2 - (\nabla U) \cdot \nabla_z + \rho \partial_\theta \right] g\] is a generator of a diffusion in the space \(\mathcal{H}\). In particular, it has no positive spectrum, and thus the same is true for its image under the conjugation by \(e^{-U}\), which is a unitary operator between the two Hilbert spaces. Explicitly, substituting \(g = e^U \phi\) and multiplying the resulting operator by \(-1\) for greater clarity, we obtain the Schrödinger operator \(H\) on the space \(L^2(\mathbb{R}^{d_Z})\): \[\phi \mapsto H\phi = -e^{-U}L\left(e^U \phi\right) = - \frac{1}{2} \nabla_z^2\phi + V\phi \quad \text{with} \quad V = \frac{1}{2}\left(\left|\nabla_z U\right|^2 - \nabla_z^2U\right).\] Under our assumptions on \(U\), the potential \(V\) is bounded below and goes to \(+\infty\) for \(|z| \to \infty\). It follows that \(H\) has only discrete spectrum (theorem 10.4.3 in [12]). An explicit calculation shows that \(H\) applied to the function \(e^{-U}\) is zero, so zero is its eigenvalue. Since \(H\) has the same spectrum as \(-L\), it follows that zero is its lowest eigenvalue which by theorem XIII.47 of [11] is non-degenerate. In terms of \(\phi\), the equation for \(g\) becomes \[H\phi +in\rho\phi = -fe^{-U}\] If \(n = 0\) this equation has a solution if (and only if) its right-hand side is orthogonal to the ground state, i.e. if \(\langle f, 1 \rangle_\mathcal{H} = 0\). To treat the case when \(n \not = 0\), we will first show that the spectrum of the Schrödinger operator with the complex potential \(V + in\rho\) is also discrete. Indeed, denoting the operators of multiplication by \(\rho\) and \(V\) also by \(\rho\) and \(V\), we have \[\label{eq:resolv-factors} \rho\left(bI + H\right)^{-1} = \rho (bI + V)^{-\frac{1}{2}} (bI + V)^{\frac{1}{2}} \left(bI + H\right)^{-\frac{1}{2}} \left(bI + H\right)^{-\frac{1}{2}}\tag{27}\] The product \(\rho (bI + V)^{-\frac{1}{2}}\) is bounded by assumption. The product \((bI + V)^{\frac{1}{2}} (bI + H)^{-\frac{1}{2}}\) is bounded because \(bI + V \leq bI + H\), so for any \(\phi\) \[\begin{align} \left((bI + V)^{\frac{1}{2}} (bI + H)^{-\frac{1}{2}} \phi, (bI + V)^{\frac{1}{2}} (bI + H)^{-\frac{1}{2}} \phi\right) &= \left((bI + H)^{-\frac{1}{2}}\phi, (bI + V) (bI + H)^{-\frac{1}{2}} \phi\right) \\ &\leq \left((bI + H)^{-\frac{1}{2}} \phi, (bI + H) (bI + H)^{-\frac{1}{2}} \phi\right) = \left(\phi, \phi\right) \end{align}\] Finally, since \(b > 0\) and the spectrum of \(H\) is non-negative and discrete, the factor \((bI + H)^{-\frac{1}{2}}\) in 27 is a compact operator. So the operator \(\rho (bI + H)^{-1}\) is a product of a bounded operator and a compact operator, hence is also compact. This shows that the perturbation \(in\rho\) is compact relative to \(H\) and thus, by the Weyl theorem (theorem 9.5.1 in [12]), adding it does not change the essential spectrum of \(H\). We claim that \(0\) is not an eigenvalue of the perturbed operator. For, suppose that \(\psi = \alpha + i\beta\) and \(H\psi +\rho \psi= 0\). Taking the real and imaginary parts, we obtain: \[H\alpha -\rho\beta = 0 \quad \text{ and } \quad H\beta + \rho\alpha = 0.\] Multiplying the first equation by \(\alpha\) and the second equation by \(\beta\), adding up and integrating, we get \[(\alpha, H\alpha) + (\beta, H\beta) = 0\] Since \(H\) is nonnegative-definite, this implies that \((\alpha, H\alpha) = (\beta, H\beta) = 0\). Since the zero eigenvalue of \(H\) is simple, this is only possible if \(\alpha\) and \(\beta\), and hence also \(\psi\), are multiples of the ground state of \(H\)—a contradiction. It now follows from the Fredholm alternative that the equation \[H\phi + in\rho\phi = -fe^{-U}\] has a (unique) solution. Then, since \(\rho\) is real and \(H\) is positive semi-definite, \[\mathrm{Re} \langle f, g \rangle_\mathcal{H} = \mathrm{Re} \langle -(H + in\rho) \phi, \phi \rangle = \mathrm{Re} \left(\langle -H \phi, \phi \rangle + in \langle \rho \phi, \phi \rangle\right) = \langle -H \phi, \phi \rangle \leq 0\] where \(\langle \cdot, \cdot \rangle\) is the inner product in \(L^2(\mathbb{R}^{d_Z})\). ◻

Proof of 7. Let \({Y^{(\epsilon)}} := ({X^{(\epsilon)}}, {\Theta^{(\epsilon)}}, {Z^{(\epsilon)}})\) be a solution of 12 . We regard \({Y^{(\epsilon)}}\) as a process in \(\mathbb{R}^{d_X} \times \mathbf{S}^1 \times \mathbb{R}^{d_Z}\). Then, \({Y^{(\epsilon)}}\) is a Feller-Dynkin process with generator \({A^{(\epsilon)}}\) given by \[\left({A^{(\epsilon)}} f\right)(x, \theta, z) = b(x, \theta) \cdot \nabla_x f + \frac{1}{\epsilon} \sum_{\alpha=1}^{d_B} \phi_\alpha(\theta) v_\alpha(x) \cdot \nabla_x f + \frac{1}{\epsilon^2} \rho(z) \partial_\theta f - \frac{1}{\epsilon^2} \nabla U(z) \cdot \nabla_z f + \frac{1}{2 \epsilon^2} \nabla_z^2 f\] for sufficiently differentiable \(f \in C_0(\mathbb{R}^{d_X} \times \mathbf{S}^1 \times \mathbb{R}^{d_Z})\). We wish to apply 9. We take \(R := \mathbb{R}^{d_X}\) and \(S := \mathbf{S}^1 \times \mathbb{R}^{d_Z}\). We define \(\mu_Z\) to be the probability measure on \(\mathbb{R}^{d_Z}\) obtained by normalizing \(e^{-2U(z)} \; \mathrm{d}z\). Then, we define the probability measure \(\mu := \frac{1}{2\pi} \; \mathrm{d}\theta \times \mu_Z\) on \(S = \mathbf{S}^1 \times \mathbb{R}^{d_Z}\) and take \(\mathcal{B} := L^2(S, \mu)\) as well as \[\Omega := \mathrm{Trig}(\mathbf{S}^1) \otimes L^2(\mathbb{R}^{d_Z}, \mu_Z) \quad \text{where} \quad \mathrm{Trig}(\mathbf{S}^1) := \left\{\phi \in C_b(\mathbf{S}^1) \,:\, \phi \text{ is a trigonometric polynomial} \right\}.\] By 11 for \(p = 2\), the semigroup for \({Y^{(\epsilon)}}\) is strongly continuous on \(C_0(R; \mathcal{B})\). Next, we decompose \({A^{(\epsilon)}}\) as \[\begin{align} {A^{(\epsilon)}} f = L f + \frac{1}{\epsilon} &J f + \frac{1}{\epsilon^2} K f \quad \text{for sufficiently differentiablefwhere} \\ L := b(x, \theta) \cdot \nabla_x \qquad J := \sum_{\alpha=1}^{d_B} &\phi_\alpha(\theta) v_\alpha(x) \cdot \nabla_x \qquad K := \frac{1}{2} \nabla_z^2 - \nabla U(z) \cdot \nabla_z + \rho(z) \partial_\theta \end{align}\] while \(\eta = 1\) and \(E = 0\). We also define the bilinear form \(\langle \cdot, \cdot \rangle_{\Theta,Z}\) for any \(\phi, \psi \in \ker(\omega) \cap \Omega\) as \[\label{eq:integ-noise-form} \langle \phi, \psi \rangle_{\Theta,Z} := -\omega(\phi^* \cdot K^{-1} \psi) = \langle \phi, \psi \rangle_K.\tag{28}\] Note that, since \(H\) in the proof of 13 is self-adjoint, the form \(\langle \cdot, \cdot \rangle_{\Theta,Z}\) is symmetric. Moreover, because of the negativity conditions in 13, the form \(\langle \cdot, \cdot \rangle_{\Theta,Z}\) is positive semi-definite. Now, we verify the conditions for 9.

1. Observe that \(K(1) = 0\) and, by 13, since the elements \(f e^{n\theta i}\) for \(f \in L^2(\mathbb{R}^{d_Z}, \mu_Z)\) span \(\Omega\), \(K\) is right-invertible on \(\Omega\).

2. For all \(\alpha\) and \(\beta\), \(\nabla_{v_\alpha} (\eta^{-1/2}) = \nabla_{v_\alpha} (1) = 0\). Since \(\mathrm{Trig}(\mathbf{S}^1)\) is an algebra with \(\phi_\alpha \in \mathrm{Trig}(\mathbf{S}^1) \otimes 1\) and \(K^{-1} \phi_\beta \in \mathrm{Trig}(\mathbf{S}^1) \otimes L^2(\mu_Z)\), it follows that \(\phi_\alpha \cdot K^{-1} \phi_\beta \in \mathrm{Trig}(\mathbf{S}^1) \otimes L^2(\mu_Z) = \Omega\).

3. Since \(E = 0\), it fulfills all of the necessary conditions.

4. The elements of \(D\) have first derivatives in \(C_0(R)\) so \(D \subseteq \mathcal{D}\left(L\right)\). By the assumptions on \(b\), \(L f = b \cdot \nabla_x f \in C_0(R) \otimes \Omega\). Since \(\nabla_x\) is \(L^2(S, \mu)\)-linear so is \(L\).

Then, by 10, we see that the solution \(X\) of the SDE (?? ) has an infinitesimal generator \(A\) fulfilling 15 . Thus, by 9, the solutions \({X^{(\epsilon)}}\) to the SDE (12 ) converge in law to \(X\). ◻

Proof of 8. We will apply 7. We have \(\rho(z) = z\), \(U(z) = \frac{1}{2} z^2\), and \(\phi_1(\theta) = \cos(\theta)\). Thus, we know \(V = |\nabla_z U|^2 - \nabla_z^2 U = z^2 - 1\) so \(V + 2 = z^2 + 1 \geq 0\) and \(|\rho|^2 = z^2 \leq z^2 + 1\). Let \(\{H_n\}_{n=0}^\infty\) be the Hermite polynomials. \(A_{\Theta,Z}^{-1} e^{ik\theta}\) can be calculated by expanding the sought function in terms of Hermite polynomials, substituting into the equation, and solving for the coefficients of the expansion. This is standard so we just give the result. Recall that \(H_n' = 2zH_n - H_{n+1} = 2n H_{n-1}\) so \(H_n'' - 2zH_n' = 2n (H_{n-1}' - 2z H_{n-1}) = -2n H_n\). Now, we define \(\Phi_k : \mathbb{R} \times \mathbb{R} \to \mathbb{C}\) as \[\Phi_k(\theta, z) := e^{3k^2/4} e^{ik(\theta + z)} \sum_{n=0}^\infty \frac{(-ik)^n}{(n + k^2 / 2) 2^n n!} H_n(z - ik).\] The infinitesimal generator for \((\Theta, Z)\) (the driving process without time scaling) is \(A_{\Theta,Z} = \frac{1}{2} \partial_z^2 - z \partial_z + z \partial_\theta\) so \[e^{-ik(\theta + z)} A_{\Theta,Z} e^{ik(\theta + z)} = \frac{1}{2} (\partial_z + ik)^2 - z (\partial_z + ik) + z(\partial_\theta + ik) = \frac{1}{2} \partial_z^2 - (z - ik) \partial_z - \frac{k^2}{2} + z \partial_\theta\] and \(\left(\frac{1}{2} \partial_z^2 - (z - ik) \partial_z\right) H_n(z - ik) = -n H_n(z - ik)\). Thus, using the generating function for the Hermite polynomials, namely \(e^{2zt - t^2} = \sum_{n=0}^\infty H_n(z) \frac{t^n}{n!}\), \(A_{\Theta,Z}\) will act on \(\Phi_k(\theta, z)\) as \[\begin{align} A_{\Theta,Z} \Phi_k &= e^{3k^2/4} e^{ik(\theta + z)} \sum_{n=0}^\infty \frac{(-ik)^n}{(n + k^2 / 2) 2^n n!} \left(\frac{1}{2} \partial_z^2 - (z - ik) \partial_z - \frac{k^2}{2} + z \partial_\theta\right) H_n(z - ik) \\ &= - e^{3k^2/4} e^{ik(\theta + z)} \sum_{n=0}^\infty \frac{(-ik)^n}{2^n n!} H_n(z - ik) = - e^{3k^2/4} e^{ik(\theta + z)} e^{2(z - ik)(-ik/2) - (-ik/2)^2} = - e^{ik\theta}. \end{align}\] Therefore, \(- \Phi_k = A_{\Theta,Z}^{-1} e^{ik\theta}\). We will let \(\mathcal{H} = L^2\left(\mathbb{R}, \frac{1}{\sqrt{\pi}} e^{-z^2} \; \mathrm{d}z\right)\) which is the space in which \(\left\{(2^n n!)^{-1/2} H_n \right\}\) is an orthogonal basis. Then, for any \(k, \ell \in \mathbb{Z}\), we can calculate the value of the form \[\begin{align} \langle e^{i \ell \theta}, e^{i k \theta} \rangle_{\Theta,Z} &= - \int_{-\infty}^\infty \int_0^{2\pi} e^{-i \ell \theta} (- \Phi_k(\theta, z)) \frac{1}{2\pi} \frac{e^{-z^2}}{\sqrt{\pi}} \; \mathrm{d}\theta \; \mathrm{d}z \\ &= e^{3k^2/4} \sum_{n=0}^\infty \frac{(-ik)^n}{(n + k^2/2) 2^n n!} \left(\frac{1}{2\pi}\int_0^{2\pi} e^{i(k -\ell) \theta} \; \mathrm{d}m\right) \left(\frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-ik(z - ik)} H_n(z - ik) e^{-z^2 + 2ikz + k^2} \; \mathrm{d}z\right) \\ &= e^{3k^2/4} \delta_{k \ell} \sum_{n=0}^\infty \frac{(-ik)^n}{(n + k^2/2) 2^n n!} \frac{1}{\sqrt{\pi}} \int_{-\infty - ik}^{\infty - ik} e^{-ikz} H_n(z) e^{-z^2} \; \mathrm{d}z \end{align}\] By the holomorphy of \(e^{- z^2 - ikz} H_n(z)\) and because of the rapid decay of \(e^{-z^2}\), we can use the generating function and orthogonality of the Hermite polynomials to get \[\begin{align} \frac{1}{\sqrt{\pi}} \int_{-\infty - ik}^{\infty - ik} e^{2 \left(-\frac{ik}{2}\right) z} H_n(z) e^{-z^2} \; \mathrm{d}z &= e^{\left(-\frac{ik}{2}\right)^2} \sum_{j=0}^\infty \frac{(-ik)^j}{2^j j!} \int_{-\infty}^\infty H_j(z) H_n(z) \frac{1}{\sqrt{\pi}} e^{-z^2} \; \mathrm{d}z \\ &= e^{-k^2/4} \sum_{j=0}^\infty \frac{(-ik)^j}{2^j j!} 2^n n! \delta_{nj} = e^{-k^2/4} (-ik)^n \end{align}\] and so \(\langle e^{i\ell \theta}, e^{ik \theta} \rangle_{\Theta,Z}\) becomes \[\langle e^{i \ell \theta}, e^{ik \theta} \rangle_{\Theta,Z} = e^{3k^2/4} \delta_{k\ell} \sum_{n=0}^\infty \frac{(-ik)^n}{(n + k^2/2) 2^n n!} e^{-k^2/4} (-ik)^n = e^{\frac{k^2}{2}} \delta_{k\ell} \sum_{n=0}^\infty \frac{(-1)^n k^{2n}}{(n + k^2/2) 2^n n!}\] completing the proof. ◻