Simultaneous Small Noise Limit for Singularly Perturbed Slow-Fast Coupled Diffusions

Abstract

We consider a simultaneous small noise limit for a singularly perturbed coupled diffusion described by

$$\begin{aligned} dX^{\varepsilon }_t= & {} b(X^{\varepsilon }_t, Y^{\varepsilon }_t)dt + \varepsilon ^{\alpha }dB_t, \\ dY^{\varepsilon }_t= & {} - \frac{1}{\varepsilon } \nabla _yU(X^{\varepsilon }_t, Y^{\varepsilon }_t)dt + \frac{s(\varepsilon )}{\sqrt{\varepsilon }} dW_t, \end{aligned}$$

where $B_t, W_t$ are independent Brownian motions on ${\mathbb R}^d$ and ${\mathbb R}^m$ respectively, $b : \mathbb {R}^d \times \mathbb {R}^m \rightarrow \mathbb {R}^d$, $U : \mathbb {R}^d \times \mathbb {R}^m \rightarrow \mathbb {R}$ and $s :(0,\infty ) \rightarrow (0,\infty )$. We impose regularity assumptions on b, U and let $0< \alpha < 1.$ When $s(\varepsilon )$ goes to zero slower than a prescribed rate as $\varepsilon \rightarrow 0$, we characterize all weak limit points of $X^{\varepsilon }$, as $\varepsilon \rightarrow 0$, as solutions to a differential equation driven by a measurable vector field. Under an additional assumption on the behaviour of $U(x, \cdot )$ at its global minima we characterize all limit points as Filippov solutions to the differential equation.

Appendices

Appendix A: Existence, Uniqueness, and Gradient Estimates

In this section we show that the coupled system (4) and (5) has a unique strong solution. We begin with a technical lemma.

Lemma A.1

Under (U1), (8) and (9) in assumption (U2) there is $K_4 >0$ and $R^\prime \ge R$ such that

$$\begin{aligned} \langle \nabla _y U(x, y) , y \rangle> K_4 \Vert y\Vert ^2, \quad \Vert y \Vert > R^\prime . \end{aligned}$$

(70)

Also, there exists a nonnegative continuous function $g: (0,\infty ) \rightarrow (0,\infty )$ such that

$$\begin{aligned}&\sup _{z,y \in {\mathbb R}^m: \Vert z-y \Vert = r} - \frac{1}{r} \langle \nabla _yU(x,z) - \nabla _yU(x,y), z- y \rangle \le g(r), \text{ for } \text{ all } r > 0,\nonumber \\&\text{ with } \Gamma := \int _0^\infty g(s) ds < \infty . \end{aligned}$$

(71)

Proof

We proceed as follows. Let $\mathbb {B}_a$ denote the closed ball of radius a centred at the origin. For any y with $||y|| > R$, writing $\nabla _y U(x,y) = \nabla _y U(x,0) + \int _0^1 D^2_y U(x,ty) y ~dt$, we have

$$\begin{aligned} \langle \nabla _y U(x,y), y \rangle= & {} \langle \nabla _y U(x,0), y \rangle + \int _0^1 \langle y, D_y^2U(x,ty) y \rangle ~dt \\\ge & {} -M||y|| - \int _0^{R/||y||} M||y||^2 dt + \int _{R/||y||}^1 K_3||y||^2 dt \\= & {} -M||y|| - MR ||y|| + K_3 ||y||^2 (1 - R/||y||) \\= & {} ( K_3 ||y||^2 - (M + MR + K_3 R)) ||y|| \\\ge & {} K_4 ||y||^2 \end{aligned}$$

for any $K_4 > 0$ and $\Vert y\Vert \ge R^\prime \ge R + M(1+R)/K_3 + K_4 / K_3$. In the second inequality above, we have used (9) for the line segment joining 0 to y that lies within $\mathbb {B}_R$ and (8) for the remaining line segment. This establishes (70).

Next, for any $y,z \in {\mathbb R}^m$, define $t_0(y,z)$ to be the fractional length of the line segment joining y to z that is within $\mathbb {B}_R$. Take $R_1 = R(1+2Mm/K_3)$. With $r = ||y-z||$, we can write

$$\begin{aligned}&\frac{1}{r} \langle \nabla _yU(x,z) - \nabla _yU(x,y), z- y \rangle \nonumber \\&\quad = \frac{1}{r} \int _0^1 \langle (z-y), D_y^2U(x,y+t(z-y)) (z-y) \rangle ~dt \nonumber \\&\quad = \frac{1}{r} \int _0^1 \langle (z-y), D_y^2U(x,y+t(z-y)) (z-y) \rangle ~1_{\mathbb {B}_R}(y+t(z-y)) ~ dt \nonumber \\&\qquad + \frac{1}{r} \int _0^1 \langle (z-y), D_y^2U(x,y+t(z-y)) (z-y) \rangle ~ 1_{\mathbb {B}_R^c}(y+t(z-y)) ~ dt \nonumber \\&\quad \ge -\frac{1}{r} t_0(y,z) Mmr^2 + \frac{1}{r}(1 - t_0(y,z)) K_3 r^2 \end{aligned}$$

(72)

$$\begin{aligned}&= r (K_3 - t_0(y,z) (Mm+K_3)) \nonumber \\&\ge {\left\{ \begin{array}{ll} -Mmr &{} \text{ if } y,z \in \mathbb {B}_{R_1} \\ 0 &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

(73)

The inequality in (72) follows because:

• (a) from (9), on account of $||D^2_yU(x,y+t(z-y))|| \le M$ when $y+t(z-y) \in \mathbb {B}_R$, we easily obtain the simple inequality $\langle (z-y), D_y^2(x, y+t(z-y))(z-y) \rangle \ge -Mmr^2$ using which the first term is obtained; and

• (b) from (8), $\langle (z-y), D_y^2(x, y+t(z-y))(z-y) \rangle \ge K_3 r^2$ when $y+t(z-y)$ is outside $\mathbb {B}_R$.

The inequality in (73) follows from the easily verifiable fact

$$\begin{aligned} t_0(y,z) \le 2R/(R+R_1) = K_3/(K_3 + Mm) \end{aligned}$$

(74)

for every y, z such that one of them is outside $\mathbb {B}_{R_1}$.

From (73), it is clear that we may take $g(\cdot )$ to be any continuous function that dominates the function $Mmr \cdot \mathbf{1}\{ r \le 2R_1\}$, and there is at least one such $g(\cdot )$ that satisfies $\int _0^{\infty } g(s) ds < \infty$. $\square$

Given Brownian motion $B_t$ on ${\mathbb R}^d$ and an independent Brownian motion $W_t$ on a filtered probability space $(\Omega , \mathcal{F}, {\mathbb {P}})$, a strong solution to the coupled system (4) and (5) is a continuous process $(X^{\varepsilon }_t, Y^{\varepsilon }_t)$ that is adapted to the complete filtration generated by B, W, and satisfies (4) and (5). We say that strong uniqueness holds for the coupled system (4) and (5) if whenever $(X^{\varepsilon }_t, Y^{\varepsilon }_t)$ and $(\tilde{X}^{\varepsilon }_t, \tilde{Y}^{\varepsilon }_t)$ are two strong solutions of the coupled system (4) and (5) with the common initial condition $x_0,y_0$, then ${\mathbb {P}}((X^{\varepsilon }_t, Y^{\varepsilon }_t) = (\tilde{X}^{\varepsilon }_t, \tilde{Y}^{\varepsilon }_t) \text{ for } \text{ all } t \ge 0) = 1.$

Lemma A.2

Assume (B1), (U1) and (U2). Let $\varepsilon >0, 0< \alpha < 1$ and $s(\varepsilon ) >0$ be given. The coupled system given by (4) and (5) has a unique strong solution.

Proof

By assumptions (B1) and (U1), we know that $b: {\mathbb R}^d \times {\mathbb R}^m { \rightarrow \mathbb {R}^d}$ and $\nabla _yU: {\mathbb R}^d \times {\mathbb R}^m { \rightarrow \mathbb {R}^m}$ are locally Lipschitz functions. By [26, page 178 Theorem 3.1] there exist a unique strong solution, $({X}^{\varepsilon }_t, {Y}^{\varepsilon }_t)_{0 \le t < \zeta }$ where

$$\begin{aligned} \zeta = \inf \{ t \ge 0: \Vert {X}^{\varepsilon }_t\Vert ^2 + \Vert {Y}^{\varepsilon }_t\Vert ^2 = \infty \}. \end{aligned}$$

We will now establish nonexplosiveness of the process. Let $f : {\mathbb R}^d \times {\mathbb R}^m \rightarrow [0, \infty )$ be given by $f(x,y) = ~\Vert x \Vert ^2 + \Vert y \Vert ^2$. Let

$$\begin{aligned} \sigma _n =\inf \{ t \ge 0: \Vert {X}^{\varepsilon }_t\Vert ^2 + \Vert {Y}^{\varepsilon }_t\Vert ^2 = n\}. \end{aligned}$$

Clearly $\sigma _n \le \zeta$ almost surely for all $n \ge 1$. Let $t >0$ be given. Applying Ito’s formula at time $\sigma _n \wedge t$, we obtain that

$$\begin{aligned}&{\mathbb E}[f(X^{\varepsilon }_{\sigma _n \wedge t},Y^{\varepsilon }_{\sigma _n \wedge t})] = f(x_0,y_0) \nonumber \\&\quad + {\mathbb E}\int _0^{\sigma _n \wedge t}2 \left( \langle X^{\varepsilon }_r , b( X^{\varepsilon }_r, Y^{\varepsilon }_r) \rangle - \langle Y^{\varepsilon }_r , \frac{1}{\varepsilon }\nabla _yU( X^{\varepsilon }_r, Y^{\varepsilon }_r) \rangle \right) dr \nonumber \\&\quad + (m s(\varepsilon )^2/\varepsilon + d \varepsilon ^{2\alpha }) {\mathbb E}(\sigma _n \wedge t). \end{aligned}$$

(75)

Using the fact that b is bounded from assumption (B1) we have, for $r >0$,

$$\begin{aligned} \langle X^{\varepsilon }_r , b( X^{\varepsilon }_r, Y^{\varepsilon }_r) \rangle \le c_1 d \Vert X^{\varepsilon }_r\Vert \Vert b \Vert _\infty . \end{aligned}$$

(76)

Using (9) from assumption (U2) and (70) derived above we have, for $r >0$,

$$\begin{aligned} - \langle Y^{\varepsilon }_r , \frac{1}{\varepsilon }\nabla _yU( X^{\varepsilon }_r, Y^{\varepsilon }_r) \rangle < {\left\{ \begin{array}{ll} c_2 (M,R) &{}\text{ if } { \Vert Y_r\Vert \le R}\\ 0 &{}\text{ if } { \Vert Y_r\Vert > R}. \end{array}\right. } \end{aligned}$$

(77)

Substituting (76) and (77) in (75) we have

$$\begin{aligned} {\mathbb E}(f(X^{\varepsilon }_{\sigma _n \wedge t},Y^{\varepsilon }_{\sigma _n \wedge t}))\le & {} f(x_0,y_0) + {\mathbb E}\int _0^{\sigma _n \wedge t} (2c_1 d \Vert X^{\varepsilon }_r\Vert \Vert b \Vert _\infty + 2 c_2(M,R)) dr \\&+ (m s(\varepsilon )^2/\varepsilon + d \varepsilon ^{2\alpha }) {\mathbb E}(\sigma _n \wedge t) \\\le & {} f(x_0,y_0) + c_3\int _0^{t} E\Vert X^{\varepsilon }_r\Vert dr + c_4 t \\\le & {} f(x_0,y_0) + c_5\int _0^{t} (1+r) dr + c_4 t \\\le & {} c_6 + c_7t +c_8 t^2, \end{aligned}$$

where the penultimate inequality uses $c_3 {\mathbb E}[\Vert X^{\varepsilon }_r\Vert ] \le c_5(1+r)$ for a suitable $c_5$, a fact that follows from (4) and the boundedness assumption on b in (U1). As $\sigma _n \rightarrow \zeta$ almost surely, the above would imply

$$\begin{aligned} {\mathbb E}[f(X^{\varepsilon }_{\zeta \wedge t},Y^{\varepsilon }_{\zeta \wedge t})] \le c_6 + c_7t +c_8 t^2. \end{aligned}$$

(78)

Thus if $\zeta < t$ then we have a contradiction, as the left-hand side is infinity and the right-hand side is finite. As $t >0$ was arbitrary, we have $\zeta = \infty$ almost surely. This establishes nonexplosiveness of the process and completes the proof of strong uniqueness. $\square$

Appendix B: Nonlinear Filtering Equation

Let $x_0 \in {\mathbb R}^d$, $y_0\in {\mathbb R}^m$, $\sigma _1 >0$ and $\sigma _2 >0$. On the probability space $(\Omega , \mathcal{F}, {\mathbb {P}})$ let $\{B_t\}_{t \ge 0}$ and $\{W_t\}_{t \ge 0}$ be Brownian motions on ${\mathbb R}^d$ and ${\mathbb R}^m$ respectively. In this section, we consider the coupled diffusion $(X_t, Y_t)_{t \in [0,T]}$ on $\mathbb {R}^d\times \mathbb {R}^m$ described by

$$\begin{aligned} X_t= & {} x_0 + \int _0^t b_1(X_s, Y_s)ds + \sigma _1 B_t, \end{aligned}$$

(79)

$$\begin{aligned} Y_t= & {} y_0 + \int _0^t b_2(X_s, Y_s)ds + \sigma _2 W_t, \end{aligned}$$

(80)

where $0 \le t \le T$, $b_1 : \mathbb {R}^d \times \mathbb {R}^m \rightarrow \mathbb {R}^d$ and $b_2 : \mathbb {R}^d \times \mathbb {R}^m \rightarrow \mathbb {R}^m.$ We will make the following assumptions:

• $b_1 \in C_b({\mathbb R}^d \times {\mathbb R}^m)$ is locally Lipschitz continuous in y-variable and is uniformly (w.r.t. y) Lipschitz continuous in x-variable, i.e. $\exists K_1 > 0$ such that $\forall \ x, x' \in {\mathbb R}^d, y \in {\mathbb R}^m$

  $$\begin{aligned} \parallel b_1(x, y) - b_1(x', y)\parallel ~ \le K_1 \parallel x-x'\parallel . \end{aligned}$$

• $b_2 \in C^1(\mathbb {R}^d \times \mathbb {R}^m).$ Further, $b_2(x, y)$ is uniformly (w.r.t. y) Lipschitz continuous in x-variable, i.e. $\exists K_2 > 0$ such that $\forall \ x, x' \in {\mathbb R}^d$, $y \in {\mathbb R}^m$,

  $$\begin{aligned} \parallel b_2(x, y) - b_2(x', y)\parallel ~ \le K_2 \parallel x-x'\parallel . \end{aligned}$$

• There exists $R>0, M>0$ such that for all $x \in {\mathbb R}^d$

  $$\begin{aligned} \sup _{\Vert y \Vert \le R} \Vert b_2(x,y)\Vert \le M. \end{aligned}$$

  Further, there exist $K_4 >0$ and $R^\prime \ge R$ such that for all $x \in {\mathbb R}^d$

  $$\begin{aligned} - \langle b_2(x,y) , y \rangle> K_4 \Vert y\Vert ^2, \quad \Vert y \Vert > R^\prime . \end{aligned}$$

Using the above assumptions in the same proof as in Lemma A.2, it is standard to see that the above coupled system has a unique strong solution. With $f \in C_b^2({\mathbb R}^m)$, for $y \in {\mathbb R}^m$ let

$$\begin{aligned} { {\mathcal L}^x_2 f }(y) = \frac{\sigma _2^2}{2} \Delta f (y) + \langle b_2(x,\cdot ), \nabla f (y) \rangle , \end{aligned}$$

where x is treated as a parameter. Let ${\mathcal F}_t^X = \sigma (X_s, s \le t)$, ${\mathcal F}^X = \bigvee _{t \ge 0} {\mathcal F}_t^X$. Define ${\mathcal F}_t^{X,Y}$ and ${\mathcal F}^{X,Y}$ analogously. Define $\pi _t(dy)$ as the conditional law of $Y_t$ given ${\mathcal F}_t^X$ so that

$$\begin{aligned} \pi _t(f) := \ E[f(Y_t) | {\mathcal F}_t^X] \quad \text{ for } f \in C_b^2({\mathbb R}^m). \end{aligned}$$

Proposition B.1

(Nonlinear Filtering Equation) The measure valued process $\pi$ is the unique solution to the (Fujisaki-Kallianpur-Kunita) nonlinear filtering equation

$$\begin{aligned} \pi _t(f)= & {} f(y_0) + \ \int _0^t \pi _s({{\mathcal L}^{X_s}_2 f)} ds + {\frac{1}{\sigma _1}} \int _0^t\langle \pi _s(f b_1(X_s, \cdot ))\nonumber \\&\quad - \pi _s (f) \pi _s(b_1(X_s, \cdot )), d \tilde{B}_s \rangle , \quad f \in C_b^2({\mathbb R}^m), \end{aligned}$$

(81)

where $\tilde{B}$ is a standard Brownian motion.

Remark B.2

${\tilde{B}}$ is explicitly defined later in the proof. It is called the ‘innovations process’ and, under mild technical conditions, is known to generate the same increasing $\sigma$-fields as B [3].

We will closely mimic the arguments in [6], Chapter 3] proved for the case when $b_1(x,y)$ is a function of the first argument alone. In our setting, $b_1(x,y)$ is a function of both arguments.

Set

$$\begin{aligned} \Lambda _s \ = \ \exp \left\{ -\frac{1}{\sigma _1} \int ^s_0 \langle b_1(X_u , Y_u), d B_u \rangle - \frac{1}{2\sigma _1^2} \int ^s_0 \parallel b_1(X_u, Y_u)\parallel ^2 du \right\} , \ s \ge 0, \end{aligned}$$

(82)

and

$$\begin{aligned} \bar{B}_s \ = \ B_s + \frac{1}{\sigma _1} \int ^s_0 b_1(X_u , Y_u) du , \ s \ge 0. \end{aligned}$$

Define the probability measure Q by

$$\begin{aligned} \frac{d Q\Big |_{{\mathcal F}_s^{X,Y}}}{dP\Big |_{{\mathcal F}_s^{X,Y}}} \ = \ \Lambda _s, s > 0. \end{aligned}$$

This consistently defines Q on ${\mathcal F}^{X,Y}$. As $b_1$ is bounded, by the Cameron-Martin-Girsanov theorem, it follows that $\bar{B}_\cdot$ is an $\mathbb {R}^d$-valued standard Brownian motion under Q. Under Q, the joint process (X, Y) given by (79) - (80) takes the form

$$\begin{aligned} X_t= & {} x_0 + \sigma _1 \bar{B}_t, \nonumber \\ Y_t= & {} y_0 + \int _0^tb_2(X_s, Y_s) ds + \sigma _2 W_t. \end{aligned}$$

(83)

Before we begin the proof we need some preliminary lemmas.

Lemma B.3

For $t > 0$, let Z be a Q-integrable ${\mathcal F}_t^{X,Y}$-measurable $\mathbb {R}^d$-valued random variable. Then

$$\begin{aligned} E^Q [Z | {\mathcal F}_t^X] \ = \ E^Q [Z | {\mathcal F}^X]. \end{aligned}$$

Proof

Set

$$\begin{aligned} \tilde{\mathcal F}^X_t \ = \ \sigma (X_{t+s}- X_t, s \ge 0). \end{aligned}$$

Then ${\mathcal F}^X = \tilde{\mathcal F}^X_t\vee {\mathcal F}^X_t$, and since $X_s = \sigma _1 \bar{B}_s$, an $\{{\mathcal F}_s^X\}$-Wiener process under Q, $\tilde{\mathcal F}^X_t$ is independent of ${\mathcal F}_t^X$ under Q. Hence

$$\begin{aligned} \begin{array}{lll} E^Q [ Z | {\mathcal F}_t^X ] &{} = &{} E^Q [ Z | \tilde{\mathcal F}^X_t\vee {\mathcal F}_t^X ] \\ &{} = &{} E^Q[ Z | {\mathcal F}^X] . \\ \end{array} \end{aligned}$$

This completes the proof of the lemma. $\square$

Lemma B.4

Let $\{{\alpha }_t, ~t \ge 0\}$ be an $\{{\mathcal F}_t^{X,Y}\}$-progressively measurable $\mathbb {R}$-valued process such that

$$\begin{aligned} E^Q \Big [ \int ^t_0 {\alpha }^2_s ds \Big ] < \infty \ \forall \ t > 0. \end{aligned}$$

Then

$$\begin{aligned} E^Q \Big [ \int ^t_0 {\alpha }_s dX_s \Big | {\mathcal F}^X \Big ] \ = \ \int ^t_0 E^Q[ {\alpha }_s | {\mathcal F}^X] dX_s . \end{aligned}$$

Proof

Using Lemma B.3, it follows that

$$\begin{aligned} E^Q \Big [ \int ^t_0 {\alpha }_s dX_s \Big | {\mathcal F}^X \Big ], \ E^Q [{\alpha }_t | {\mathcal F}^X] \end{aligned}$$

are ${\mathcal F}^X_t$-measurable. Hence using the ‘density result’ of Krylov and Rozovskii, see [6], Lemma B.39, p.355], it is enough to show

$$\begin{aligned} E^Q \Big [ \beta _t E^Q \Big [ \int ^t_0 {\alpha }_s dX_s \Big | {\mathcal F}^X \Big ] \Big ]= & {} E^Q \Big [ \beta _t \int ^t_0 E^Q [{\alpha }_s | {\mathcal F}^X] dX_s \Big ] \end{aligned}$$

(84)

for all process $\beta (\cdot )$ of the form

$$\begin{aligned} \beta _t = 1 + \int ^t_0 i \langle \beta _s r_s, dX_s \rangle \end{aligned}$$

for a deterministic $r \in L^{\infty }([0, t] ; \mathbb {R}^d)$. Consider

$$\begin{aligned}&E^Q \Big [ \beta _t E^Q \Big [ \int ^t_0 {\alpha }_s dX_s \Big | {\mathcal F}^X \Big ] \Big ] \\&\quad = E^Q \Big [ \beta _t \int ^t_0 {\alpha }_s dX_s \Big ] \\&\quad = E^Q \Big [ \int ^t_0 {\alpha }_s dX_s \Big ] + E^Q \Big [ \Big ( \int ^t_0 i \langle \beta _s r_s, dX_s \rangle \Big ) \Big ( \int ^t_0 {\alpha }_s dX_s \Big ) \Big ]\\&\quad = \sigma _1^2 E^Q \Big [\int ^t_0i \beta _s r_s {\alpha }_s ds \Big ]\\&\quad = \sigma _1^2 E^Q \Big [E^Q \Big [\int ^t_0i \beta _s r_s {\alpha }_s ds \Big | {\mathcal F}^X \Big ] \Big ] \\&\quad = E^Q \Big [ \Big ( \int ^t_0 i \langle \beta _s r_s, dX_s \rangle \Big ) \Big ( \int ^t_0 E^Q [{\alpha }_s | {\mathcal F}^X] dX_s \Big ) \Big ]\\&\quad = E^Q \Big [ \beta _t \int ^t_0 E^Q [{\alpha }_s | {\mathcal F}^X] dX_s \Big ]. \\ \end{aligned}$$

This completes the proof of the lemma.

Lemma B.5

Let $x \in {\mathbb R}^d$. Let $\{{\alpha }_t, ~t \ge 0\}$ be $\{{\mathcal F}_t^{X,Y}\}$-progressively measurable process such that

$$\begin{aligned} E^Q \Big [ \int ^t_0 {\alpha }^2_s d \langle M^f \rangle _s \Big ] < \infty , \ f \in C^2_b(\mathbb {R}^m), \ t \ge 0, \end{aligned}$$

where

$$\begin{aligned} M^f_t \, = \, f(Y_t) - f({y_0}) - \int ^t_0 {\mathcal L}_2^x f(Y_s) ds, \end{aligned}$$

and $\langle M^f \rangle _t$ is its quadratic variation. Then

$$\begin{aligned} E^Q \Big [ \int ^t_0 {\alpha }_s d M^f_s \Big | {\mathcal F}^X \Big ] = 0. \end{aligned}$$

Proof

Via Itô’s formula, we first obtain

$$\begin{aligned} dM_t^f = \sigma _2 \langle \nabla f(Y_t), dW_t \rangle . \end{aligned}$$

(85)

Under P, this is driven by a Brownian motion independent of $B_t$, which leads to

$$\begin{aligned} \langle M^f, X \rangle _t = 0, \ P- \text{ almost } \text{ surely } \end{aligned}$$

and hence $Q-$almost surely. Using this, the proof follows along the lines of the proof of Lemma B.4.

Set

$$\begin{aligned} \tilde{\Lambda }_t = \Lambda ^{-1}_t, \ t \ge 0, \end{aligned}$$

and for $g \in C^2(\mathbb {R}^d \times \mathbb {R}^m)$ with a little abuse of notation denote

$$\begin{aligned} \pi _t(g) := \pi _t(g(X_t, \cdot )) = E [ g(X_t, Y_t) | {\mathcal F}_t^X]. \end{aligned}$$

We then have the following.

Lemma B.6

(Kallianpur–Striebel formula) For $g \in C^2(\mathbb {R}^d \times \mathbb {R}^m)$,

$$\begin{aligned} \pi _t(g) = \ \frac{E^Q [ \tilde{\Lambda }_t g | {\mathcal F}^X]}{E^Q[\tilde{\Lambda }_t | {\mathcal F}^X]}. \end{aligned}$$

Proof

In view of Lemma B.3, it is enough to show that

$$\begin{aligned} \pi _t(g) E^Q[\tilde{\Lambda }_t | {\mathcal F}_t^X] \ = \ E^Q[\tilde{\Lambda }_t g | {\mathcal F}_t^X]. \end{aligned}$$

Since both left and right sides are ${\mathcal F}_t^X$-measurable, it is enough to show that

$$\begin{aligned} E^Q \left[ \beta \pi _t(g) E^Q \left[ \tilde{\Lambda }_t | {\mathcal F}_t^X \right] \right] \ = \ E^Q \left[ \beta \tilde{\Lambda }_t g \right] \end{aligned}$$

(86)

for all ${\mathcal F}_t^X$-measurable $\beta$. This is now easily verified since, for such $\beta$, we have

$$\begin{aligned} E^Q \left[ \beta \pi _t(g) E^Q \left[ \tilde{\Lambda }_t | {\mathcal F}_t^X \right] \right]= & {} E^Q \left[ \beta \pi _t(g) \tilde{\Lambda }_t \right] = E \left[ \beta \pi _t(g) \right] \\= & {} E \left[ \beta E \left[ g | {\mathcal F}_t^X \right] \right] = E \left[ E \left[ \beta g | {\mathcal F}_t^X \right] \right] \\= & {} E \left[ \beta g \right] = E^Q \left[ \tilde{\Lambda }_t \beta g \right] . \end{aligned}$$

This completes the proof of the lemma. $\square$

We are now ready to prove Proposition B.1. We shall derive first the Zakai equation solved by certain unnormalized conditional laws. Then we shall show existence to the Fujisaki-Kallianpur-Kunita) nonlinear filtering equation (81), followed by uniqueness.

Proof of Proposition B.1

Observe that $\{\Lambda _t, t \ge 0\}$ is given by the solution of the SDE

$$\begin{aligned} {\Lambda _t \ =\ 1 - \int _0^t\Lambda _s \sigma _1^{-1} \langle b_1(X_s,Y_s), dB_s \rangle ,} \end{aligned}$$

for $t \ge 0$. Hence by a routine application of It$\hat{\mathrm{o}}$’s formula it follows that

$$\begin{aligned} {\tilde{\Lambda }_t \ = \ 1 + \int _0^t\tilde{\Lambda }_s \sigma _1^{-2} \langle b_1(X_s, Y_s), dX_s \rangle .} \end{aligned}$$

(87)

From this, since $X_t$ is driven by $B_t$ and $Y_t$ is driven by $W_t$, for $f \in C_b^2({\mathbb R}^m)$, the cross-variation $\langle \tilde{\Lambda }, f(Y_{\cdot }) \rangle _t = 0$ P-a.s. and hence Q-a.s. Using It$\hat{\mathrm{o}}$’s formula again, we get

$$\begin{aligned} {\tilde{\Lambda }_t f(Y_t) = f(y_0) + \int _0^t\tilde{\Lambda }_s [ { {\mathcal L}^{X_s}_2 f (Y_s) }ds + \sigma _2 \langle \nabla f(Y_s), dW_s \rangle ] + \int _0^t f(Y_s) d \tilde{\Lambda }_s ,} \end{aligned}$$

and hence, using (85) and (87), we get

$$\begin{aligned} \tilde{\Lambda }_t f(Y_t)&= f(y_0) + \int ^t_0 \tilde{\Lambda }_s {{\mathcal L}^{X_s}_2 f (Y_s) } ds + \int ^t_0 \tilde{\Lambda }_s d M^f_s \nonumber \\&\quad + \sigma _1^{-2} \int ^t_0 \tilde{\Lambda }_s f(Y_s) \langle b_1(X_s, Y_s), dX_s \rangle . \end{aligned}$$

(88)

Taking conditional expectation $E^Q[ \ \cdot \ | {\mathcal F}^X]$ in (88) we have using Lemma B.5 we have

$$\begin{aligned} E^Q\left[ \tilde{\Lambda }_t f(Y_t)| {\mathcal F}^X \right]&= f(y_0) + E^Q\left[ \int ^t_0 \tilde{\Lambda }_s( {{\mathcal L}^{X_s}_2 f(Y_s) ) } ds| {\mathcal F}^X \right] \nonumber \\&\quad + \sigma _1^{-2}E^Q\left[ \int ^t_0 \tilde{\Lambda }_sf((Y_s) \langle b_1(X_s, Y_s)), d X_s \rangle | {\mathcal F}^X \right] , \nonumber \\&\text{ and } \text{ using } \text{ Lemma }~\hbox {B}.4~\text{ we } \text{ have } \text{ the } \text{ above } \text{ equal } \text{ to }\nonumber \\&=f(y_0) + \int ^t_0 E^Q\left[ \tilde{\Lambda }_s( {{\mathcal L}^{X_s}_2 f(Y_s) ) } | {\mathcal F}^X \right] ds \nonumber \\&\quad + \sigma _1^{-2} \int ^t_0 \langle E^Q\left[ \tilde{\Lambda }_sf((Y_s) b_1(X_s, Y_s))| {\mathcal F}^X \right] , d X_s \rangle . \end{aligned}$$

(89)

For $g \in C({\mathbb R}^d\times {\mathbb R}^m)$, denoting

$$\begin{aligned} \rho _t(g) \ = \ \pi _t(g) E^Q[ \tilde{\Lambda }_t | {{\mathcal F}^X} ] \end{aligned}$$

in (88) and using Lemma B.6, we arrive at the Zakai equation

$$\begin{aligned} \rho _t(f) = f(y_0) + \int ^t_0 \rho _s( {{\mathcal L}^{X_s}_2 f ) } ds + \sigma _1^{-2} \int ^t_0 \langle \rho _s( f b_1(X_s, \cdot )), d X_s \rangle . \end{aligned}$$

(90)

For $\mathbf 1 :=$ the constant function identically equal to 1, we see that $\rho _t(\mathbf 1 ) = E^Q[ \tilde{\Lambda }_t | {{\mathcal F}^X} ]$, and hence

$$\begin{aligned} \pi _t(f) = \frac{\rho _t(f)}{\rho _t(\mathbf 1 )}. \end{aligned}$$

(91)

The nonnegative measure valued process $\{\rho _t\}_{t \ge 0}$ is called the process of unnormalized conditional laws in view of (91).

Now we are ready to prove the existence theorem for the Fujisaki-Kallianpur-Kunita (FKK) equation, (81). From the Zakai equation (90) we get

$$\begin{aligned} \rho _t(f) \, = \, f(y_0) + \int ^t_0 \rho _s(\mathbf 1 )\pi _s( {{\mathcal L}^{X_s}_2 f ) } ds + \sigma _1^{-2} \int ^t_0 {\rho _s(\mathbf 1 )}\langle \pi _s( f b_1(X_s, \cdot )), d X_s \rangle , \end{aligned}$$

(92)

In particular, one can deduce that

$$\begin{aligned} \rho _t(\mathbf 1 ) = 1 + \sigma _1^{-2} \int _0^t \rho _s(\mathbf 1 ) \langle \pi _s( b_1(X_s, \cdot )), dX_s \rangle . \end{aligned}$$

(93)

Using It$\hat{\mathrm{o}}$’s formula, we get

$$\begin{aligned} \frac{1}{\rho _t(\mathbf 1 )} = 1 - \sigma _1^{-2}\int _0^t\frac{1}{\rho _s(\mathbf 1 )} \langle \pi _s( b_1(X_s, \cdot )), dX_s \rangle + \sigma _1^{-2}\int _0^t\frac{1}{\rho _s(\mathbf 1 )} \Vert \pi _s(b_1(X_s,\cdot ))\Vert ^2 ds,\end{aligned}$$

(94)

Note that the cross-variation

$$\begin{aligned} \langle \rho (f) , \frac{1}{\rho (\mathbf 1 )} \rangle _t \, = \, -\int _0^t \sigma _1^{-2} \langle \pi _s( b_1), \pi _s( b_1 f) \rangle ds, \end{aligned}$$

(95)

From It$\hat{\mathrm{o}}$’s formula for the product of $\rho _t(f)$ and $\frac{1}{\rho _t(\mathbf 1 )}$ we get

$$\begin{aligned} \frac{\rho _t(f)}{\rho _t(\mathbf 1 )}&= f(y_0) + \int _0^t\rho _s(f) d \frac{1}{\rho _s(\mathbf 1 )} + \int _0^t \frac{1}{\rho _s(\mathbf 1 )} d\rho _s(f) + \langle \rho (f), \frac{1}{\rho (\mathbf 1 )}\rangle _t \end{aligned}$$

Substituting (92),(94), and (95) in the above we have

$$\begin{aligned} \frac{\rho _t(f)}{\rho _t(\mathbf 1 )}&= f(y_0) + \int _0^t\rho _s(f) \left[ -\frac{1}{\rho _s(\mathbf 1 )} \sigma _1^{-2}\langle \pi _s( b_1(X_s, \cdot )), dX_s \rangle \right. \\&\quad \left. + \sigma _1^{-2}\frac{1}{\rho _s(\mathbf 1 )} \Vert \pi _s(b_1(X_s,\cdot ))\Vert ^2 ds, \right] \\&\quad + \int _0^t \frac{1}{\rho _s(\mathbf 1 )}\left[ \rho _s(\mathbf 1 )\pi _s( {{\mathcal L}^{X_s}_2 f ) } ds \right. \\&\quad \left. + \sigma _1^{-2}{\rho _s(\mathbf 1 )}\langle \pi _s( f b_1(X_s, \cdot )), d X_s \rangle \right] \\&\quad - \int _0^t \sigma _1^{-2} \langle \pi _s( b_1), \pi _s( b_1 f) \rangle ds. \end{aligned}$$

From (91), using some simple algebra in the above we obtain

$$\begin{aligned} \pi _t(f)&=f(y_0) + \int _0^t \pi _s( {{\mathcal L}^{X_s}_2 f ) } ds + \sigma _1^{-2}\int _0^t \langle \pi _s( f b_1)\nonumber \\&\quad - \pi _s(f) \pi _s( b_1), d X_s -\pi _s(b_1)ds\rangle \end{aligned}$$

(96)

Let

$$\begin{aligned} I_t = X_t - \int _0^t\pi _s( b_1) ds, \end{aligned}$$

the so called ‘innovation process’. For $0 \le s < t$, we have

$$\begin{aligned} \begin{array}{lll} E [ I_t - I_s | \mathcal {F}_s^X] &{} = &{} \displaystyle { E \Big [ \int ^t_s E[ b_1(X_u, Y_u) - \pi _u( b_1(X_u, \cdot )) \ | \ {\mathcal F}_u^X ] du \Big | \mathcal {F}_s^X\Big ]}\\ &{} = &{} \displaystyle { \int ^t_s E \Big [ b_1(X_u, Y_u) - E [ b_1(X_u, Y_u) \ | \ {\mathcal F}_u^X]| \mathcal {F}_s^X \Big ] du} \\ &{} = &{} 0.\\ \end{array} \end{aligned}$$

Thus $\{{I}_t | t \ge 0\}$ is an $\{{\mathcal F}_t^X\}$-martingale with mean 0 and quadratic variation $\sigma _1^2t$. Thus by Levy’s characterization, I is a scaled Brownian motion. Define

$$\begin{aligned} \tilde{B}_t := \sigma _1^{-1}I_t, \ t \ge 0. \end{aligned}$$

(97)

So $\tilde{B}_t$ is a $\{{\mathcal F}_t^X\}$-adapted standard Brownian motion under P. Therefore we have shown that,

$$\begin{aligned} \pi _t(f)&=f(y_0) + \int _0^t \pi _s( {{\mathcal L}^{X_s}_2 f ) } ds + {\sigma _1^{-1}}\int _0^t \langle \pi _s( f b_1)- \pi _s(f) \pi _s( b_1), d\tilde{B}_s\rangle , \end{aligned}$$

(98)

with $\tilde{B}_s$ being a standard Brownian motion. Thus we have shown existence of a solution to the FKK equation. Uniqueness of the FKK equation in the sense of martingale problem follows from Theorem 3.3 of Kurtz and Ocone [29]. Note that while Kurtz and Ocone [29] cite nonlinear filtering as an example of this theorem, they consider the classical formulation (see [29, Theorem 4.1]) which is more restrictive than ours. However the aforementioned theorem ([29, Theorem 3.3]) is general enough to cover our problem. $\square$

From (93) we have

$$\begin{aligned} \rho _t(\mathbf 1 ) = \exp \left\{ \sigma _1^{-{2}}\int _0^t\langle \pi _s(b_1(X_s, \cdot )), dX_s\rangle - \frac{\sigma _1^{-{4}}}{2}\int _0^t\Vert \pi _s(b_1(X_s, \cdot ))\Vert ^2ds \right\} . \end{aligned}$$

Since $\pi _t(f) = \frac{\rho _t(f)}{\rho _t(\mathbf 1 )},$

$$\begin{aligned} \rho _t(f)= & {} \pi _t(f)\rho _t(\mathbf 1 ) = \pi _t(f) \exp \left\{ \sigma _1^{-{2}}\int _0^t\langle \pi _s(b_1(X_s, \cdot )), dX_s\rangle \right. \\&\left. - \frac{\sigma _1^{-{4}}}{2}\int _0^t\Vert \pi _s(b_1(X_s, \cdot ))\Vert ^2ds\right\} . \end{aligned}$$

Thus solutions $\pi , \rho$ of FKK, resp. Zakai equations are in one-one correspondence and uniqueness of one implies that of the other.

Remark B.7

It is interesting to note that some of the earlier uniqueness arguments for the classical framework such as one using multiple Wiener integral expansion due to [31] or via the Clark-Davis ‘pathwise’ filter as in [23], do not work for our case. (The latter would work only if $b_1(x, \cdot ) = \nabla F(x, \cdot )$ for a suitable F.)

Appendix C: Laplace’s Principle

We now characterize weak limit points of the sequence of invariant measures for the fast process $\nu ^{\varepsilon _n, x^{\varepsilon _n}}$ when $\varepsilon _n \rightarrow 0$ and for deterministic $x^{\varepsilon _n} \rightarrow x$. This is used in the proof of Proposition 2.4(c,d).

Lemma C.1

Let $n \ge 1, 0< \varepsilon _n < 1, x^{\varepsilon _n} \in {\mathbb R}^d$, and $x \in \mathbb {R}^d$. Suppose $\varepsilon _n \rightarrow 0$ and $x^{\varepsilon _n} \rightarrow x$ as $n\rightarrow \infty$.

• (a) Then the sequence of measures $\nu ^{\varepsilon _n, x^{\varepsilon _n}}$ is tight and any limit point is supported on $\arg \min \{U(x, \cdot )\}$.

• (b) Assume (U4) and let $x \in D_L^\circ$ for some $L \ge 1$. Then $\nu ^{\varepsilon _n, x^{\varepsilon _n}}$ converges weakly to $\nu ^{0,x}$, where $\nu ^{0,x}$ is given by

  $$\begin{aligned} \sum _{i=1}^{L} \delta _{y_i(x)}\frac{\left( \text{ Det }\left[ D_y^2 U(x,y_i(x))\right] \right) ^{-\frac{1}{2}}}{\sum _{j=1}^{L} \left( \text{ Det }\left[ D_y^2 U(x,y_j(x))\right] \right) ^{-\frac{1}{2}}}. \end{aligned}$$

Proof

(a) Using (8) and (9) it is easy to see (see Lemma A.1 in Appendix A) that there is $K_4 >0$ and an $R^\prime \ge R$ such that

$$\begin{aligned} \langle \nabla _y U(x, y) , y \rangle> K_4 \Vert y\Vert ^2, \quad \Vert y\Vert > R^\prime . \end{aligned}$$

(99)

Let $h: {\mathbb R}^m \rightarrow {\mathbb R}$ be given by

$$\begin{aligned} h(y) ~ = ~ \parallel y \parallel ^2. \end{aligned}$$

Using (99), there is $R^\prime >0$ such that

$$\begin{aligned} {{\mathcal L}^{\varepsilon _n, x^{\varepsilon _n}}( h )} (y)= & {} \frac{s(\varepsilon _n)^{2}}{2}\Delta h(y) - \langle \nabla _yU(x^{\varepsilon _n}, y), \nabla h(y) \rangle \nonumber \\= & {} m s(\varepsilon _n)^{2} - 2 \langle \nabla _yU(x^{\varepsilon _n}, y), y \rangle \nonumber \\< & {} m s(\varepsilon _n)^{2} - 2K_4 \parallel y \parallel ^2 < 0 , \end{aligned}$$

(100)

for all $\parallel y \parallel > R^\prime$ and $n \ge 1.$ We can assume without loss of generality that $\max \{\Vert x^{\varepsilon _n} \Vert , \Vert x\Vert \} \le R^\prime$. So by regularity assumption on U from (U1) we have

$$\begin{aligned} \int _{\parallel y \parallel \le R^\prime } \mid m s(\varepsilon _n)^{2} - 2 \langle \nabla _yU(x^{\varepsilon _n}, y), y \rangle \mid \nu ^{\varepsilon _n, x^{\varepsilon _n}}(dy) \le K \end{aligned}$$

(101)

for some $K \equiv K(m,s,R^\prime ,U)>0$. Using (100), (101) along with Proposition 2.4 in [33] and its proof, we have for all $n \ge 1$

$$\begin{aligned} \nu ^{\varepsilon _n, x^{\varepsilon _n}}( \mid {\mathcal L}^{\varepsilon _n}(x^{\varepsilon _n} , h ) \mid ) < 2K. \end{aligned}$$

(102)

Define $g: {\mathbb R}^m \rightarrow {\mathbb R}$ by $g(y) = 2K_4 \parallel y \parallel ^2 - m s(\varepsilon )^{2}$. Using (100) and (102) we have

$$\begin{aligned} \int _{\parallel y \parallel > R^\prime } g(y) \nu ^{\varepsilon _n, x^{\varepsilon _n}}(dy) \le \nu ^{\varepsilon _n, x^{\varepsilon _n}}( \mid {\mathcal L}^{\varepsilon _n}(x^{\varepsilon _n} , h ) \mid ) < 2K. \end{aligned}$$

(103)

As $g(y) \rightarrow \infty$ when $\parallel y \parallel \rightarrow \infty$ we can conclude that the sequence of measures $\{\nu ^{\varepsilon _n, x_n}\}_{n \ge 1}$ is tight. We will now show that any limit point $\nu$ is supported on $\arg \min U(x, \cdot )$.

Let $z \in {\mathbb R}^m, z \not \in \arg \min \{U(x, \cdot ) \}.$ As $U(x^{\varepsilon _n},\cdot )$ converges to $U(x,\cdot )$ uniformly on compact sets, there exists $\delta >0$ and $r >0$ such that

$$\begin{aligned} U(x^{\varepsilon _n},y) > U(x,y_i(x)) + \frac{\delta }{2}, \,\, \forall y \in B(z,r) \end{aligned}$$

and

$$\begin{aligned} U(x^{\varepsilon _n},y) < U(x,y_i(x)) + \frac{\delta }{4}, \,\, \forall y \in B(y_i(x),r). \end{aligned}$$

Therefore, for $n \ge 1$,

$$\begin{aligned} \frac{\nu ^{\varepsilon _{n}, x^{\varepsilon _n}}(B(z,r))}{\nu ^{\varepsilon _{n}, x^{\varepsilon _n}}(B(y_i(x), r))}= & {} \frac{\int _{B(z,r)} e^{-2\frac{U(x^{\varepsilon _n},y)}{s(\varepsilon _n)^2}} dy}{\int _{B(y_i(x),r)} e^{-2\frac{U(x^{\varepsilon _n},y)}{s(\varepsilon _n)^2}}dy} = \frac{\int _{B(z,r)} e^{-2\frac{U(x^{\varepsilon _n},y) -U(x,y_i(x))}{s(\varepsilon _n)^2}} dy}{\int _{B(y_i(x),r)} e^{-2\frac{U(x^{\varepsilon _n},y) -U(x,y_i(x))}{s(\varepsilon _n)^2}}dy}\\\le & {} \frac{\mid B(z,r)\mid e^{-\frac{\delta }{ s(\varepsilon _n)^2}}}{\mid B(y_i(x),r)\mid e^{-\frac{\delta }{2s(\varepsilon _n)^2}}} = e^{-\frac{\delta }{2 s(\varepsilon _n)^2}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle \lim _{n \rightarrow \infty } \nu ^{\varepsilon _{n}, x^{\varepsilon _n}}(B(z,r)) = 0. \end{aligned}$$

Hence any limit point $\nu$ is supported on the $\arg \min \{ U(x,\cdot )\}.$

(b) Let $x \in D_L^\circ$ for some $L \ge 1$. Under (U4) the global minima $y_i(x), 1 \le i \le L$ are nondegenerate, i.e., the matrix $D^2_yU(x, y_i(x))$ is positive definite for $1 \le i \le L$. Since $D_L^\circ$ is open, using $U \in C^2({\mathbb R}^m \times {\mathbb R}^d)$ in (U1), with a suitable relabelling of the minima if necessary, we have $y_i(x^{\varepsilon _n}) \rightarrow y_i(x) \ \forall 1 \le i \le L$ as $n \rightarrow \infty$. Let $B_i$ be the ball centered at $y_i(x)$ with radius 1 for each i. Let n be sufficiently large so that $y_i(x^{\varepsilon _n})$ are in a ball centered at $y_i(x)$ with radius $\frac{1}{2}$. Let $B^n_i$ be ball centered at $y_i(x^{\varepsilon _n})$ with radius $\frac{1}{4}$ for each i. Note that $\nabla _yU(x^{\varepsilon _n}, y_i(x^{\varepsilon _n})) = 0$ and $U(x^{\varepsilon _n}, y_i(x^{\varepsilon _n})) = \min U(x^{\varepsilon _n}, \cdot ):= u_{\min }$ (say) as it does not depend on i. Using Taylor’s expansion up to second order, we have that for each $y \in B_i$ there is a $\tilde{y}_i(x^{\varepsilon _n}) \in B_i$ such that

$$\begin{aligned} U(x^{\varepsilon _n}, y)= & {} U(x^{\varepsilon _n}, y_i(x^{\varepsilon _n})) + \frac{1}{2}(y - y_i(x^{\varepsilon _n}))^TD^2_yU(x^{\varepsilon _n}, \tilde{y}_i(x^{\varepsilon _n}))(y - y_i(x^{\varepsilon _n}))\\= & {} u_{\min } + \frac{1}{2} (y - y_i(x^{\varepsilon _n}))^TD^2_yU(x^{\varepsilon _n}, \tilde{y}_i(x^{\varepsilon _n}))(y - y_i(x^{\varepsilon _n})). \end{aligned}$$

The above and standard fact about Gaussian random variables implies:

$$\begin{aligned} \int _{B_i}e^{-2\frac{U(x^{\varepsilon _n}, y)}{s(\varepsilon _n)^2}}dy&\le e^{-2\frac{u_{\min }}{s(\varepsilon _n)^2}} \int _{{\mathbb R}^m}e^{-2\frac{(y - y_i(x^{\varepsilon _n}))^TD^2_yU(x^{\varepsilon _n}, \tilde{y}_i(x^{\varepsilon _n}))(y - y_i(x^{\varepsilon _n}))}{2s(\varepsilon _n)^2}}dy \nonumber \\&= e^{-2\frac{u_{\min }}{s(\varepsilon _n)^2}} \left( \frac{(2\pi )^m s(\varepsilon _n)^2}{2}\text{ Det }\left( D^2_yU(x^{\varepsilon _n}, y_i(x^{\varepsilon _n}))^{-1}\right) \right) ^{\frac{1}{2}}; \end{aligned}$$

(104)

and

$$\begin{aligned} \int _{B_i}e^{-2\frac{U(x^{\varepsilon _n}, y)}{s(\varepsilon _n)^2}}dy&\ge e^{-2\frac{u_{\min }}{s(\varepsilon _n)^2}} \int _{B^n_i}e^{-2\frac{(y - y_i(x^{\varepsilon _n}))^TD^2_yU(x^{\varepsilon _n}, \tilde{y}_i(x^{\varepsilon _n}))(y - y_i(x^{\varepsilon _n}))}{2s(\varepsilon _n)^2}}dy\nonumber \\&= e^{-2\frac{u_{\min }}{s(\varepsilon _n)^2}} \left( \frac{(2\pi )^ms(\varepsilon _n)^2}{2}\text{ Det }\left( D^2_yU(x^{\varepsilon _n}, y_i(x^{\varepsilon _n}))^{-1}\right) \right) ^{\frac{1}{2}} \nonumber \\&\quad P\left( Z^m \in \frac{1}{s(\varepsilon _n)} A_{i,n}\right) , \end{aligned}$$

(105)

where $A_{i,n} = \left\{ z \in {\mathbb R}^m : \left\| \left( D^2_yU(x^{\varepsilon _n}, y_i(x^{\varepsilon _n})) \right) ^{-1/2} z \right\| \le \frac{1}{2\sqrt{2}} \right\}$ and $Z^m$ is a standard $m-$dimensional Gaussian random variable. Note that $A_{i,n}$ is a bounded set in ${\mathbb R}^m$ and by (U1), $U \in C^2({\mathbb R}^d \times {\mathbb R}^m)$. So as $n \rightarrow \infty$ we have

$$\begin{aligned}&P \left( Z^m \in \frac{1}{s(\varepsilon _n)} A_{i,n} \right) \rightarrow 1 \text{ and } \left( \text{ Det }\left( D^2_yU(x^{\varepsilon _n}, y_i(x^{\varepsilon _n}))^{-1}\right) \right) ^{\frac{1}{2}}\\&\quad \rightarrow \left( \text{ Det }\left( D^2_yU(x, y_i(x))^{-1}\right) \right) ^{\frac{1}{2}} \end{aligned}$$

for all i. Therefore using a standard sandwich argument we can conclude that, for balls $B_i$ and $B_j$, $1 \le i, j \le L$,

$$\begin{aligned} \frac{\nu ^{\varepsilon _n,x^{\varepsilon _n}}(B_i)}{\nu ^{\varepsilon _n,x^{\varepsilon _n}}(B_j)}= \frac{\int _{B_i}e^{-2\frac{U(x^{\varepsilon _n}, y)}{s(\varepsilon _n)^2}}dy}{\int _{B_j}e^{-2\frac{U(x^{\varepsilon _n}, y)}{s(\varepsilon _n)^2}}dy} \rightarrow \frac{\left( \text{ Det }\left( D^2_yU(x, y_i(x))^{-1}\right) \right) ^{\frac{1}{2}}}{\left( \text{ Det }\left( D^2_yU(x, y_j(x))^{-1}\right) \right) ^{\frac{1}{2}}} \,\, \text{ as } n \rightarrow \infty . \end{aligned}$$

(106)

From (a) we know that the sequence of measures $\{\nu ^{\varepsilon _n,x^{\varepsilon _n}}\}_{n \ge 1}$ are tight and all limit points are measures supported on the $\arg \min U(x, \cdot )$. Consequently by (106) we have that any limit point $\nu ^{0,x}$ is given by

$$\begin{aligned} \nu ^{0,x}(\cdot ) = \sum _{i=1}^{L}\frac{\left( \text{ Det }\left[ D_y^2 U(x,y_i(x))\right] \right) ^{-\frac{1}{2}}}{\sum _{j=1}^{L} \left( \text{ Det }\left[ D_y^2 U(x,y_j(x))\right] \right) ^{-\frac{1}{2}}}\delta _{y_i(x)}( \cdot ). \end{aligned}$$

Since all subsequential limit points are the same we have the result. $\square$