1 Introduction

Consider the following obstacle problems for quasilinear stochastic partial differential equations (SPDEs) in \(\mathbb {R}^d\):

$$\begin{aligned}&dU(t,x)+\frac{1}{2}\Delta U(t,x)+\sum _{i=1}^d\partial _ig_i(t,x,U(t,x),\nabla U(t,x))dt\nonumber \\&\quad +f(t,x,U(t,x),\nabla U(t,x))dt\nonumber \\&\quad +\, \sum _{j=1}^{\infty } h_j(t,x,U(t,x),\nabla U(t,x))dB^j_t =-R(dt,dx), \end{aligned}$$
(1.1)
$$\begin{aligned}&U(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&U(T,x)=\Phi (x), \quad \quad x\in \mathbb {R}^d, \end{aligned}$$
(1.2)

where \(B^j_t, j=1,2,\ldots \) are independent real-valued standard Brownian motions, the stochastic integral against Brownian motions is interpreted as the backward Ito integral, \(\Delta \) is the Laplacian operator, \(f, g_i, h_j\) are appropriate measurable functions specified later, L(tx) is the given barrier/obstacle function, R(dtdx) is a random measure which is a part of the solution pair (UR). The random measure R plays a similar role as a local time which prevents the solution U(tx) from falling below the barrier L.

Such SPDEs appear in various applications like pathwise stochastic control problems, the Zakai equations in filtering and stochastic control with partial observations. Existence and uniqueness of the above stochastic obstacle problems were established in [13] based on an analytical approach. Existence and uniqueness of the obstacle problems for quasi-linear SPDEs on the whole space \(\mathbb {R}^d\) and driven by finite dimensional Brownian motions were studied in [20] using the approach of backward stochastic differential equations (BSDEs). Obstacle problems for nonlinear stochastic heat equations driven by space-time white noise were studied by several authors, see [23, 28] and references therein.

In this paper, we are concerned with the small noise large deviation principle(LDP) of the following obstacle problems for quasilinear SPDEs:

$$\begin{aligned}&dU^{\varepsilon }(t,x)+\frac{1}{2}\Delta U^{\varepsilon }(t,x)+\sum _{i=1}^d\partial _ig_i(t,x,U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dt \nonumber \\&\quad +\, f(t,x,U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dt\nonumber \\&\quad +\, \sqrt{\varepsilon }\sum _{j=1}^{\infty } h_j(t,x,U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dB^j_t =-R^{\varepsilon }(dt,dx), \end{aligned}$$
(1.3)
$$\begin{aligned}&U^{\varepsilon }(t,x)\ge L(t,x), \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&U^{\varepsilon }(T,x)= \Phi (x), \quad x\in \mathbb {R}^d. \end{aligned}$$
(1.4)

Large deviations for stochastic evolution equations and stochastic partial differential equations driven by Brownian motions have been investigated in many papers, see e.g. [3, 5, 6, 8, 11, 18, 19, 25, 27] and references therein.

To obtain the large deviation principle, we will adopt the weak convergence approach introduced by Budhiraja, Dupuis and Maroulas in [2,3,4]. We refer the reader to [2, 11, 18, 19, 3, 25] for large deviation principles of various dynamical systems driven by Gaussian noises.

In order to apply the weak convergence method to the obstacle problems, because of the singularity introduced by the reflection/local time, it seems difficult to directly use the criteria in [2]. We therefore first need to provide a sufficient condition to verify the criteria of Budhiraja–Dupuis–Maroulas. This sufficient condition turns out to be particularly suitable for stochastic dynamics generated by stochastic differential equations and stochastic partial differential equations with reflection. The advantage of the new sufficient condition is to shift the difficulty of proving the tightness of the perturbations of stochastic differential(partial differential) equations to a study of the continuity (with respect to the driving signals) of deterministic skeleton equations associated with the stochastic equations. This new sufficient condition is recently successfully applied to obtain a large deviation principle for stochastic conservation laws (Ref. [9]), which otherwise could not (at least very difficult) be established using the original form of the criteria in [2].

The important part of the current work is to study the continuity of the deterministic obstacle problems driven by the elements in the Cameron–Martin space of the driving Brownian motions. We need to show that if the driving signals converge weakly in the Cameron–Martin space, then the corresponding solutions of the skeleton equations converge in the appropriate state space. This turns out to be hard because of the singularity caused by the obstacle. To overcome the difficulties, we have to appeal to the penalized approximation of the skeleton equation and to establish some uniform estimate for the solutions of the approximating equations with the help of the backward stochastic differential equation representation of the solutions. This is purely due to the technical reason because primarily the LDP problem has not much to do with backward stochastic differential equations.

The rest of the paper is organized as follows. In Sect. 2, we introduce the stochastic obstacle problem and the precise framework. In Sect. 3, we recall the weak convergence approach of large deviations and present a sufficient condition. Section 4 is devoted to the study of skeleton obstacle problems. We will show that the solution of the skeleton problem is continuous with respect to the driving signal. The proof of the large deviation principle is in Sect. 5.

2 The Framework

2.1 Obstacle Problems

Let \(H:=\mathbf {L}^2(\mathbb {R}^d)\) be the Hilbert space of square integrable functions with respect to the Lebesgue measure on \(\mathbb {R}^d\). The associated scalar product and the norm are denoted by

$$\begin{aligned} (u,v)=\displaystyle \int _{\mathbb {R}^d}u\left( x\right) v\left( x\right) dx, \quad |u|=\left( \displaystyle \int _{\mathbb {R}^d}u^2(x)dx\right) ^{1/2}. \end{aligned}$$

Let \(V:=H(\mathbb {R}^d)\) denote the first order Sobolev space, endowed with the norm and the inner product:

$$\begin{aligned} \Vert u\Vert= & {} \left( \displaystyle \int _{\mathbb {R}^d}|\nabla u|^2(x)dx+\displaystyle \int _{\mathbb {R}^d}|u|^2(x)dx\right) ^{1/2}, \\ \langle u, v \rangle= & {} \displaystyle \int _{\mathbb {R}^d}(\nabla u)\cdot (\nabla v)(x)dx+\displaystyle \int _{\mathbb {R}^d}u(x)v(x)dx. \end{aligned}$$

\(V^*\) will denote the dual space of V. When causing no confusion, we also use \(\langle u,v\rangle \) to denote the dual pair between V and \(V^*\).

Our evolution problem will be considered over a fixed time interval [0, T]. Now we introduce the following assumptions.

Assumption 2.1

  1. (i)

    \(f:[0,T]\times \mathbb {R}^{d}\times \mathbb {R}\times \mathbb {R}^{d}\rightarrow \mathbb {R}\), \(h=(h_1,\ldots ,h_i,\ldots ):[0,T]\times \mathbb {R}^{d}\times \mathbb {R}\times \mathbb {R}^{d}\rightarrow \mathbb {R}^{\infty }\) and \(g=(g_1,\ldots ,g_d):[0,T]\times \mathbb {R}^{d}\times \mathbb {R}\times \mathbb {R}^{d}\rightarrow \mathbb {R}^{d}\) are measurable in (txyz) and satisfy \( f^0, h^0, g^0 \in \mathbf {L}^2\left( [0,T] \times \mathbb {R}^{d}\right) \cap \mathbf {L}^{\infty }\left( [0,T] \times \mathbb {R}^{d}\right) \) where \(f^0(t,x) := f(t,x,0, 0)\), \(h^0(t,x) := (\sum _{j=1}^{\infty }h_j(t, ,x,0, 0)^2)^{\frac{1}{2}}\) and \(g^0(t,x) := (\sum _{j=1}^{d}g_j(t, ,x,0, 0)^2)^{\frac{1}{2}}\).

  2. (ii)

    There exist constants \(c>0\)\(0<\alpha <1\) and \(0<\beta <1\) such that for any \((t,x)\in [0,T]\times \mathbb {R}^d~;~(y_1,z_1),(y_2,z_2)\in \mathbb {R}\times \mathbb {R}^{d}\)

    $$\begin{aligned} |f(t,x,y_1,z_1)-f(t,x,y_2,z_2)|\le & {} c\big (|y_1-y_2|+|z_1-z_2|\big ) \\ \left( \sum _{i=1}^{\infty } |h_i(t,x,y_1,z_1)-h_i(t,x,y_2,z_2)|^2\right) ^{1/2}\le & {} c|y_1-y_2|+\beta |z_1-z_2|\\ \left( \sum _{i=1}^{d} |g_i(t,x,y_1,z_1)-g_i(t,x,y_2,z_2)|^2\right) ^{1/2}\le & {} c|y_1-y_2|+\alpha |z_1-z_2|. \end{aligned}$$
  3. (iii)

    There exists a function \(\bar{h}\in L^2(\mathbb {R}^{d})\cap L^{\infty }(\mathbb {R}^{d})\) such that for \((t,x,y,z) \in [0,T]\times \mathbb {R}^{d}\times \mathbb {R}\times \mathbb {R}^{d}\),

    $$\begin{aligned} \left( \sum _{i=1}^{\infty } |h_i(t,x,y,z)|^2\right) ^{1/2}\le \bar{h}(x). \end{aligned}$$
  4. (iv)

    The contract property: \(\alpha +\displaystyle \frac{\beta ^2}{2}<\displaystyle \frac{1}{2}\).

  5. (v)

    The barrier function \(L(t,x):\mathbb {R}^d\rightarrow \mathbb {R}\) satisfies

    $$\begin{aligned} \frac{\partial L(t,x)}{\partial t}, \quad \nabla L(t,x), \quad \Delta L(t,x)\in L^2([0,T]\times \mathbb {R}^d)\cap L^{\infty }([0,T]\times \mathbb {R}^d), \end{aligned}$$

    where the gradient \(\nabla \) and the Laplacian \(\Delta \) act on the space variable x.

Let \(H_T:=C([0, T], H)\cap L^2([0, T], V)\) be the Banach space endowed with the norm

$$\begin{aligned} \Vert u\Vert _{H_T}=\sup _{0\le t\le T}|u_s|+\left( \int _0^T\Vert u_s\Vert ^2ds\right) ^{1/2}. \end{aligned}$$

We denote by \( {\mathcal H}_T\) the space of predictable, processes \((u_t, t\ge 0 ) \) such that \(u\in H_T\) and that

$$\begin{aligned} \mathbb {E}\, \left[ \underset{ 0 \le s \le T}{\displaystyle \sup } |u_s |_2^2 + \displaystyle \int _0^T \Vert u_s\Vert ^2 ds\right] <\infty . \end{aligned}$$

The space of test functions is \(\mathcal {D}= \mathcal {C}_c^{\infty } (\mathbb {R}^+) \otimes \mathcal {C}_c^{\infty } (\mathbb {R}^d)\), where \(\mathcal {C}_c^{\infty } (\mathbb {R}^+)\) denotes the space of real-valued infinitely differentiable functions with compact supports in \(\mathbb {R}^+\) and \(\mathcal {C}_c^{\infty }(\mathbb {R}^d)\) is the space of infinitely differentiable functions with compact supports in \(\mathbb {R}^d\).

Let \(B^j_t, j=1,2,\ldots \) be a sequence of independent real-valued standard Brownian motions on a complete filtrated probability space \((\Omega , \mathcal {F}, \mathcal {F}_t, P)\). We now precise the definition of solutions for the reflected quasilinear SPDE (1.1):

Definition 2.1

We say that a pair (UR) is a solution of the obstacle problem (1.1) if

  1. (1)

    \(U\in {\mathcal H}_T\), \(U(t,x)\ge L(t,x)\), \(dP\otimes dt\otimes dx\)-a.e. and \(U(T,x)=\Phi (x)\), \(dx-a.e.\)

  2. (2)

    R is a random regular measure on \([0,T)\times \mathbb {R}^d\),

  3. (3)

    for every \(\varphi \in \mathcal D\)

    $$\begin{aligned}&(U_t,\varphi _t)-(\Phi ,\varphi _T)+\int _t^T(U_s, \partial _s\varphi _s)ds+\frac{1}{2}\int _t^T\langle \nabla U_s, \nabla \varphi _s \rangle ds \nonumber \\&\quad =\int _t^T(f_s(U_s,\nabla U_s),\varphi _s)ds+\sum _{j=1}^{\infty }\int _t^T (h_s^j(U_s, \nabla U_s), \varphi _s)dB_s^j\ \nonumber \\&\quad \quad -\sum _{i=1}^d \int _t^T(g_s^i(U_s, \nabla U_s), \partial _i\varphi _s)ds+\int _t^T\int _{\mathbb {R}^d}\varphi _s(x)R(dx,ds), \end{aligned}$$
    (2.1)
  4. (4)

    U admits a quasi-continuous version \(\tilde{U}\), and

    $$\begin{aligned} \int _0^T\int _{\mathbb {R}^d}(\tilde{U}(s,x)-L(s,x))R(dx,ds)=0\quad \quad a.s. \end{aligned}$$

Remark 2.1

We refer the reader to [13] for the precise definition of regular measures and quasi-continuity of functions on the space \([0, T]\times \mathbb {R}^d\).

Let us recall the following result from [13, 20].

Theorem 2.1

Let Assumption 2.1 hold and assume \(\Phi (x)\ge L(T,x)\) dx-a.e.. Then there exists a unique solution (UR) to the obstacle problem (1.1).

2.2 The Measures \(\mathbb {P}^m\)

The operator \( \partial _t + \frac{1}{2} \Delta \), which represents the main linear part in the equation (1.11.2), is associated with the Bownian motion in \(\mathbb {R}^d\). The sample space of the Brownian motion is \( \Omega ' = \mathcal {C }([0, \infty ); \mathbb {R}^d)\), the canonical process \((W_t)_{t \ge 0}\) is defined by \( W_t (\omega ) = \omega (t)\), for any \( \omega \in \Omega '\), \(t \ge 0\) and the shift operator, \( \theta _t \, : \, \Omega ' \longrightarrow \Omega '\), is defined by \( \theta _t (\omega ) (s) = \omega (t+s)\), for any \(s \ge 0\) and \( t \ge 0\). The canonical filtration \( \mathcal {F}_t^W = \sigma \left( W_s; s \le t \right) \) is completed by the standard procedure with respect to the probability measures produced by the transition function

$$\begin{aligned} P_t (x, dy) = q_t (x-y) dy, \quad t >0, \quad x \in \mathbb {R}^d, \end{aligned}$$

where \( q_t (x) = (2\pi t)^{- \frac{d}{2}} \exp (-|x|^2/2t)\) is the Gaussian density. Thus we get a continuous Hunt process \((\Omega ', W_t, \theta _t, \mathcal {F}, \mathcal {F}^W_t, \mathbb {P}^x)\). We shall also use the backward filtration of the future events \( \mathcal {F}'_t = \sigma \left( W_s; \; \, s \ge t \right) \) for \(t\ge 0\). \(\mathbb {P}^0\) is the Wiener measure, which is supported by the set \( \Omega '_0 = \{ \omega \in \Omega ', \; \, \omega (0) =0 \}\). We also set \( \Pi _0 (\omega ) (t) = \omega (t) - \omega (0),\, t \ge 0\), which defines a map \( \Pi _0 \, : \, \Omega ' \rightarrow \Omega '_0\). Then \(\Pi = (W_0, \Pi _0 ) \, : \, \Omega ' \rightarrow \mathbb {R}^d \times \Omega '_0\) is a bijection. For each probability measure \(\mu \) on \(\mathbb {R}^d\), the probability \(\mathbb {P}^{\mu }\) of the Brownian motion started with the initial distribution \(\mu \) is given by

$$\begin{aligned} \mathbb {P}^{\mu } = \Pi ^{-1} \left( \mu \otimes \mathbb {P}^0 \right) . \end{aligned}$$

In particular, for the Lebesgue measure in \(\mathbb {R}^d\), which we denote by \( m = dx\), we have

$$\begin{aligned} \mathbb {P}^{m} = \Pi ^{-1} \left( dx\otimes \mathbb {P}^0 \right) . \end{aligned}$$

Notice that \(\{W_{t-r}, \mathcal {F}'_{t-r}, r\in [0, t]\}\) is a backward local martingale under \(\mathbb {P}^m\). Let \(J(\cdot ,\cdot ): [0, \infty )\times \mathbb {R}^d \rightarrow \mathbb {R}^d\) be a measurable function such that \(J\in \mathbf {L}^2([0,T] \times \mathbb {{R}}^d\rightarrow \mathbb {{R}}^d) \) for every \(T>0\). We recall the forward and backward stochastic integral defined in [20, 27] under the measure \(\mathbb {P}^m\).

$$\begin{aligned} \displaystyle \int _s^t J(r,W_{r})*dW_r=\displaystyle \int _s^t \langle J(r,W_{r}), dW_r\rangle +\displaystyle \int _s^t \langle J(r,W_{r}), d\overleftarrow{W}_r\rangle . \end{aligned}$$

When J is smooth, one has

$$\begin{aligned} \displaystyle \int _s^t J(r,W_{r})*dW_r=-2\displaystyle \int _s^t div(J(r,\cdot ))(W_r)dr. \end{aligned}$$
(2.2)

We refer the reader to [20, 27] for more details.

3 A Sufficient Condition for LDP

In this section we recall the criteria obtained in [2] for proving a large deviation principle and we will provide a sufficient condition to verify the criteria.

Let \(\mathcal {E}\) be a Polish space with the Borel \(\sigma \)-field \(\mathcal {B}(\mathcal {E})\). Recall

Definition 3.1

(Rate function) A function \(I: \mathcal {E}\rightarrow [0,\infty ]\) is called a rate function on \(\mathcal {E}\), if for each \(M<\infty \), the level set \(\{x\in \mathcal {E}:I(x)\le M\}\) is a compact subset of \(\mathcal {E}\).

Definition 3.2

(Large deviation principle) Let I be a rate function on \(\mathcal {E}\). A family \(\{X^\varepsilon \}\) of \(\mathcal {E}\)-valued random elements is said to satisfy a large deviation principle on \(\mathcal {E}\) with rate function I if the following two claims hold.

  1. (a)

    (Upper bound) For each closed subset F of \(\mathcal {E}\),

    $$\begin{aligned} \limsup _{\varepsilon \rightarrow 0}\varepsilon \log \mathbb {P}(X^\varepsilon \in F)\le - \inf _{x\in F}I(x). \end{aligned}$$
  2. (b)

    (Lower bound) For each open subset G of \(\mathcal {E}\),

    $$\begin{aligned} \liminf _{\varepsilon \rightarrow 0}\varepsilon \log \mathbb {P}(X^\varepsilon \in G)\ge - \inf _{x\in G}I(x). \end{aligned}$$

3.1 A Criteria of Budhiraja–Dupuis

The Cameron–Martin space associated with the Brownian motion \(\{B_t=(B_t^1,\ldots ,B^j_t,\ldots ), t\in [0,T]\}\) is isomorphic to the Hilbert space \(K:=L^2([0,T]; l^2)\) with the inner product:

$$\begin{aligned} \langle h_1, h_2\rangle _{K}:=\int _0^T\langle h_1(s), h_2(s)\rangle _{l^2}ds, \end{aligned}$$

where

$$\begin{aligned} l^2= \left\{ a=(a_1,\ldots ,a_j,\ldots ); \sum _{i=1}^{\infty }a_i^2<\infty \right\} . \end{aligned}$$

\(l^2\) is a Hilbert space with inner product \(\langle a, b\rangle _{l^2}=\sum _{i=1}^{\infty }a_ib_i\) for \(a,b\in l^2\).

Let \(\tilde{K}\) denote the class of \(l^2\)-valued \(\{\mathcal {F}_t\}\)-predictable processes \(\phi \) that belong to the space K a.s.. Let \(S_N=\{k\in K; \int _0^T\Vert k(s)\Vert _{l^2}^2ds\le N\}\). The set \(S_N\) endowed with the weak topology is a compact Polish space. Set \(\tilde{S}_N=\{\phi \in \tilde{K};\phi (\omega )\in S_N, \mathbb {P}\text {-a.s.}\}\).

The following result was proved in [2].

Theorem 3.1

For \(\varepsilon >0\), let \(\Gamma ^\varepsilon \) be a measurable mapping from \(C([0,T];\mathbb {R}^\infty )\) into \(\mathcal {E}\). Set \(X^\varepsilon :=\Gamma ^\varepsilon (B(\cdot ))\). Suppose that there exists a measurable map \(\Gamma ^0:C([0,T];\mathbb {R}^\infty )\rightarrow \mathcal {E}\) such that

  1. (a)

    for every \(N<+\infty \) and any family \(\{k^\varepsilon ;\varepsilon >0\}\subset \tilde{S}_N\) satisfying that \(k^\varepsilon \) converges in law as \(S_N\)-valued random elements to some element k as \(\varepsilon \rightarrow 0\), \(\Gamma ^\varepsilon \left( B(\cdot )+\frac{1}{\sqrt{\varepsilon }}\int _0^{\cdot }k^\varepsilon (s)ds\right) \) converges in law to \(\Gamma ^0(\int _0^{\cdot }k(s)ds)\) as \(\varepsilon \rightarrow 0\);

  2. (b)

    for every \(N<+\infty \), the set

    $$\begin{aligned} \left\{ \Gamma ^0\left( \int _0^{\cdot }k(s)ds\right) ; k\in S_N\right\} \end{aligned}$$

    is a compact subset of \(\mathcal {E}\).

Then the family \(\{X^\varepsilon \}_{\varepsilon >0}\) satisfies a large deviation principle in \(\mathcal {E}\) with the rate function I given by

$$\begin{aligned} I(g):=\inf _{\{k\in K;g=\Gamma ^0(\int _0^{\cdot }k(s)ds)\}}\left\{ \frac{1}{2}\int _0^T\Vert k(s)\Vert _{l^2}^2ds\right\} ,\ g\in \mathcal {E}, \end{aligned}$$
(3.1)

with the convention \(\inf \{\emptyset \}=\infty \).

3.2 A Sufficient Condition

Here is a sufficient condition for verifying the assumptions in Theorem 3.1.

Theorem 3.2

For \(\varepsilon >0\), let \(\Gamma ^\varepsilon \) be a measurable mapping from \(C([0,T];\mathbb {R}^\infty )\) into \(\mathcal {E}\). Set \(X^\varepsilon :=\Gamma ^\varepsilon (B(\cdot ))\). Suppose that there exists a measurable map \(\Gamma ^0:C([0,T];\mathbb {R}^\infty )\rightarrow \mathcal {E}\) such that

  1. (i)

    for every \(N<+\infty \), any family \(\{k^\varepsilon ;\varepsilon >0\}\subset \tilde{S}_N\) and any \(\delta >0\),

    $$\begin{aligned} \lim _{\varepsilon \rightarrow 0}P(\rho \left( Y^\varepsilon , Z^\varepsilon \right) >\delta )=0, \end{aligned}$$

    where \(Y^\varepsilon =\Gamma ^\varepsilon (B(\cdot )+\frac{1}{\sqrt{\varepsilon }}\int _0^{\cdot }k^\varepsilon (s)ds)\), \(Z^\varepsilon =\Gamma ^0\left( \int _0^{\cdot }k^\varepsilon (s)ds\right) \) and \(\rho (\cdot , \cdot )\) stands for the metric in the space \(\mathcal {E}\)

  2. (ii)

    for every \(N<+\infty \) and any family \(\{k^\varepsilon ;\varepsilon >0\}\subset {S}_N\) satisfying that \(k^\varepsilon \) converges weakly to some element k as \(\varepsilon \rightarrow 0\), \(\Gamma ^0\left( \int _0^{\cdot }k^\varepsilon (s)ds\right) \) converges to \(\Gamma ^0(\int _0^{\cdot }k(s)ds)\) in the space \(\mathcal {E}\).

Then the family \(\{X^\varepsilon \}_{\varepsilon >0}\) satisfies a large deviation principle in \(\mathcal {E}\) with the rate function I given by

$$\begin{aligned} I(g):=\inf _{\{k\in K;g=\Gamma ^0(\int _0^{\cdot }k(s)ds)\}}\left\{ \frac{1}{2}\int _0^T\Vert k(s)\Vert _{l^2}^2ds\right\} ,\ g\in \mathcal {E}, \end{aligned}$$
(3.2)

with the convention \(\inf \{\emptyset \}=\infty \).

Remark 3.1

When proving a small noise large deviation principle for stochastic differential equations/stochastic partial differential equations, condition (i) is usually not difficult to check because the small noise disappears when \(\varepsilon \rightarrow 0\).

Proof

We will show that the conditions in Theorem 3.1 are fulfilled. Condition (b) in Theorem 3.1 follows from condition (ii) because \(S_N\) is compact with respect to the weak topology. Condition (i) implies that for any bounded, uniformly continuous function \(G(\cdot )\) on \(\mathcal {E}\),

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}E[|G(Y^\varepsilon )-G(Z^\varepsilon )|]=0. \end{aligned}$$

Thus, condition (a) will be satisfied if \(Z^\varepsilon \) converges in law to \(\Gamma ^0(\int _0^{\cdot }k(s)ds)\) in the space \(\mathcal {E}\). This is indeed true since the mapping \(\Gamma ^0\) is continuous by condition (ii) and \(k^\varepsilon \) converge in law as \(S_N\)-valued random elements to k. The proof is complete. \(\square \)

4 Skeleton Equations

Recall \(K:=L^2([0, T], l^2)\). Let \(k=(k^1,\ldots ,k^j,\ldots )\in K\) and consider the deterministic obstacle problem:

$$\begin{aligned}&du^k (t,x)+\frac{1}{2}\Delta u^k(t,x)+\sum _{i=1}^d\partial _ig_i(t,x, u^k(t,x), \nabla u^k(t,x))dt\nonumber \\&\quad +f(t,x,u^k(t,x), \nabla u^k(t,x))dt\nonumber \\&\quad + \sum _{j=1}^{\infty } h_j(t,x,u^k(t,x), \nabla u^k(t,x))k^j_tdt =-\nu ^k(dt,dx), \end{aligned}$$
(4.1)
$$\begin{aligned}&u^k(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&u^k(T,x)= \Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(4.2)

The existence and uniqueness of the solution of the deterministic obstacle problem (4.1) can be obtained similarly as the random obstacle problem (1.1) ( but simpler). We refer the reader to [13] for more details. Denote by \(u^{k^{\varepsilon }}\) the solution of equation (4.1) with \(k^{\varepsilon }\) replacing k. The main purpose of this section is to show that \(u^{k^{\varepsilon }}\) converges to \(u^k\) in the space \(H_T\) if \(k^{\varepsilon }\rightarrow k\) weakly in the Hilbert space K. To this end, we first need to establish a number of preliminary results.

Consider the penalized equation:

$$\begin{aligned}&du^{k,n} (t,x)+\frac{1}{2}\Delta u^{k,n}(t,x)+\sum _{i=1}^d\partial _ig_i(t,x, u^{k,n}(t,x), \nabla u^{k,n}(t,x))dt \nonumber \\&\quad +\, f(t,x,u^{k,n}(t,x), \nabla u^{k,n}(t,x))dt\nonumber \\&\quad + \, \sum _{j=1}^{\infty } h_j(t,x,u^{k,n}(t,x), \nabla u^{k,n}(t,x))k^j_tdt =-n(u^{k,n}(t,x)-L(t,x))^-dt , \nonumber \\\end{aligned}$$
(4.3)
$$\begin{aligned}&u^{k,n}(T,x)=\Phi (x), \quad x\in \mathbb {R}^d. \end{aligned}$$
(4.4)

It is known that \(u^{k,n}\rightarrow u^k\) as \(n\rightarrow \infty \) for a fixed \(k\in K\) (please see [13]). For later use, we need to show that for any \(M>0\), \(u^{k,n}\rightarrow u^k\) uniformly over the bounded subset \(\{k; \Vert k\Vert _K\le M\}\) as \(n\rightarrow \infty \). For this purpose, it turns out that we have to appeal to the BSDE representation of the solutions. Let \(Y^{k,n}_t:=u^{k,n}(t, W_t)\), \(Z^{k,n}_t=\nabla u^{k,n}(t, W_t)\). Then it was shown in [20] that \((Y^{k,n}, Z^{k,n})\) is the solution of the backward stochastic differential equation under \(\mathbb {P}^m\):

$$\begin{aligned} Y_{t}^{k,n}= & {} \Phi (W_{T})+\displaystyle \int _{t}^{T}f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr+\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&+\, n\displaystyle \int _t^T (Y^{k,n}_r-S_r)^-dr +\frac{1}{2}\displaystyle \int _s^T g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})*dW_r-\displaystyle \int _{t}^{T}Z_{r}^{k,n}dW_{r}.\nonumber \\ \end{aligned}$$
(4.5)

where \(S_r=L(r, W_r)\) satisfies

$$\begin{aligned} dS_r=\frac{\partial L}{\partial r}(r,W_r)dr+ \frac{1}{2}\Delta L(r,W_r)dr+\nabla L(r,W_r)dW_{r}. \end{aligned}$$
(4.6)

The following result is a uniform estimate for \((Y^{k,n}, Z^{k,n})\).

Lemma 4.1

For \(M>0\), we have the following estimate:

$$\begin{aligned}&\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\underset{n}{\displaystyle \sup }\left\{ \mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }|Y_{t}^{k,n}|^2\right] +\mathbb {E}^m\left[ \displaystyle \int _0^T|Z_t^{k,n}|^2dt\right] \right. \nonumber \\&\quad \quad \left. +\, \mathbb {E}^m\left[ \left( n\displaystyle \int _0^T(Y_t^{k,n}-S_t)^-dt\right) ^2\right] \right\} \nonumber \\&\quad \le \, c_M \left[ |\Phi |^2+\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }|S_{t}|^2\right] +\displaystyle \int _{\mathbb {R}^d}\displaystyle \int _0^T[|f^0(t,x)|^2\right. \nonumber \\&\quad \quad \left. +\, |g^0(t,x)|^2+|h^0(t,x)|^2]dtdx\right] \end{aligned}$$
(4.7)

The proof of this lemma is a repeat of the proof of Lemma 6 in [20]. One just needs to notice that when applying the Grownwall’s inequality, the constant \(c_M\) on on right of (4.7) only depends on the norm of k which is bounded by M.

We also need the following estimate.

Lemma 4.2

$$\begin{aligned} \underset{n}{\displaystyle \sup }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ n\displaystyle \int _0^T[(Y_{t}^{k,n}-S_t)^-]^2dt\right] \le C_M. \end{aligned}$$
(4.8)

Proof

Let \(F(z)=z^2\). Applying the Ito’s formula (see [20]) we have

$$\begin{aligned} F(Y_{t}^{k,n}-S_t)= & {} F(\Phi (W_{T})-S_T)+\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&+\,\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&+\, n\displaystyle \int _t^T F^{\prime }(Y_{r}^{k,n}-S_r)(Y^{k,n}_r-S_r)^-dr\nonumber \\&+\,\frac{1}{2}\displaystyle \int _s^T F^{\prime }(Y_{r}^{k,n}-S_r)g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})*dW_r\nonumber \\&+\,\displaystyle \int _t^T \Big \langle \nabla (F^{\prime }(u^{k,n}(r, \cdot )-L(r, \cdot ))), g(r,\cdot ,u^{k,n}(r, \cdot ),\nabla u^{k,n}(r, \cdot )) \Big \rangle (W_r)dr\nonumber \\&-\,\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)Z_{r}^{k,n}dW_{r}+\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)\frac{\partial L}{\partial r}(r,W_r)dr\nonumber \\&+\,\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)\frac{1}{2}\Delta L(r,W_r)dr+\displaystyle \int _{t}^{T}F^{\prime }(Y_{r}^{k,n}-S_r)\nabla L(r,W_r)dW_{r}\nonumber \\&-\,\frac{1}{2}\int _t^TF^{\prime \prime }(Y_{r}^{k,n}-S_r)|Z_r^{k,n}-\nabla L(r,W_r)|^2dr. \end{aligned}$$
(4.9)

Rearranging the terms we get

$$\begin{aligned}&(Y_{t}^{k,n}-S_t)^2 +\int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|^2dr+2n\displaystyle \int _t^T [(Y^{k,n}_r-S_r)^-]^2dr\nonumber \\&\quad =(\Phi (W_{T})-S_T)^2+2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&\quad \quad +\,2\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&\quad \quad +\,\displaystyle \int _t^T (Y_{r}^{k,n}-S_r)g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})*dW_r\nonumber \\&\quad \quad +\,2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})\Big \rangle dr\nonumber \\&\quad \quad -2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)Z_{r}^{k,n}dW_{r}+2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\frac{\partial L}{\partial r}(r,W_r)dr\nonumber \\&\quad \quad +\,\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\frac{1}{2}\Delta L(r,W_r)dr+2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\nabla L(r,W_r)dW_{r}.\nonumber \\ \end{aligned}$$
(4.10)

Using the conditions on h in the Assumption 2.1, for any given positive constant \(\varepsilon _1\) we have

$$\begin{aligned}&2\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&\quad =2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\sum _{j=1}^{\infty }(h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-h_j(r,W_{r},S_r,\nabla L(r,W_r)))k_r^jdr\nonumber \\&\quad \quad +\,2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\sum _{j=1}^{\infty }(h_j(r,W_{r},S_r,\nabla L(r,W_r))-h_j(r,W_{r},0,0))k_r^jdr\nonumber \\&\quad \quad +\,2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)\sum _{j=1}^{\infty }h_j(r,W_{r},0,0)k_r^jdr\nonumber \\&\quad \le 2\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-S_r|\left( \sum _{j=1}^{\infty }(h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-h_j(r,W_{r},S_r,\nabla L(r,W_r)))^2\right) ^{\frac{1}{2}}\Vert k_r\Vert _{l^2}dr\nonumber \\&\quad \quad +\,2\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-S_r|\left( \sum _{j=1}^{\infty }(h_j(r,W_{r},S_r,\nabla L(r,W_r))-h_j(r,W_{r},0,0))^2\right) ^{\frac{1}{2}}\Vert k_r\Vert _{l^2}dr\nonumber \\&\quad \quad +\, 2\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-S_r|\left( \sum _{j=1}^{\infty }h_j(r,W_{r},0,0)^2\right) ^{\frac{1}{2}}\Vert k_r\Vert _{l^2}dr\nonumber \\&\quad \le C\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-S_r|^2\Vert k_r\Vert _{l^2}^2dr+\varepsilon _1\displaystyle \int _{t}^{T}|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _{t}^{T}[L(r,W_r)^2+|\nabla L(r,W_r)|^2 +h^0(r,W_{r})^2]dr. \end{aligned}$$
(4.11)

By the assumptions on g, for any given positive constant \(\varepsilon _2\) we have

$$\begin{aligned}&2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})\Big \rangle dr\nonumber \\&\quad = 2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})-g(r,W_{r},S_r,\nabla L(r,W_r))\Big \rangle dr\nonumber \\&\quad \quad +\,2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r,W_{r},S_r,\nabla L(r,W_r))-g(r,W_{r},0,0)\Big \rangle dr\nonumber \\&\quad \quad +\,2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r,W_{r},0,0) \Big \rangle dr\nonumber \\&\quad \le 2C \displaystyle \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)| |Y^{k,n}_r-S_r|dr +2\alpha \displaystyle \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|[|L(r,W_r)|+|\nabla L(r,W_r)| +g^0(r,W_{r})]dr\nonumber \\&\quad \le C \displaystyle \int _t^T|Y^{k,n}_r-S_r|^2dr +(2\alpha +\varepsilon _2)\displaystyle \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _t^T[|L(r,W_r)|^2+|\nabla L(r,W_r)|^2 +g^0(r,W_{r})^2]dr. \end{aligned}$$
(4.12)

By a similar calculation, we have for any given \(\varepsilon _3>0\),

$$\begin{aligned}&2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&\quad \le C \displaystyle \int _t^T|Y^{k,n}_r-S_r|^2dr +\varepsilon _3\displaystyle \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _t^T[|L(r,W_r)|^2+|\nabla L(r,W_r)|^2 +f^0(r,W_{r})^2]dr. \end{aligned}$$
(4.13)

Substitute (4.11), (4.12) and (4.13) back into to (4.10), choose \(\varepsilon _1, \varepsilon _2, \varepsilon _3\) sufficiently small to obtain

$$\begin{aligned}&\mathbb {E}^m[(Y_{t}^{k,n}-S_t)^2] +\mathbb {E}^m\left[ \int _t^T|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\right] +2n\mathbb {E}^m\left[ \displaystyle \int _t^T [(Y^{k,n}_r-S_r)^-]^2dr\right] \nonumber \\&\quad \le C\mathbb {E}^m[(\Phi (W_{T})-S_T)^2]+C\mathbb {E}^m\left[ \displaystyle \int _{t}^{T}\{f^0(r,W_{r})^2+h^0(r,W_{r})^2+g^0(r,W_{r})^2\}dr\right] \nonumber \\&\quad \quad +\, C\displaystyle \int _{t}^{T}\mathbb {E}^m[(Y_{r}^{k,n}-S_r)^2]\Vert k_r\Vert _{l^2}^2dr+C\mathbb {E}^m\left[ \displaystyle \int _{t}^{T}(Y_{r}^{k,n}-S_r)^2dr\right] \nonumber \\&\quad \quad +\, C\mathbb {E}^m\left[ \displaystyle \int _{t}^{T}\left[ \left( \frac{\partial L}{\partial r}(r,W_r)+\Delta L(r,W_r)\right) ^2+|L(r,W_r)|^2+|\nabla L(r,W_r)|^2\right] \right. dr,\qquad \qquad \end{aligned}$$
(4.14)

where the condition on \(\alpha \) in the Assumption 2.1 was used. Now the desired conclusion (4.8) follows from the Grownwall’s inequality. \(\square \)

Lemma 4.3

For \(M>0\), we have

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m[\underset{0\le t\le T}{\displaystyle \sup }[(Y_{t}^{k,n}-S_t)^-]^4] =0. \end{aligned}$$
(4.15)

Proof

Let \(G(z)=(z^-)^4\). By the Ito’s formula we have

$$\begin{aligned} G(Y_{t}^{k,n}-S_t)= & {} \displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&+\,\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&+\,n\displaystyle \int _t^T G^{\prime }(Y_{r}^{k,n}-S_r)(Y^{k,n}_r-S_r)^-dr\nonumber \\&+\,\frac{1}{2}\displaystyle \int _s^T G^{\prime }(Y_{r}^{k,n}-S_r)g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})*dW_r\nonumber \\&+\,\displaystyle \int _t^T \Big \langle \nabla (G^{\prime }(u^{k,n}(r, \cdot )-L(r, \cdot ))), g(r,\cdot ,u^{k,n}(r,\cdot ),\nabla u^{k,n}(r,\cdot ))\Big \rangle (W_r)dr\nonumber \\&-\,\displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)Z_{r}^{k,n}dW_{r}+\displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)\frac{\partial L}{\partial r}(r,W_r)dr\nonumber \\&+\,\displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)\frac{1}{2}\Delta L(r,W_r)dr+\displaystyle \int _{t}^{T}G^{\prime }(Y_{r}^{k,n}-S_r)\nabla L(r,W_r)dW_{r}\nonumber \\&-\frac{1}{2}\int _t^TG^{\prime \prime }(Y_{r}^{k,n}-S_r)|Z_r^{k,n}-\nabla L(r,W_r)|^2dr. \end{aligned}$$
(4.16)

Rearrange the terms in the above equation to get

$$\begin{aligned}&[(Y_{t}^{k,n}-S_t)^-]^4+6\int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}\nonumber \\&\quad \quad -\, \nabla L(r,W_r)|^2dr+4n\displaystyle \int _t^T [(Y^{k,n}_r-S_r)^-]^4dr\nonumber \\&\quad =\, -4\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&\quad \quad -\, 4\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3 h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&\quad \quad -\, 2\displaystyle \int _t^T [(Y_{r}^{k,n}-S_r)^-]^3g(r,W_{r}(x),Y_{r}^{k,n},Z_{r}^{k,n})*dW_r\nonumber \\&\quad \quad +\, 12\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2\Big \langle Z^{k,n}_r-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})\Big \rangle dr\nonumber \\&\quad \quad +\, 4\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3Z_{r}^{k,n}dW_{r}-4\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3\frac{\partial L}{\partial r}(r,W_r)dr\nonumber \\&\quad \quad -\, 2\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3\Delta L(r,W_r)dr-4\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3\nabla L(r,W_r)dW_{r}.\nonumber \\ \end{aligned}$$
(4.17)

By Assumption 2.1, for any given positive constant \(\varepsilon _1\) we have

$$\begin{aligned}&12\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2\Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})\Big \rangle dr\nonumber \\&\quad =\, 12\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2\langle Z_r^{k,n}-\nabla L(r,W_r), g(r, W_r, Y^{k,n}_r, Z_r^{k,n})\nonumber \\&\quad \quad -\, g(r,W_{r},S_r,\nabla L(r,W_r))\rangle dr\nonumber \\&\quad \quad +\, 12\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2 \langle Z_r^{k,n}-\nabla L(r,W_r), g(r,W_{r},S_r,\nabla L(r,W_r))\nonumber \\&\quad \quad -\, g(r,W_{r},0,0)\rangle dr\nonumber \\&\quad \quad +\, 12\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2 \Big \langle Z_r^{k,n}-\nabla L(r,W_r), g(r,W_{r},0,0)\Big \rangle dr\nonumber \\&\quad \le \, C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^3|Z_r^{k,n}-\nabla L(r,W_r)| dr \nonumber \\&\quad \quad +\, 12\alpha \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|[|L(r,W_r)|+|\nabla L(r,W_r)|]dr\nonumber \\&\quad \quad +\, C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|g^0(r,W_{r})dr\nonumber \\&\quad \le \, \varepsilon _1\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2 dr \nonumber \\&\quad \quad +\, 12\alpha \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\nonumber \\&\quad \quad +\, C\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4[|L(r,W_r)|^2+|\nabla L(r,W_r)|^2\nonumber \\&\quad \quad +\, g^0(r,W_{r})^2]dr+C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4dr\nonumber \\&\quad \le \, (\varepsilon _1+12\alpha )\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2 dr \nonumber \\&\quad \quad +\, C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4dr \end{aligned}$$
(4.18)

Using again Assumption 2.1 and the similar computation as above we can show that for any constants \(\varepsilon _2>0, \varepsilon _3>0\),

$$\begin{aligned}&-\,4\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})k_r^jdr\nonumber \\&\quad \le \,\varepsilon _2\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2 dr +C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4dr\nonumber \\&\qquad +\,C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4\Vert k_r\Vert _{l^2}^2dr+C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2dr, \end{aligned}$$
(4.19)

and

$$\begin{aligned}&-\,4\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})dr\nonumber \\&\quad \le \varepsilon _3\displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2 dr +C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^4dr\nonumber \\&\quad \quad +\,C \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2dr. \end{aligned}$$
(4.20)

Put (4.20), (4.19), (4.18) and (4.17) together, select the constants \(\varepsilon _1\), \(\varepsilon _2\) and \(\varepsilon _3\) sufficiently small, and take expectation to get

$$\begin{aligned}&\mathbb {E}^m[[(Y_{t}^{k,n}-S_t)^-]^4]+\mathbb {E}^m\left[ \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}- \nabla L(r,W_r)|^2dr\right] \nonumber \\&\quad \quad +\, 4n\mathbb {E}^m\left[ \displaystyle \int _t^T [(Y^{k,n}_r-S_r)^-]^4dr\right] \nonumber \\&\quad \le +\, C \displaystyle \int _t^T\mathbb {E}^m[[(Y_{r}^{k,n}-S_r)^-]^4]dr\nonumber \\&\quad \quad +\, C \displaystyle \int _t^T\mathbb {E}^m[[(Y_{r}^{k,n}-S_r)^-]^4]\Vert k_r\Vert _{l^2}^2dr+C \mathbb {E}^m\left[ \displaystyle \int _t^T[(Y_{r}^{k,n}-S_r)^-]^2dr\right] \qquad \qquad \end{aligned}$$
(4.21)

Applying the Grownwall’s inequality and Lemma 4.2 we obtain

$$\begin{aligned}&\underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\underset{0\le t\le T}{\displaystyle \sup }\mathbb {E}^m[[(Y_{t}^{k,n}-S_t)^-]^4]\nonumber \\&\quad \le \, C_M \underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \displaystyle \int _{0}^{T}[(Y_{r}^{k,n}-S_r)^-]^2dr\right] =0, \end{aligned}$$
(4.22)

and

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \int _0^T[(Y_{r}^{k,n}-S_r)^-]^2|Z_r^{k,n}-\nabla L(r,W_r)|^2dr\right] =0. \end{aligned}$$
(4.23)

Observe that by the assumptions on the function g,

$$\begin{aligned}&2\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }|\displaystyle \int _t^T [(Y_{r}^{k,n}-S_r)^-]^3g(r,W_{r}(x),Y_{r}^{k,n},Z_{r}^{k,n})*dW_r|\right] \nonumber \\&\quad \le \, C\mathbb {E}^m\left[ \left( \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^6|g|^2(r,W_{r}(x),Y_{r}^{k,n},Z_{r}^{k,n})dr\right) ^{\frac{1}{2}}\right] \nonumber \\&\quad \le \, \frac{1}{4}\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }[(Y_{r}^{k,n}-S_r)^-]^4\right] \nonumber \\&\quad \quad +\, C \mathbb {E}^m\left[ \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^2|g|^2(r,W_{r}(x),Y_{r}^{k,n},Z_{r}^{k,n})dr\right] \nonumber \\&\quad \le \, \frac{1}{4}\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }[(Y_{r}^{k,n}-S_r)^-]^4\right] +C \mathbb {E}^m\left[ \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^4dr\right] \nonumber \\&\quad \quad +\, C \mathbb {E}^m\left[ \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^2|Z_{r}^{k,n}-\nabla L(r,W_r)|^2dr\right] \nonumber \\&\quad \quad +\, C \mathbb {E}^m\left[ \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^2dr\right] , \end{aligned}$$
(4.24)

and

$$\begin{aligned}&4\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }|\displaystyle \int _{t}^{T}[(Y_{r}^{k,n}-S_r)^-]^3 \langle Z_{r}^{k,n}-\nabla L(r,W_r), dW_{r} \rangle |\right] \nonumber \\&\quad \le \, C\mathbb {E}^m\left[ \left( \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^6|Z_{r}^{k,n}-\nabla L(r,W_r)|^2dr\right) ^{\frac{1}{2}}\right] \nonumber \\&\quad \le \, \frac{1}{4}\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }[(Y_{r}^{k,n}-S_r)^-]^4\right] \nonumber \\&\quad \quad +\, C \mathbb {E}^m\left[ \displaystyle \int _0^T [(Y_{r}^{k,n}-S_r)^-]^2|Z_{r}^{k,n}-\nabla L(r,W_r)|^2dr\right] . \end{aligned}$$
(4.25)

Using (4.23)–(4.25) and taking supremum over the interval [0, T] in (4.17) we further deduce that

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }[(Y_{t}^{k,n}-S_t)^-]^4\right] =0. \end{aligned}$$

completing the proof. \(\square \)

Proposition 4.1

For any \(M>0\), we have

$$\begin{aligned} \underset{n \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }|u^{k,n}-u^k|_{H_T}=0. \end{aligned}$$
(4.26)

Proof

We note that for any \(n, q\ge 1\),

$$\begin{aligned}&|u^{k,n}-u^{k,q}|_{H_T}^2\nonumber \\&\quad \le \mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }(Y_{r}^{k,n}-Y_r^{k,q})^2\right] +C \mathbb {E}^m\left[ \displaystyle \int _0^T |Z_r^{k,n}-Z_{r}^{k,q}|^2dr\right] . \end{aligned}$$
(4.27)

(4.27) follows from the fact that the law of \(W_t\) under \(\mathbb {P}^m\) is the Lebesgue measure m for any \(t\ge 0\). Please also consult [20] (Theorem 3, Corollary 1) for details. Recall that for each \(k\in K\), \(u^{k,n}\rightarrow u^k\) as \(n\rightarrow \infty \). Thus, to prove (4.26), it is sufficient to show

$$\begin{aligned} \underset{n, q \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }(Y_{t}^{k,n}-Y_t^{k,q})^2\right] =0, \end{aligned}$$
(4.28)

and

$$\begin{aligned} \underset{n, q \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \displaystyle \int _0^T |Z_r^{k,n}-Z_{r}^{k,q}|^2dr\right] =0. \end{aligned}$$
(4.29)

We will achieve this with the help of backward stochastic differential equations satisfied by \(Y^{k,n}_t=u^{k,n}(t, W_t)\). Applying Ito’s formula we have

$$\begin{aligned}&(Y_{t}^{k,n}-Y_{t}^{k,q})^2+\int _t^T|Z_r^{k,n}-Z_r^{k,q}|^2dr\nonumber \\&\quad =\, 2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-Y_r^{k,q})(f(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-f(r,W_{r},Y_{r}^{k,q},Z_{r}^{k,q}))dr\nonumber \\&\quad \quad +\, 2\sum _{j=1}^{\infty }\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-Y_r^{k,q})(h_j(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-h_j(r,W_{r},Y_{r}^{k,q},Z_{r}^{k,q}))k_r^jdr\nonumber \\&\quad \quad +\, 2n\displaystyle \int _t^T (Y_{r}^{k,n}-Y_r^{k,q})(Y^{k,n}_r-S_r)^-dr-2q\displaystyle \int _t^T (Y_{r}^{k,n}-Y_r^{k,q})(Y^{k,q}_r-S_r)^-dr\nonumber \\&\quad \quad +\, \displaystyle \int _t^T (Y_{r}^{k,n}-Y_r^{k,q})(g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-g(r,W_{r},Y_{r}^{k,m},Z_{r}^{k,q}))*dW_r\nonumber \\&\quad \quad +\, 2\displaystyle \int _t^T \Big \langle Z_r^{k,n}-Z_r^{k,q}, g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-g(r,W_{r},Y_{r}^{k,q},Z_{r}^{k,q})\Big \rangle dr\nonumber \\&\quad \quad -\, 2\displaystyle \int _{t}^{T}(Y_{r}^{k,n}-Y_r^{k,q})\big \langle Z_{r}^{k,n}-Z_r^{k,q},dW_{r}\big \rangle \nonumber \\&\quad := \, I_1^{k,n,q}(t)+I_2^{k,n,q}(t)+I_3^{k,n,q}(t)+I_4^{k,n,q}(t)+I_5^{k,n,q}(t)+I_6^{k,n,q}(t)+I_7^{k,n,q}(t).\nonumber \\ \end{aligned}$$
(4.30)

Note that

$$\begin{aligned}&I_3^{k,n,q}(t)+I_4^{k,n,q}(t)\nonumber \\&\quad =\, 2n\displaystyle \int _t^T (Y_{r}^{k,n}-Y_r^{k,q})(Y^{k,n}_r-S_r)^-dr-2q\displaystyle \int _t^T (Y_{r}^{k,n}-Y_r^{k,q})(Y^{k,q}_r-S_r)^-dr\nonumber \\&\quad \le \, 2n\displaystyle \int _t^T (Y_r^{k,q}-S_r)^-(Y^{k,n}_r-S_r)^-dr+2q\displaystyle \int _t^T (Y_{r}^{k,n}-S_r)^-(Y^{k,q}_r-S_r)^-dr\nonumber \\&\quad \le \, 2\underset{0\le r\le T}{\displaystyle \sup }(Y_r^{k,q}-S_r)^- n\displaystyle \int _0^T(Y^{k,n}_r-S_r)^-dr\nonumber \\&\quad \quad +\, 2\underset{0\le r\le T}{\displaystyle \sup }(Y_r^{k,n}-S_r)^- q\displaystyle \int _0^T(Y^{k,q}_r-S_r)^-dr. \end{aligned}$$
(4.31)

By Young’s inequality, we have for any \(\delta _1>0\),

$$\begin{aligned} I_1^{k,n,q}(t)\le \delta _1\displaystyle \int _{t}^{T}|Z_{r}^{k,n}-Z_r^{k,q}|^2dr+ C\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-Y_r^{k,q}|^2dr. \end{aligned}$$
(4.32)

Moreover for any \(\delta _2>0\), we have

$$\begin{aligned} I_2^{k,n,q}(t)\le \delta _2\displaystyle \int _{t}^{T}|Z_{r}^{k,n}-Z_r^{k,q}|^2dr+ C\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-Y_r^{k,q}|^2(1+\Vert k_r\Vert _{l^2}^2)dr. \end{aligned}$$
(4.33)

Using Young’s inequality again, we have for any \(\delta _3>0\),

$$\begin{aligned} I_6^{k,n,q}(t)\le (\delta _3+2\alpha )\displaystyle \int _{t}^{T}|Z_{r}^{k,n}-Z_r^{k,q}|^2dr+ C\displaystyle \int _{t}^{T}|Y_{r}^{k,n}-Y_r^{k,q}|^2dr. \end{aligned}$$
(4.34)

Substitute (4.31)–(4.34) back to (4.30), choose constants \(\delta _i, i=1,2,3 \) sufficiently small and take expectation to obtain

$$\begin{aligned}&\mathbb {E}^m[(Y_{t}^{k,n}-Y_{t}^{k,q})^2]+\mathbb {E}^m\left[ \int _t^T|Z_r^{k,n}-Z_r^{k,q}|^2dr\right] \nonumber \\&\quad \le \, C\mathbb {E}^m\left[ \displaystyle \int _{t}^{T}|Y_{r}^{k,n}-Y_r^{k,q}|^2(1+\Vert k_r\Vert _{l^2}^2)dr\right] \nonumber \\&\quad \quad +\, C\left( \mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }[(Y_r^{k,q}-S_r)^-]^2\right] \right) ^{\frac{1}{2}} \left( \mathbb {E}^m\left[ (n\displaystyle \int _0^T(Y^{k,n}_r-S_r)^-dr)^2\right] \right) ^{\frac{1}{2}}\nonumber \\&\quad \quad +\, C\left( \mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }[(Y_r^{k,n}-S_r)^-]^2\right] \right) ^{\frac{1}{2}} \left( \mathbb {E}^m\left[ \left( q\displaystyle \int _0^T(Y^{k,q}_r-S_r)^-dr\right) ^2\right] \right) ^{\frac{1}{2}}\nonumber \\ \end{aligned}$$
(4.35)

Using Lemmas 4.1, 4.3 and applying the Grownwall’s inequality we deduce that

$$\begin{aligned} \underset{n, q \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\underset{0\le t\le T}{\displaystyle \sup }\mathbb {E}^m[(Y_{t}^{k,n}-Y_t^{k,q})^2] =0, \end{aligned}$$
(4.36)

and

$$\begin{aligned} \underset{n, q \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \int _0^T|Z_{t}^{k,n}-Z_t^{k,q}|^2dt\right] =0. \end{aligned}$$
(4.37)

Next we will strengthen the convergence in (4.36) to

$$\begin{aligned} \underset{n, q \rightarrow \infty }{\lim }\underset{\{k\in K; \Vert k\Vert _K\le M\}}{\displaystyle \sup }\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }(Y_{t}^{k,n}-Y^{k,q})^2\right] =0. \end{aligned}$$
(4.38)

We notice that by the Burkhölder’s inequality, for any \(\delta _4>0\) we have

$$\begin{aligned}&\mathbb {E}^m[ \underset{0\le t\le T}{\displaystyle \sup }|I_5^{k,n,q}(t)|]\nonumber \\&\quad \le \, C\mathbb {E}^m\left[ \left( \displaystyle \int _0^T (Y_{r}^{k,n}-Y_r^{k,q})^2|g(r,W_r,Y_{r}^{k,n},Z_{r}^{k,n})-g(r,W_{r},Y_{r}^{k,q},Z_{r}^{k,q})|^2dr\right) ^{\frac{1}{2}}\right] \nonumber \\&\quad \le \, \delta _4\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }(Y_{r}^{k,n}-Y_r^{k,q})^2\right] \nonumber \\&\quad \quad +\, C \mathbb {E}^m\left[ \displaystyle \int _0^T |g(r,W_{r},Y_{r}^{k,n},Z_{r}^{k,n})-g(r,W_{r},Y_{r}^{k,q},Z_{r}^{k,q})|^2dr\right] \nonumber \\&\quad \le \, \delta _4\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }(Y_{r}^{k,n}-Y_r^{k,q})^2\right] +C\mathbb {E}^m\left[ \displaystyle \int _0^T |Z_{r}^{k,n}-Z_{r}^{k,q}|^2dr\right] \nonumber \\&\quad \quad +\, C\mathbb {E}^m\left[ \displaystyle \int _0^T |Y_{r}^{k,n}-Y_{r}^{k,q}|^2dr\right] . \end{aligned}$$
(4.39)

Similarly, we have for \(\delta _5>0\)

$$\begin{aligned}&\mathbb {E}^m\left[ \underset{0\le t\le T}{\displaystyle \sup }|I_7^{k,n,q}(t)|\right] \nonumber \\&\quad \le C\mathbb {E}^m\left[ \left( \displaystyle \int _0^T (Y_{r}^{k,n}-Y_r^{k,q})^2|Z_r^{k,n}-Z_r^{k,q}|^2dr\right) ^{\frac{1}{2}}\right] \nonumber \\&\quad \le \delta _5\mathbb {E}^m\left[ \underset{0\le r\le T}{\displaystyle \sup }(Y_{r}^{k,n}-Y_r^{k,q})^2\right] +C \mathbb {E}^m\left[ \displaystyle \int _0^T |Z_r^{k,n}-Z_{r}^{k,q}|^2dr\right] .\qquad \end{aligned}$$
(4.40)

Now use the above two estimates (4.39) and (4.40) and the already proved (4.36) to obtain (4.38). This completes the proof.\(\square \)

Theorem 4.1

Let Assumptions 2.1 hold. Assume that \(k^{\varepsilon }\rightarrow k\) weakly in the Hilbert space K as \(\varepsilon \rightarrow 0\). Then \(u^{k^{\varepsilon }}\) converges to \(u^k\) in the space \(H_T\), where \(u^{k^{\varepsilon }}\) denotes the solution of equation (4.1) with \(k^{\varepsilon }\) replacing k.

Proof

We will first prove a similar convergence result for the corresponding penalized PDEs and then combined with the uniform convergence proved in Proposition 4.1 we complete the proof of Theorem 4.1. Let \(u^{k^{\varepsilon }, n}\) be the solution to the following penalized PDE:

$$\begin{aligned}&du^{k^{\varepsilon }, n} (t,x)+\frac{1}{2}\Delta u^{k^{\varepsilon }, n}(t,x)+\sum _{i=1}^d\partial _ig_i(t,x, u^{k^{\varepsilon }, n}(t,x), \nabla u^{k^{\varepsilon }, n}(t,x))dt\nonumber \\&\quad +\, f(t,x,u^{k^{\varepsilon }, n}(t,x), \nabla u^{k^{\varepsilon }, n}(t,x))dt\nonumber \\&\quad +\, \sum _{j=1}^{\infty } h_j(t,x,u^{k^{\varepsilon }, n}(t,x), \nabla u^{k^{\varepsilon }, n}(t,x))k^{\varepsilon ,j}_tdt \nonumber \\&\quad \quad =\, -n(u^{k^{\varepsilon }, n}(t,x)-L(t,x))^-dt , \end{aligned}$$
(4.41)
$$\begin{aligned}&u^{k^{\varepsilon }, n}(T,x)=\Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(4.42)

We first fix the integer n and show \(\lim _{\varepsilon \rightarrow 0}\Vert u^{k^{\varepsilon },n}-u^{k,n}\Vert _{H_T}=0\), \(u^{k,n}\) is the solution of equation (4.41) with \(k^{\varepsilon }\) replaced by k. To this end, we first prove that the family \(\{ u^{k^{\varepsilon }, n}, \varepsilon >0\}\) is tight in the space \(L^2([0, T], L_{loc}^2(\mathbb {R}^d))\). Using the chain rule and Gronwall’s inequality, as in Lemma 4.1 , we can show that

$$\begin{aligned} \underset{\varepsilon }{\displaystyle \sup }\Vert u^{k^{\varepsilon }, n}\Vert _{H_T}^2=\underset{\varepsilon }{\displaystyle \sup }\left\{ \sup _{0\le t\le T}|u^{k^{\varepsilon }, n}(t)|^2+\int _0^T\Vert u^{k^{\varepsilon }, n}(t)\Vert ^2dt\right\} <\infty . \end{aligned}$$
(4.43)

For \(\beta \in (0,1)\), recall that \(W^{\beta ,2}([0,T], V^*)\) is the space of mappings \(v(\cdot ): [0, T]\rightarrow V^*\) that satisfy

$$\begin{aligned} \Vert v\Vert _{W^{\beta ,2}([0,T], V^*)}^2=\int _0^T\Vert v(t)\Vert _{V^*}^2+\int _0^T\int _0^T\frac{\Vert v(t)-v(s)\Vert _{V^*}^2}{|t-s|^{1+2\beta }} <\infty . \end{aligned}$$
(4.44)

It is well known (see e.g. [15]) that the imbedding

$$\begin{aligned} L^2([0,T], V)\cap W^{\beta ,2}([0,T], V^*)\hookrightarrow L^2([0,T], L_{loc}^2(\mathbb {R}^d)) \end{aligned}$$

is compact. As an equation in \(V^*\), we have

$$\begin{aligned} u^{k^{\varepsilon }, n} (t)= & {} \Phi +\frac{1}{2}\int _t^T\Delta u^{k^{\varepsilon }, n}(s)ds+\int _t^T\sum _{i=1}^d\partial _ig_i(s,x, u^{k^{\varepsilon }, n}(s,x), \nabla u^{k^{\varepsilon }, n}(s,x))ds \nonumber \\&+\int _t^Tf(s,x,u^{k^{\varepsilon }, n}(s,x), \nabla u^{k^{\varepsilon }, n}(s,x))ds\nonumber \\&+ \sum _{j=1}^{\infty } \int _t^Th_j(s,x,u^{k^{\varepsilon }, n}(s,x), \nabla u^{k^{\varepsilon }, n}(s,x))k^{\varepsilon ,j}_sds \nonumber \\&+\,n\int _t^T(u^{k^{\varepsilon }, n}(s,x)-L(s,x))^-ds \nonumber \\:= & {} \Phi +I_1(t)+I_2(t)+I_3(t)+I_4(t)+I_5(t). \end{aligned}$$
(4.45)

In view of (4.43), we have

$$\begin{aligned} \Vert I_1(t)-I_1(s)\Vert _{V^*}^2\le & {} C\int _s^t\Vert \Delta u^{k^{\varepsilon }, n}(r)\Vert _{V^*}^2dr (|t-s|)\nonumber \\\le & {} C\int _0^T\Vert u^{k^{\varepsilon }, n}(r)\Vert ^2dr (|t-s|)\le C|t-s|. \end{aligned}$$
(4.46)

Using the condition (iii) in Assumption 2.1, we have

$$\begin{aligned} \Vert I_4(t)-I_4(s)\Vert _{V^*}^2\le C\left( \int _0^T\Vert k^{\varepsilon }_r\Vert _{l^2}^2dr\right) |\bar{h}|^2 |t-s|\le C|t-s|. \end{aligned}$$
(4.47)

By (4.43) and the similar calculations as above we also have

$$\begin{aligned} \Vert I_i(t)-I_i(s)\Vert _{V^*}^2\le C|t-s|, \quad i=2,3,5. \end{aligned}$$
(4.48)

Thus, for \(\beta \in (0,\frac{1}{2})\), it follows from (4.45) –(4.48) that

$$\begin{aligned} \underset{\varepsilon }{\displaystyle \sup }\Vert u^{k^{\varepsilon }, n}\Vert _{W^{\beta ,2}([0,T], V^*)}^2<\infty . \end{aligned}$$
(4.49)

Combining (4.49) with (4.43), we conclude that \(\{ u^{k^{\varepsilon }, n}, \varepsilon >0\}\) is tight in the space \(L^2([0, T], L_{loc}^2(\mathbb {R}^d))\). Now, applying the chain rule, we obtain

$$\begin{aligned}&|u^{k^{\varepsilon }, n}(t)-u^{k,n}(t)|^2\nonumber \\&\quad =\, -\int _t^T|\nabla (u^{k^{\varepsilon },n}(s)-u^{k,n}(s))|^2ds\nonumber \\&\qquad -\, 2\int _t^T\Big \langle g(s,\cdot , u^{k^{\varepsilon },n}(s,\cdot ), \nabla u^{k^{\varepsilon },n}(s,\cdot ))\nonumber \\&\qquad -\, g(s,\cdot , u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot )), \nabla (u^{k^{\varepsilon }, n}(s)-u^{k,n}(s))\Big \rangle ds\nonumber \\&\qquad +\, 2\int _t^T\Big \langle f(s,\cdot ,u^{k^{\varepsilon }, n}(s,\cdot ), \nabla u^{k^{\varepsilon }, n}(s,\cdot ))\nonumber \\&\qquad -\, f(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot )),u^{k^{\varepsilon }, n}(s)-u^{k,n}(s)\Big \rangle ds\nonumber \\&\qquad +\, 2\int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }(h_j(s,\cdot ,u^{k^{\varepsilon }, n}(s,\cdot ), \nabla u^{k^{\varepsilon }, n}(s,\cdot ))\nonumber \\&\qquad -\,h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot )))k^{\varepsilon ,j}_s\Big \rangle ds \nonumber \\&\qquad +\, 2\int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon ,j}_s-k^{j}_s)\Big \rangle ds \nonumber \\&\qquad +\, 2n\int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), (u^{k^{\varepsilon }, n}(s)-L(s,\cdot ))^--(u^{k,n}(s,\cdot )-L(s, \cdot ))^-\Big \rangle ds\nonumber \\ \end{aligned}$$
(4.50)

By the assumptions on \(h_j\) and Young’s inequality, we see that for any given \(\delta _1>0\),

$$\begin{aligned}&2\int _t^T \Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }(h_j(s,\cdot ,u^{k^{\varepsilon }, n}(s,\cdot ), \nabla u^{k^{\varepsilon }, n}(s,\cdot ))\nonumber \\&\qquad -h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot )))k^{\varepsilon ,j}_s\Big \rangle ds \nonumber \\&\quad \le \,\delta _1 \int _t^T |\nabla (u^{k^{\varepsilon }, n}(s)-u^{k,n}(s))|^2ds+C\int _t^T|u^{k^{\varepsilon }, n}(s)-u^{k,n}(s)|^2(1+\Vert k_s^{\varepsilon }\Vert _{l^2}^2)ds.\nonumber \\ \end{aligned}$$
(4.51)

Using the assumptions on f, g and (4.51) it follows from (4.50) that there exist positive constants \(\delta \), C such that

$$\begin{aligned}&|u^{k^{\varepsilon }, n}(t)-u^{k,n}(t)|^2+\delta \int _t^T|\nabla (u^{k^{\varepsilon }, n}(s)-u^{k,n}(s))|^2ds\nonumber \\&\quad \le \, C\int _t^T|u^{k^{\varepsilon }, n}(s)-u^{k,n}(s)|^2(1+\Vert k_s^{\varepsilon }\Vert _{l^2}^2)ds\nonumber \\&\quad \quad +\, 2\int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon ,j}_s-k^{j}_s)\Big \rangle ds.\nonumber \\ \end{aligned}$$
(4.52)

By Gronwall’s inequality, (4.52) yields that

$$\begin{aligned}&\underset{0\le t\le T}{\displaystyle \sup }\left| u^{k^{\varepsilon }, n}(t)-u^{k,n}(t)\right| ^2\nonumber \\&\quad \le \, \exp (C\int _0^T(1+\Vert k_s^{\varepsilon }\Vert _{l^2}^2)ds)\underset{0\le t\le T}{\displaystyle \sup }\left| \int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)\right. \nonumber \\&\left. \qquad -\,u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon ,j}_s-k^{j}_s)\Big \rangle ds\right| \nonumber \\&\quad \le \, C\underset{0\le t\le T}{\displaystyle \sup }\left| \int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \right. \nonumber \\&\left. \qquad \times \, \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon ,j}_s-k^{j}_s)\Big \rangle ds\right| . \end{aligned}$$
(4.53)

To show \(\lim _{\varepsilon \rightarrow 0}\Vert u^{k^{\varepsilon },n}-u^{k,n}\Vert _{H_T}=0\), in view of (4.52) and (4.53), it suffices to prove

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\underset{0\le t\le T}{\displaystyle \sup }\left| \int _t^T\Big \langle u^{k^{\varepsilon }, n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon ,j}_s-k^{j}_s)\Big \rangle ds\right| =0. \end{aligned}$$
(4.54)

This will be achieved if we show that for any sequence \(\varepsilon _m\rightarrow 0\), one can find a subsequence \(\varepsilon _{m_i}\rightarrow 0\) such that

$$\begin{aligned} \lim _{i\rightarrow \infty }\underset{0\le t\le T}{\displaystyle \sup }\left| \int _t^T\Big \langle u^{k^{\varepsilon _{m_i}} , n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\right| =0 \end{aligned}$$
(4.55)

Now fix a sequence \(\varepsilon _m\rightarrow 0\). Since \(\{ u^{k^{\varepsilon _{m}}, n}, m\ge 1\}\) is tight in \(L^2([0,T], L_{loc}^2(\mathbb {R}^d))\), there exist a subsequence \(m_i, i\ge 1\) and a mapping \(\tilde{u}\) such that \(u^{k^{\varepsilon _{m_i}},n}\rightarrow \tilde{u}\) in \(L^2([0,T], L_{loc}^2(\mathbb {R}^d))\). Moreover, because of the uniform bound of \(u^{k^{\varepsilon _{m_i}},n}\) in (4.43), \(\tilde{u}\) belongs to \(L^2([0,T], H)\). Now,

$$\begin{aligned}&\underset{0\le t\le T}{\displaystyle \sup }\Big |\int _t^T\Big \langle u^{k^{\varepsilon _{m_i}} , n}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\nonumber \\&\quad \le \, \underset{0\le t\le T}{\displaystyle \sup }\Big |\int _t^T\Big \langle u^{k^{\varepsilon _{m_i}} , n}(s)-\tilde{u}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\nonumber \\&\quad \quad +\,\underset{0\le t\le T}{\displaystyle \sup }\Big |\int _t^T\Big \langle \tilde{u}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds.\nonumber \\ \end{aligned}$$
(4.56)

Since \(k^{\varepsilon _{m_i}}\rightarrow k\) weakly in \(L^2([0,T], l^2)\), for every \(t>0\), it holds that

$$\begin{aligned} \lim _{i\rightarrow \infty }\int _t^T\Big \langle \tilde{u}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds=0. \end{aligned}$$
(4.57)

On the other hand, using the assumption on h, for \(0<t_1<t_2\le T\), we have

$$\begin{aligned}&\left| \int _{t_1}^{t_2}\left\langle \tilde{u}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\right\rangle ds\right| \nonumber \\&\quad \le \, C\left( \int _{t_1}^{t_2}|\tilde{u}(s)-u^{k,n}(s)|^2ds\right) ^{\frac{1}{2}}\left( \int _{t_1}^{t_2}\Vert k^{\varepsilon _{m_i}}_s-k_s\Vert _{l^2}^2ds\right) ^{\frac{1}{2}}\nonumber \\&\quad \le \,C\left( \int _{t_1}^{t_2}|\tilde{u}(s)-u^{k,n}(s)|^2ds\right) ^{\frac{1}{2}}. \end{aligned}$$
(4.58)

Combing (4.57) and (4.58) we deduce that

$$\begin{aligned} \lim _{i\rightarrow \infty }\underset{0\le t\le T}{\displaystyle \sup }\left| \int _t^T\Big \langle \tilde{u}(s)-u^{k,n}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\right| =0. \end{aligned}$$
(4.59)

By Hölder’s inequality and the assumption on h, we have

$$\begin{aligned}&\left| \int _t^T\Big \langle u^{k^{\varepsilon _{m_i}} , n}(s)-\tilde{u}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ), \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\right| \nonumber \\&\quad \le \, \int _0^T\int _{\mathbb {R}^d}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)| \nonumber \\&\quad \quad \times \,\left( \sum _{j=1}^{\infty }h_j^2(s,\cdot ,u^{k,n}(s,x), \nabla u^{k,n}(s,x))^2\right) ^{\frac{1}{2}}dx (\Vert k^{\varepsilon _{m_i}}_s\Vert _{l^2}+\Vert k_s\Vert _{l^2})ds)\nonumber \\&\quad \le \, \left( \int _0^T(\Vert k^{\varepsilon _{m_i}}_s\Vert _{l^2}^2+\Vert k_s\Vert _{l^2}^2)ds\right) ^{\frac{1}{2}} \left( \int _0^Tds\left( \int _{\mathbb {R}^d}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)| \bar{h}(x)dx\right) ^2\right) ^{\frac{1}{2}}\nonumber \\&\quad \le \, C \left( \int _0^Tds\left( \int _{\mathbb {R}^d}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)| \bar{h}(x)dx\right) ^2\right) ^{\frac{1}{2}}. \end{aligned}$$
(4.60)

For any \(M>0\), denote by \(B_M\) the ball in \(\mathbb {R}^d\) centered at zero with radius M. We can bound the right side of (4.60) as follows:

$$\begin{aligned}&\int _0^Tds\left( \int _{\mathbb {R}^d}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)| \bar{h}(x)dx\right) ^2\nonumber \\&\quad \le \, C\int _0^Tds\left( \int _{B_M}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)|^2dx\right) \left( \int _{\mathbb {R}^d}\bar{h}^2(x)dx\right) \nonumber \\&\quad \quad +\, C\int _0^Tds\left( \int _{\mathbb {R}^d}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)|^2dx\right) \left( \int _{B_M^c}\bar{h}^2(x)dx\right. \nonumber \\&\quad \le \, C\int _0^Tds\left( \int _{B_M}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)|^2dx\right) +C\int _{B_M^c}\bar{h}^2(x)dx,\qquad \end{aligned}$$
(4.61)

where the uniform \(L^2([0, T]\times \mathbb {R}^d)\)-bound of \(u^{k^{\varepsilon _{m_i}} , n}\) has been used. Now given any constant \(\delta >0\), we can pick a constant M such that \(C\int _{B_M^c}\bar{h}^2(x)dx\le \delta \). For the chosen constant M, we have

$$\begin{aligned} \lim _{i\rightarrow \infty }\int _0^Tds\left( \int _{B_M}|u^{k^{\varepsilon _{m_i}} , n}(s,x)-\tilde{u}(s,x)|^2dx\right) =0. \end{aligned}$$

Thus, it follows from (4.60), (4.61) that

$$\begin{aligned}&\lim _{i\rightarrow \infty }\underset{0\le t\le T}{\displaystyle \sup }\Big |\int _t^T\Big \langle u^{k^{\varepsilon _{m_i}} , n}(s)-\tilde{u}(s), \sum _{j=1}^{\infty }h_j(s,\cdot ,u^{k,n}(s,\cdot ),\nonumber \\&\quad \nabla u^{k,n}(s,\cdot ))(k^{\varepsilon _{m_i},j}_s-k^{j}_s)\Big \rangle ds\le \delta ^{\frac{1}{2}}. \end{aligned}$$
(4.62)

Since \(\delta \) is arbitrary, (4.55) follows from (4.56), (4.59) and (4.62). Hence we have proved \(\lim _{\varepsilon \rightarrow 0}\Vert u^{k^{\varepsilon },n}-u^{k,n}\Vert _{H_T}=0\).

Now we are ready to complete the last step of the proof. For any \(n\ge 1\), we have

$$\begin{aligned}&\Vert u^{k^{\varepsilon }}-u^k\Vert _{H_T}\nonumber \\&\quad \le \Vert u^{k^{\varepsilon }}-u^{k^{\varepsilon }, n}\Vert _{H_T}+\Vert u^{k^{\varepsilon }, n}-u^{k,n}\Vert _{H_T}+\Vert u^{k,n}-u^k\Vert _{H_T}. \end{aligned}$$
(4.63)

For any given \(\delta >0\), by Proposition 4.1 there exists an integer \(n_0\) such that \(\sup _{\varepsilon }\Vert u^{k^{\varepsilon }}-u^{k^{\varepsilon }, n_0}\Vert _{H_T}\le \frac{\delta }{2}\) and \(\Vert u^k-u^{k,n_0}\Vert _{H_T}\le \frac{\delta }{2}\). Replacing n in (4.63) by \(n_0\) we get

$$\begin{aligned} \Vert u^{k^{\varepsilon }}-u^k\Vert _{H_T}\le \delta +\Vert u^{k^{\varepsilon }, n_0}-u^{k,n_0}\Vert _{H_T}. \end{aligned}$$

As we just proved

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\Vert u^{k^{\varepsilon }, n_0}-u^{k,n_0}\Vert _{H_T}=0, \end{aligned}$$

we obtain that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\Vert u^{k^{\varepsilon }}-u^k\Vert _{H_T}\le \delta . \end{aligned}$$

Since the constant \(\delta \) is arbitrary, the proof is complete. \(\square \)

5 Large Deviations

After the preparations in Sect. 4, we are ready to state and to prove the large deviation result. Recall that \(U^{\varepsilon }\) is the solution of the obstacle problem:

$$\begin{aligned}&dU^{\varepsilon }(t,x)+\frac{1}{2}\Delta U^{\varepsilon }(t,x)+\sum _{i=1}^d\partial _ig_i(t,x,U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dt \nonumber \\&\quad +\, f(t,x, U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dt\nonumber \\&\quad +\, \sqrt{\varepsilon }\sum _{j=1}^{\infty } h_j(t,x, U^{\varepsilon }(t,x),\nabla U^{\varepsilon }(t,x))dB^j_t =-R^{\varepsilon }(dt,dx), \end{aligned}$$
(5.1)
$$\begin{aligned}&U^{\varepsilon }(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&U^{\varepsilon }(T,x)=\Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(5.2)

For \(k\in K=L^2([0,T], l^2)\), denote by \(u^k\) the solution of the following deterministic obstacle problem:

$$\begin{aligned}&du^k(t,x)+\frac{1}{2}\Delta u^k(t,x)+\sum _{i=1}^d\partial _ig_i(t,x, u^k(t,x), \nabla u^k(t,x))dt \nonumber \\&\quad +\, f(t,x,u^k(t,x),\nabla u^k(t,x))dt\nonumber \\&\quad +\, \sum _{j=1}^{\infty } h_j(t,x,u^k(t,x),\nabla u^k(t,x))k^j_tdt =-\nu ^k(dt,dx), \end{aligned}$$
(5.3)
$$\begin{aligned}&u^k(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&u^k(T,x)= \Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(5.4)

Define a measurable mapping \(\Gamma ^0:C([0,T];\mathbb {R}^\infty )\rightarrow H_T\) by

$$\begin{aligned} \Gamma ^0\left( \int _0^{\cdot }k_sds\right) :=u^k\quad \quad \quad \quad \text{ for }\quad \quad \quad k\in K, \end{aligned}$$

where \(u^k\) is the solution of (5.3). Here is the main result:

Theorem 5.1

Let the Assumption 2.1 hold. Then the family \(\{U^\varepsilon \}_{\varepsilon >0}\) satisfies a large deviation principle on the space \(H_T\) with the rate function I given by

$$\begin{aligned} I(g):=\inf _{\{k\in K;g=\Gamma ^0(\int _0^{\cdot }k_sds)\}}\left\{ \frac{1}{2}\int _0^T\Vert k_s\Vert _{l^2}^2ds\right\} ,\ g\in H_T, \end{aligned}$$
(5.5)

with the convention \(\inf \{\emptyset \}=\infty \).

Proof

The existence of a unique strong solution of the obstacle problem (5.1) implies that for every \(\varepsilon >0\), there exists a measurable mapping \(\Gamma ^\varepsilon \left( \cdot \right) :C([0,T];\mathbb {R}^\infty )\rightarrow H_T\) such that

$$\begin{aligned} U^{\varepsilon }=\Gamma ^\varepsilon \left( B(\cdot )\right) . \end{aligned}$$

To prove the theorem, we are going to show that the conditions (i) and (ii) in Theorem 3.2 are satisfied. Condition (ii) is exactly the statement of Theorem 4.1. It remains to establish the condition (i) in Theorem 3.2. Recall the definitions of the spaces \(S_N\) and \(\tilde{S}_N\) given in Sect. 3. Let \(\{k^{\varepsilon }, \varepsilon >0\}\subset \tilde{S}_N\) be a given family of stochastic processes. Applying Girsanov theorem it is easy to see that \(U^{\varepsilon ,k^\varepsilon }=\Gamma ^\varepsilon \left( B(\cdot )+\frac{1}{\sqrt{\varepsilon }}\int _0^{\cdot }k^\varepsilon (s)ds\right) \) is the solution of the stochastic obstacle problem:

$$\begin{aligned}&dU^{\varepsilon ,k^\varepsilon }(t,x)+\frac{1}{2}\Delta U^{\varepsilon ,k^\varepsilon }(t,x)+\sum _{i=1}^d\partial _ig_i(t,x,U^{\varepsilon ,k^\varepsilon }(t,x),\nabla U^{\varepsilon ,k^\varepsilon }(t,x))dt \nonumber \\&\quad +\, f(t,x, U^{\varepsilon ,k^\varepsilon }(t,x),\nabla U^{\varepsilon ,k^\varepsilon }(t,x))dt\nonumber \\&\quad +\, \sqrt{\varepsilon }\sum _{j=1}^{\infty } h_j(t,x, U^{\varepsilon ,k^\varepsilon }(t,x),\nabla U^{\varepsilon ,k^\varepsilon }(t,x))dB^j_t\nonumber \\&\quad + \, \sum _{j=1}^{\infty } h_j(t,x, U^{\varepsilon ,k^\varepsilon }(t,x),\nabla U^{\varepsilon ,k^\varepsilon }(t,x))k^{\varepsilon , j}_tdt =-\nu ^{\varepsilon }(dt,dx), \end{aligned}$$
(5.6)
$$\begin{aligned}&U^{\varepsilon ,k^\varepsilon }(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&U^{\varepsilon ,k^\varepsilon }(T,x)=\Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(5.7)

Moreover, \(V^{k^\varepsilon }=\Gamma ^0\left( \int _0^{\cdot }k^\varepsilon (s)ds\right) \) is the solution of the random obstacle problem:

$$\begin{aligned}&dV^{k^\varepsilon }(t,x)+\frac{1}{2}\Delta V^{k^\varepsilon }(t,x)+\sum _{i=1}^d\partial _ig_i(t,x,V^{k^\varepsilon }(t,x),\nabla V^{k^\varepsilon }(t,x))dt \nonumber \\&\quad +\, f(t,x, V^{k^\varepsilon }(t,x),\nabla V^{k^\varepsilon }(t,x))dt\nonumber \\&\quad +\, \sum _{j=1}^{\infty } h_j(t,x, V^{k^\varepsilon }(t,x),\nabla V^{k^\varepsilon }(t,x))k^{\varepsilon , j}_tdt =-\mu ^{\varepsilon }(dt,dx), \end{aligned}$$
(5.8)
$$\begin{aligned}&V^{k^\varepsilon }(t,x)\ge L(t,x), \quad \quad (t,x)\in \mathbb {R}^+\times \mathbb {R}^d,\nonumber \\&V^{k^\varepsilon }(T,x)=\Phi (x), \quad \quad x\in \mathbb {R}^d. \end{aligned}$$
(5.9)

The condition (ii) in Theorem 3.2 will be satisfied if we prove

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\left\{ E[\underset{0\le t\le T}{\displaystyle \sup }|U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t|^2]+E[\int _0^T\Vert U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t\Vert ^2dt]\right\} =0, \end{aligned}$$
(5.10)

here \(U^{\varepsilon ,k^\varepsilon }_t=U^{\varepsilon ,k^\varepsilon }(t, \cdot )\) and \(V^{k^\varepsilon }_t=V^{k^\varepsilon }(t, \cdot )\). The rest of the proof is to establish (5.10). By Ito formula, we have

$$\begin{aligned}&|U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t|^2+\int _t^T|\nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s)|^2ds \nonumber \\&\quad =\, -2\int _t^T\Big \langle \nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s), g(s,\cdot ,U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)-g(s,\cdot ,V^{k^\varepsilon }_s,\nabla V^{k^\varepsilon }_s)\Big \rangle ds\nonumber \\&\quad \quad \, +2\int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, f(s,\cdot ,U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)-f(s,\cdot ,V^{k^\varepsilon }_s,\nabla V^{k^\varepsilon }_s)\Big \rangle ds\nonumber \\&\quad \quad +\, 2\int _t^T\sum _{j=1}^{\infty }<U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)-h_j(s,\cdot , V^{k^\varepsilon }_s,\nabla V^{k^\varepsilon }_s)>k^{\varepsilon , j}_sds\nonumber \\&\quad \quad +\, 2\sqrt{\varepsilon }\sum _{j=1}^{\infty } \int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)\Big \rangle dB^j_s\nonumber \\&\quad \quad +\, 2\int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, d\nu _s^{\varepsilon }- d\mu _s^{\varepsilon }\Big \rangle +\varepsilon \sum _{j=1}^{\infty } \int _t^T|h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)|^2ds\nonumber \\&\quad :=I_1(t)+I_2(t)+I_3(t)+I_4(t)+I_5(t)+I_6(t). \end{aligned}$$
(5.11)

Here

$$\begin{aligned}&\int _t^T\langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, d\nu _s^{\varepsilon }- d\mu _s^{\varepsilon }\rangle \\&\quad =\int _t^T\int _{\mathbb {R}^d}(U^{\varepsilon ,k^\varepsilon }(s,x)-V^{k^\varepsilon }(s,x))[\nu ^{\varepsilon }(ds,dx)- \mu ^{\varepsilon }(ds,dx)]. \end{aligned}$$

With the assumptions on g in mind, applying Young’s inequality we have for any \(\delta _1>0\)

$$\begin{aligned} I_1(t)\le (\delta _1+2\alpha )\int _t^T|\nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s)|^2ds+C\int _0^t|U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s|^2ds. \end{aligned}$$
(5.12)

By the assumption on f, for any \(\delta _2>0\), we have

$$\begin{aligned} I_2(t)\le \delta _2\int _t^T|\nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s)|^2ds+C\int _0^t|U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s|^2ds. \end{aligned}$$
(5.13)

Using the assumption on h, given any \(\delta _3>0\), we also have

$$\begin{aligned} I_3(t)\le \delta _3\int _t^T|\nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s)|^2ds+C\int _0^t|U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s|^2(1+\Vert k^{\varepsilon }_s\Vert _{l^2}^2) ds. \end{aligned}$$
(5.14)

For the term \(I_5\) in (5.11), because \(U^{\varepsilon ,k^\varepsilon }_s-L(s,\cdot )\ge 0\), \(V^{\varepsilon ,k^\varepsilon }_s-L(s,\cdot )\ge 0\) and because that the random measures \(\nu _s^{\varepsilon }\), \(\mu _s^{\varepsilon }\) are positive, we have

$$\begin{aligned} I_5(t)=2\int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-L(s,\cdot ) + L(s,\cdot )-V^{k^\varepsilon }_s, d\nu _s^{\varepsilon }- d\mu _s^{\varepsilon }\Big \rangle \le 0. \end{aligned}$$
(5.15)

Substituting (5.12)–(5.15) back into (5.11), choosing \(\delta _1, \delta _2, \delta _3\) sufficiently small and rearranging terms we can find a positive constant \(\delta >0\) such that

$$\begin{aligned}&|U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t|^2+\delta \int _t^T|\nabla (U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s)|^2ds\nonumber \\&\quad \le C\int _t^T|U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s|^2(1+\Vert k^{\varepsilon }_s\Vert _{l^2}^2) ds\nonumber \\&\quad \quad +\,2\sqrt{\varepsilon }\sum _{j=1}^{\infty } \int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, h_j(s,x, U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)\Big \rangle dB^j_s\nonumber \\&\quad \quad +\, \varepsilon \sum _{j=1}^{\infty } \int _t^T|h_j(s,x, U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)|^2ds. \end{aligned}$$
(5.16)

By the Gronwall’s inequality it follows that

$$\begin{aligned}&\underset{0\le t\le T}{\displaystyle \sup }|U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t|^2+\int _0^T\Vert U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t\Vert ^2dt \nonumber \\&\quad \le (M_1^\varepsilon +M_2^\varepsilon ) \exp \left( C \int _0^T(1+\Vert k^{\varepsilon }_s\Vert _{l^2}^2)ds\right) \le C_M (M_1^\varepsilon +M_2^\varepsilon ), \end{aligned}$$
(5.17)

where

$$\begin{aligned} M_1^\varepsilon= & {} \underset{0\le t\le T}{\displaystyle \sup }\left| 2\sqrt{\varepsilon }\sum _{j=1}^{\infty } \int _t^T\Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)\Big \rangle dB^j_s\right| ,\\ M_2^\varepsilon= & {} \varepsilon \sum _{j=1}^{\infty } \int _0^T|h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)|^2ds. \end{aligned}$$

Using Burkholder’s inequality and the boundedness of h, we see that

$$\begin{aligned} E[M_1^\varepsilon ]\le & {} C \sqrt{\varepsilon }E\left[ \left( \sum _{j=1}^{\infty } \int _0^T \Big \langle U^{\varepsilon ,k^\varepsilon }_s-V^{k^\varepsilon }_s, h_j(s,\cdot , U^{\varepsilon ,k^\varepsilon }_s,\nabla U^{\varepsilon ,k^\varepsilon }_s)\Big \rangle ^2ds\right) ^{\frac{1}{2}}\right] \nonumber \\\le & {} C \sqrt{\varepsilon } E\left[ \left( \int _0^T|U^{\varepsilon ,k^\varepsilon }_t-V^{k^\varepsilon }_t|^2dt\right) ^{\frac{1}{2}}\right] \nonumber \\&\rightarrow 0, \quad \text{ as }\quad \quad \varepsilon \rightarrow 0, \end{aligned}$$
(5.18)

where we have used the fact that \(\sup _{\varepsilon }\{E[|U^{\varepsilon ,k^\varepsilon }_t|^2]+E[|V^{k^\varepsilon }_t|^2]\}<\infty \). By the condition on h in the Assumption 2.1, it is also clear that

$$\begin{aligned} E[M_2^\varepsilon ]\le & {} C\varepsilon E\left[ \int _0^T(1+ |U^{\varepsilon ,k^\varepsilon }_s|^2+ \Vert U^{\varepsilon ,k^\varepsilon }_s\Vert ^2)ds\right] \nonumber \\&\rightarrow 0, \quad \text{ as }\quad \quad \varepsilon \rightarrow 0. \end{aligned}$$
(5.19)

Assertion (5.10) follows from (5.17) to (5.19). \(\square \)