Strategies in the principal-agent model

Abstract

In this paper, we study the principal-agent problem. It is well known that in continuous-time, it is possible to prove the existence of an equilibrium $(u^{*},S^{*})$ in a considerably general setting, and however little is known about the strategies themselves. Our goal is to present a universal way to construct these strategies as limiting strategies of a very simple type.

Appendices

Appendix I

The goal of this section is to prove Theorem 10. As we mentioned earlier, the proof is not very short so we present it here. In the first part of the appendix we solve a simpler problem and then, with the help of Ito’s Formula, we will show that this is actually general enough for our purpose.

The goal of this section is really to adapt well known constructions and proofs to our setting. We also need to use a logarithmic transform. For this reason we are going to minimize $\mathbf E \exp {y}$ rather than maximizing $\mathbf E \left[ -\exp {y}\right]$, of course the two are equivalent$.$

Let us start with

$$\begin{aligned} \left\{ \begin{array}{l} \min _{u\in \mathcal U } \log \mathbf E \left[ \exp \left\{ \int _{0}^{T}\ell (t, \mathbf X _{t},u_{t})\text{ d}t+G_{1}( \mathbf X _{T})\right\} \right] \\ \text{ d} \mathbf X _{t} = f(t, \mathbf X _{t},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t}\\ \mathbf X _{s} = x \end{array} \right. \end{aligned}$$

(89)

where we make the following assumptions on the data: ($\mathbf F$)

• F i.    $f(t,x,u):[0,T]\times \mathbf R ^{n}\times U\rightarrow \mathbf R$ is a continuous function;

• F ii.    $\forall u\in U$ and $\forall t\in [0,T] f(\cdot ,\cdot ,u):[0,T]\times \varOmega \rightarrow \mathbf R ^{d}$ is $\mathcal F _{t}$-predictable in $(t,x);$

• ${\mathbf F}$ iii.    $\left\Vert{f(u,x,t)}\right\Vert_{ \mathbf R ^{n}}\le K\left( 1+\left\Vert{x}\right\Vert_{ \mathbf R ^{n}}\right) \forall u\in U$ ,$\forall t\in [0,T],\forall x\in \mathbf R ^{n}$ and ($\mathbf C$)

• ${\mathbf C}$ i.    $\ell (t,x,u):[0,T]\times {\mathbf R }^{n}\times U\rightarrow {\mathbf R }$ is a continuous function$;$

• ${\mathbf C}$ ii.    there is a positive constant $M_{1}$ such that

  $$\begin{aligned} \ell (t,x,u)\ge -M_{1} \end{aligned}$$

  (90)

  $\forall t\in [0,T],\forall x\in \mathbf R ^{n},\forall u\in U.$

• ${\mathbf C}$ iii.    $\left\Vert{\ell (t,x_{1},u)-\ell (t,x_{2},u)}\right\Vert_{ \mathbf R ^{n}}\le K_{1}\left( 1+\left\Vert{x_{1}}\right\Vert_{ \mathbf R ^{n}}+\left\Vert{x_{2}}\right\Vert_{ \mathbf R ^{n}}\right) \left\Vert{x_{1}-x_{2}}\right\Vert_{ \mathbf R ^{n}} \forall t\in [0,T],\forall x\in \mathbf R ^{n},\forall u\in U$.

• $\mathbf C$ iv.    $\left\Vert{\ell (t,x_{1},u)}\right\Vert_{ \mathbf R ^{n}}\le K_{1}\left( 1+\left\Vert{x_{1}}\right\Vert_{ \mathbf R ^{n}}^{2}\right) \forall t\in [0,T],\forall x\in \mathbf R ^{n},\forall u\in U$ and ($\mathbf G$)

• ${\mathbf G}$ i.    $G_{1}: {\mathbf R }^{n}\rightarrow {\mathbf R }$ is a continuous function$;$

• ${\mathbf G}$ ii.    there is a positive constant $N_{1}$ such that

  $$\begin{aligned} G_{1}\left( x\right) \ge -N_{1} \end{aligned}$$

  (91)

  $\forall x\in \mathbf R ^{n}.$

• ${\mathbf G}$ iii.    $\left\Vert{G_{1}(x_{1})-G_{1}(x_{2})}\right\Vert_{ \mathbf R ^{n}}\le K_{2}\left( 1+\left\Vert{x_{1}}\right\Vert_{ \mathbf R ^{n}}+\left\Vert{x_{2}}\right\Vert_{ \mathbf R ^{n}}\right) \left\Vert{x_{1}-x_{2}}\right\Vert_{ \mathbf R ^{n}} \forall x\in \mathbf R ^{n}$.

• ${\mathbf G}$ iv.    $\left\Vert{G_{1}(x)}\right\Vert_{ {\mathbf R } ^{n}}\le K_{2}\left( 1+\left\Vert{x}\right\Vert_{ {\mathbf R }^{n}}^{2}\right) \forall x\in {\mathbf R }^{n}$

At this point, it is necessary to say a few words about the SDE we are going to use. The reader has probably already noticed that we are working with a very special class of SDE. To make things very clear, let us fix a probability space $(\varOmega , \mathcal F , \mathbf P )$ and on it we take an $n-$dimensional Brownian motion $\mathbf B =(B_{1},\ldots B_{n})$ with its natural filtration $\{ \mathcal F _{t}:t\in [0,T]\}$ such that $B_{i}$ is independent of $B_{j}$ if $i\not =j$ and such that the variance of $B_{i}(t,\cdot )$ is $t$ and $B_{i}(t,\cdot )= \mathbf E (B_{i}(T,\cdot )| \mathcal F _{t})$. The information structure is represented by the filtration $\{ \mathcal F _{t}:t\in [0,T]\}$. Let us denote with the symbol $U\subset \mathbf R ^{k}$ a compact set. Given a positive $\sigma >0$, we consider the following class of SDEs

$$\begin{aligned} \text{ d}Y_{t}=g(t,Y_{t},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \quad \text{ with} \; Y_{0}=y\in \mathbf R ^{n} \end{aligned}$$

(92)

(notice that we use $\sigma$ to also denote $\sigma I$ where $I$ is the $n\times n$ identity matrix) where $u\in \mathcal U$ and, by definition,

$$\begin{aligned} \mathcal U =\{u:[0,T]\times \varOmega \rightarrow U:u\;\text{ is}\;\mathcal F \text{-adapted}\} \end{aligned}$$

(93)

We say that the SDE has a strong solution if there exists a process $Y=\left\{ Y_{t}\right\} _{t\in [0,T]}$ such that

• 1. $Y$ is $\mathcal F _{t}\text{-adapted};$

• 2. $P(Y_{0}=y)=1;$

• 3. For any $j$ and any $t\in [0,T] \mathbf P (\int _{0}^{t}\left|{g_{j}(t,Y_{t},u_{t})}\right|\text{ d}t<\infty )=1;$

• 4. The SDE is satisfied for any $t\in [0,T].$

It is well known that an equation such as $\ \text{ d}Y_{t}=g(t,Y_{t},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t}$ with $Y_{0}=y\in \mathbf R ^{n}$ has always a strong solution as long $\sigma$ is a positive constant. Since we use only this type of SDE, we do not need to use anything more than what we have just stated.

In order to simplify we are going to introduce a few definitions. To study the problem, we introduce the following functions

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _{x,u}(s,\omega ) = \int _{s}^{T}\ell (t, \mathbf X _{t},u_{t})\text{ d}t+G_{1}( \mathbf X _{T}) \\ \\ \mathcal J (s,x,u) = \mathbf E _{s,x}\exp \left\{ \varphi _{x,u}(s,\omega )\right\} \end{array} \right. \end{aligned}$$

(94)

We define, as usual, the following

$$\begin{aligned} V(s,x)=\int \limits _{u\in U}\log \mathcal J (s,x,u) \end{aligned}$$

(95)

and

$$\begin{aligned} \varPsi (s,x)={\inf }_{u\in U} \mathcal J (s,x,u). \end{aligned}$$

(96)

We use the symbol $\mathbf X _{t}^{m}$ to denote the solution of the SDE

$$\begin{aligned} \left\{ \begin{array}{l} \text{ d} \mathbf X _{t}^{m} =\left( \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f\right) (t, \mathbf X _{t}^{m},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \\ \forall t\in [s,T] \\ \mathbf X _{s}^{m} =x\\ \end{array} \right. \end{aligned}$$

(97)

What follows is a standard argument to prove the existence of the control with obvious modifications to fit our setting. If $v,w\in \mathbf R ^{n}$ we will use the symbol $\left\langle v,w\right\rangle _{ \mathbf R ^{n}}$ to denote the inner product, if there is not any possible confusion, we will use $\left\langle v,w\right\rangle$ instead of $\left\langle v,w\right\rangle _{ \mathbf R ^{n}}$ and $\left|{v}\right|$ instead of $\left\Vert{v}\right\Vert_{ \mathbf R ^{n}}.$

Lemma 3

If we define

$$\begin{aligned} \varPsi _{m}={\inf }_{u\in U}\mathbf E _{s,x}\exp \left\{ \int _{s}^{T}\ell (t, \mathbf X _{t}^{m},u_{t})\mathrm{d}t+G_{1}( \mathbf X _{T}^{m})\right\} \end{aligned}$$

(98)

where

$$\begin{aligned} \begin{array}{ccc} \mathrm{d} \mathbf X _{t}^{m}&= \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t, \mathbf X _{t}^{m},u)\mathrm{d}t+\sigma \mathrm{d} \mathbf B _{t} \end{array} \end{aligned}$$

(99)

the following are true:

i.:

The function $\varPsi _{m}$ is in $C^{1,2}$ and it satisfies the following

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial \varPsi _{m}}{\partial \tau }+\frac{1}{2}\sigma ^{2}\Delta \varPsi _{m} = -{\min }_{u\in U}\left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(s,x,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (s,x,u)\varPsi _{m}\right] \\ \varPsi _{m}(T,x) = \exp G_{1}\left( x\right) \end{array}\right.\nonumber \\ \end{aligned}$$

(100)

ii.:

There is an admissible control $u^{*}$ such that

$$\begin{aligned} \varPsi _{m}(s,x)= \mathcal J (s,x,u^{*}) \end{aligned}$$

(101)

i.e. the inf exists.

Proof

To start we note that the equation has a solution since it is a consequence of Theorem 5.8.1 in (Ladyzenskaya et al. (1968), p. 495, 496), moreover the solution is a bounded solution in the $C^{1,2}$ class (Theorem 5.3.1 in Ladyzenskaya et al. 1968, p. 437). Since the solution of the SDE has a $L^{2}$ martingale part, we apply Ito’s Formula and we obtain the following formula

$$\begin{aligned} \varPsi _{m}(t, \mathbf X _{t}^{m})&= \varPsi _{m}(s,x)+\sigma \int \limits _{s}^{t}\nabla \varPsi _{m}(\tau , \mathbf X _{\tau }^{m})\text{ d} \mathbf B _{\tau } \\&+\int \limits _{s}^{t}\left( \frac{\partial \varPsi _{m}}{\partial \tau }(\tau , \mathbf X _{\tau }^{m})+\left\langle \nabla \varPsi _{m}(\tau , \mathbf X _{\tau }^{m}),\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t,x,u)\right\rangle \right) \text{ d}\tau \\&+\int \limits _{s}^{t}\frac{\sigma ^{2}}{2}\Delta \varPsi _{m}(\tau , \mathbf X _{\tau }^{m})\text{ d}\tau \end{aligned}$$

and, since

$$\begin{aligned} \mathbf M _{t}=\exp {\int _{s}^{t}\left( \chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (\tau , \mathbf X _{\tau }^{m},u_{\tau })\right) \text{ d}\tau } \end{aligned}$$

(102)

is a continuous semimartingale, we have

$$\begin{aligned} \mathbf M _{t}\varPsi _{m}(t, \mathbf X _{t}^{m})&= \mathbf M _{s}\varPsi _{m}(s,x)+\int \limits _{s}^{t} \mathbf M _{\tau }\varPsi _{m}(\tau , \mathbf X _{\tau }^{m})\left( \chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell \right) (\tau , \mathbf X _{\tau }^{m},u)\text{ d}\tau \\&+\int \limits _{s}^{t} \mathbf M _{\tau }\left( \frac{\partial \varPsi _{m}}{ \partial \tau }(\tau , \mathbf X _{\tau }^{m})+\left\langle \nabla \varPsi _{m}(\tau , \mathbf X _{\tau }^{m}),\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(\tau ,x,u)\right\rangle \right) \text{ d}\tau \\&+\,\sigma ^{2}\int \limits _{s}^{t} \mathbf M _{\tau }\nabla \varPsi _{m}(\tau , \mathbf X _{\tau }^{m})\text{ d} \mathbf B _{\tau }+\frac{\sigma ^{2}}{2}\int \limits _{s}^{t} \mathbf M _{\tau }\Delta \varPsi _{m}(\tau , \mathbf X _{\tau }^{m})\text{ d}\tau \end{aligned}$$

and this implies that $\varPsi _{m}(s,x)\le E_{s,x} \mathbf M _{T}\varPsi _{m}(T, \mathbf X _{T}^{m})$ and therefore, by using the fact that $\varPsi _{m}$ is the solution of the equation

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial \varPsi _{m}}{\partial \tau }+\frac{1}{2}\sigma ^{2}\Delta \varPsi _{m} = -{\min _{u\in U}}\left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(s,x,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (s,x,u)\varPsi _{m}\right] \\ \\ \varPsi _{m}(T,x) = \exp G_{1}\left( x\right) \end{array}\right.\nonumber \\ \end{aligned}$$

(103)

and taking $t=T,$ we can conclude that $\varPsi _{m}(s,x)\le J(s,x,u^{*}).$ Since the minimization problem

$$\begin{aligned} \arg \min _{u\in U} \left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(s,x,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (s,x,u)\varPsi _{m}\right] \end{aligned}$$

(104)

gives a Borel function $g(x,t)$ we can define $u_{t}(t,\widetilde{ \mathbf X }_{t})=g(t,\widetilde{ \mathbf X }_{t})$ where $\widetilde{ \mathbf X }_{t}$ is the unique strong solution of the equation

$$\begin{aligned} \text{ d}\widetilde{ \mathbf X }_{t}=\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t, \widetilde{ \mathbf X }_{t},g(t,\widetilde{ \mathbf X }_{t}))\text{ d}t+\sigma \text{ d} \mathbf B _{t}\ \end{aligned}$$

(105)

(see the discussion about the SDE at the beginning of this appendix) and this completes the proof.

At this point, we are able to state our last group of conditions. We say that the problem satisfies condition $( \mathbf H )$ when

• H i. There exists the limit $\varPsi$ of $\varPsi _{m}$.

• H ii. If

  $$\begin{aligned} g_{m}(x,t)=\arg \min _{u\in U} \left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(s,x,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (s,x,u)\varPsi _{m}\right]\nonumber \\ \end{aligned}$$

  (106)

  and

  $$\begin{aligned} g(x,t)=\arg \min _{u\in U} \left[ \left\langle f(s,x,u),\nabla \varPsi \right\rangle +\ell (s,x,u)\varPsi \right] \end{aligned}$$

  (107)

  then

  $$\begin{aligned} \lim _{m\rightarrow \infty }\sup _{\left\Vert{x}\right\Vert_{ \mathbf R ^{n}}\le K}\left|\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t,x,g_{m}(t,x))-f(t,x,g(t,x))\right|=0 \end{aligned}$$

  (108)

  for any $K>0$ and all $t\in [0,T].$

The stated condition will be very useful. It is important to notice that the family of functions $\left\{ \chi _{_{B( \mathbf R ^{n})}}^{m}\right\} _{m=1}^{\infty }$ we use has the identity function as limit.

At this point, we can prove the following

Lemma 4

There is a constant $B_{1}^{m}$such that $\left|\log \mathcal J ^{m}(s,x,u)\right|\le B_{1}^{m}(1+x^{2})$ and there is a constant $B_{1},$ independent from $m,$ such that $\left|\ln \mathcal J (s,x,u) \right|\le B_{1}(1+x^{2})$ for all $s$ and for all $x.$

Proof

To start we observe that, by using Gronwall’s Inequality (Øksendal 2010), if we define $Q_{t}=R+K(t-s)+K\int \nolimits _{s}^{t}Q_{\tau }\text{ d}\tau +\sigma \left|\mathbf B _{t}\right|$ then we have $\left|\mathbf X _{t}^{m} \right|\le Q_{t}$. In fact, we know that

$$\begin{aligned} \begin{array}{ccc} \text{ d} \mathbf X _{t}^{m}&= \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t, \mathbf X _{t}^{m},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \end{array} \end{aligned}$$

(109)

so, by definition, we have

$$\begin{aligned} \mathbf X _{t}^{m}=x+\int _{s}^{t}\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(\tau , \mathbf X _{\tau }^{m},u)\text{ d}\tau +\int _{s}^{t}\sigma \text{ d} \mathbf B _{\tau } \end{aligned}$$

(110)

and this implies that $\left|{ \mathbf X _{t}^{m}}\right|\le \left|{x}\right|+\int _{s}^{t}\left|{\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(\tau , \mathbf X _{\tau }^{m},u)}\right|\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|$ and if we observe that

$$\begin{aligned} \left|{\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(\tau , \mathbf X _{\tau }^{m},u)}\right|\le K\left( 1+\left|{ \mathbf X ^{m}}\right|\right) \end{aligned}$$

(111)

so we have

$$\begin{aligned} \left|{ \mathbf X _{t}^{m}}\right|&\le \left|{x}\right|+\int _{s}^{t}K\left( 1+\left|{ \mathbf X _{\tau }^{m}}\right|\right) \text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|\\&= \left|{x}\right|+\int _{s}^{t}K\text{ d}\tau +\int _{s}^{t}K\left|{ \mathbf X _{\tau }^{m}}\right|+\sigma \left|{ \mathbf B _{t}}\right|\end{aligned}$$

then the statement follows from Gronwall’s Inequality. In the same way, we can show that this fact implies that

$$\begin{aligned} \left|{ \mathbf X _{t}^{m}-x}\right|\le K(1+\left|{x}\right|)t+K\int _{s}^{t}\left|{ \mathbf X _{\tau }^{m}-x}\right|\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|\end{aligned}$$

(112)

and we can claim, by Gronwall’s Inequality and the Triangle Inequality, that there exists a constant $\widehat{C}$ such that

$$\begin{aligned} \left|{ \mathbf X _{t}^{m}}\right|&\le \left|{x}\right|+ \widehat{C}\left( K(1+\left|{x}\right|)T+\sigma \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\right) \\&\le (1+\left|{x}\right|)(1+\widehat{C}TK)+\widehat{C}\sigma \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\end{aligned}$$

In the same way, since $\mathbf X _{t}$ satisfies

$$\begin{aligned} \begin{array}{ccc} \text{ d} \mathbf X _{t}&= f(t, \mathbf X _{t},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \end{array} \end{aligned}$$

(113)

we have

$$\begin{aligned} \mathbf X _{t}=x+\int _{s}^{t}f(\tau , \mathbf X _{\tau },u)\text{ d}\tau +\int _{s}^{t}\sigma \text{ d} \mathbf B _{\tau } \end{aligned}$$

(114)

and this implies that $\left|{ \mathbf X _{t}}\right|\le \left|{x}\right|+\int _{s}^{t}\left|{f(\tau , \mathbf X _{\tau },u)}\right|\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|$. If we observe that

$$\begin{aligned} \left|{f(t, \mathbf X _{t},u)}\right|\le K\left( 1+\left|{ \mathbf X _{t}}\right|\right) \end{aligned}$$

(115)

then

$$\begin{aligned} \left|{ \mathbf X _{t}}\right|&\le \left|{x}\right|+\int _{s}^{t}K\left( 1+\left|{ \mathbf X _{\tau }}\right|\right) \text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|\\&= \left|{x}\right|+\int _{s}^{t}K\text{ d}\tau +\int _{s}^{t}K\left|{ \mathbf X _{\tau }}\right|\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|\end{aligned}$$

hence the statement follows from Gronwall’s inequality. In the same way, we can show that this fact implies that

$$\begin{aligned} \left|{ \mathbf X _{t}-x}\right|\le K(1+\left|{x}\right|)t+K\int _{s}^{t}\left|{ \mathbf X _{\tau }-x}\right|\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|\end{aligned}$$

(116)

and we have, by Gronwall’s Inequality and Triangle Inequality, that there exists a constant $\widehat{C}$ such that

$$\begin{aligned} \left|{ \mathbf X _{t}}\right|&\le \left|{x}\right|+ \widehat{C}\left( K(1+\left|{x}\right|)T+\sigma \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\right) \\&\le (1+\left|{x}\right|)(1+\widehat{C}TK)+\widehat{C}\sigma \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|. \end{aligned}$$

Therefore, if we choose $C=\max \left( K_{1},K_{2}\right)$ and we assume, without loss of generality, that $s=0$ we get

$$\begin{aligned} \left|{\int _{0}^{T}\ell (\tau , \mathbf X _{\iota },u_{\tau })\text{ d}\tau +G_{1}( \mathbf X _{T})}\right|&\le \int _{0}^{T}\left|{\ell (\tau , \mathbf X _{\iota },u_{\tau })}\right|\text{ d}\tau +\left|{G_{1}( \mathbf X _{T})}\right|\\&\le \int _{0}^{T}C(1+\left|{ \mathbf X _{\tau }}\right|^{2})\text{ d}\tau +C(1+\left|{ \mathbf X _{T}}\right|^{2}) \\&\le C(1+T)+C\left|{ \mathbf X _{T}}\right|^{2}+C\int _{0}^{T}\left|{ \mathbf X _{\tau }}\right|^{2}\text{ d}\tau \\&\le C(1+T)+C\left|{ \mathbf X _{T}}\right|^{2} \\&+\,C\int _{0}^{T}\left( 2(1+\left|{x}\right|)^{2}(1+\widehat{C} TK)^{2}\right) \text{ d}\tau \\&+\,C\int _{0}^{T}\left( 2\widehat{C}^{2}\sigma ^{2}\left( \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\right)^{2}\right) \text{ d}\tau \\&\le C(1+T)+C\left|{ \mathbf X _{T}}\right|^{2} \\&+C\left( 2(1+\left|{x}\right|)^{2}(1+\widehat{C}TK)^{2}\right) T \\&+2\widehat{C}^{2}C\sigma ^{2}\int _{0}^{T}\left( { \max }_{0\le t\le T}\left|{ \mathbf B _{t}}\right|^{2}\right) \text{ d}\tau \end{aligned}$$

so, after we set

$$\begin{aligned} p(\left|{x}\right|,T)=C(1+T)+C\left|{ \mathbf X _{T}}\right|^{2}+C\left( 2(1+\left|{x}\right|)^{2}(1+\widehat{C}TK)^{2}\right) T \end{aligned}$$

(117)

to simplify the notation, we can write

$$\begin{aligned} \mathbf E _{0,x}\exp \left\{ \left|{\varphi _{x,u}(t,\omega )}\right|\right\} \le \exp {p(\left|{x}\right|,T)} \mathbf E _{0,x}\exp \left\{ \left|{\widehat{C}^{2}C\sigma ^{2}\int _{0}^{T}\left( \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\right)^{2}\text{ d}t}\right|\right\} \nonumber \\ \end{aligned}$$

(118)

We need to calculate $\mathbf E _{0,x}\exp \left\{ 2\widehat{C}^{2}C\sigma ^{2}\int _{0}^{T}\left( {\max }_{0\le \tau \le T}\left|{ \mathbf B _{\tau }}\right|\right) ^{2}\text{ d}t\right\} .$ By using Doob’s maximal inequality and by using the fact that the density of $B_{t}^{(j)}$ is given by $\left( 2\pi t\right)^{-1/2}\exp (-x^{2}/2t)$ we get that, for $\sigma$ small enough, the integral is always bounded. In fact, we have the following

$$\begin{aligned} \mathbf E _{0,x}\exp \left\{ \left|{\varphi _{x,u}(t,\omega )}\right|\right\}&\le \exp {p(\left|{x}\right|,T)} \mathbf E _{0,x}\exp \left\{ \left|{\widehat{C}^{2}C\sigma ^{2}\int _{0}^{T}\left( \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\right) ^{2}\text{ d}t}\right|\right\} \\&\le \frac{2n}{\sqrt{2\pi T}}\exp {p(\left|{x}\right|,T)}\int _{-\infty }^{\infty }\exp {\left\{ 2\widehat{C}^{2}C\sigma ^{2}-\frac{1 }{2T}\right\} w^{2}}\text{ d}w. \end{aligned}$$

To complete the proof we notice that by taking the $\log$ of both sides and using the definition of $\mathcal J$ then there must exist a constant $B_{1}$ such that $\left|{\ln \mathcal J (s,x,u)}\right|\le B_{1}(1+x^{2}),$ similarly for $\mathcal J ^{m}.$

It is useful, in order to prove our next result, to simplify the notation. We set

$$\begin{aligned} \varphi _{ \mathbf X ^{m},x,u}(s,\omega ) = \int _{s}^{T}\ell (t, \mathbf X _{t}^{m},u_{t})\text{ d}t+G_{1}( \mathbf X _{T}^{m}) \end{aligned}$$

(119)

if

$$\begin{aligned} \left\{ \begin{array}{l} \text{ d} \mathbf X _{t}^{m} =\left( \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f\right)(t, \mathbf X _{t}^{m},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \\ \quad \forall t\in [s,T] \\ \mathbf X _{s}^{m} =x\\ \end{array} \right. \end{aligned}$$

(120)

and we set

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _{ \mathbf X ,x,u}(s,\omega ) = \int _{s}^{T}\ell (t, \mathbf X _{t},u_{t})\text{ d}t+G_{1}( \mathbf X _{T}) \end{array} \right. \end{aligned}$$

(121)

if

$$\begin{aligned} \left\{ \begin{array}{l} \text{ d} \mathbf X _{t} =f(t, \mathbf X _{t},u)\text{ d}t+\sigma \text{ d} \mathbf B _{t} \\ \quad \forall t\in [s,T] \\ \mathbf X _{s} =x\\ \end{array} \right. . \end{aligned}$$

(122)

We are now ready to state and prove the following

Lemma 5

We have

$$\begin{aligned} {\lim }_{m\rightarrow \infty }\left({\sup }_{u,x} \left|{ \mathbf E _{s,x}\exp \left\{ \varphi _{ \mathbf X ^{m},x,u}(s,\omega )\right\} - \mathbf E _{s,x}\exp \left\{ \varphi _{ \mathbf X ,x,u}(s,\omega )\right\} }\right|\right) =0\qquad \end{aligned}$$

(123)

i.e. we have uniform convergence on compact sets.

Proof

Let us fix a compact set $\mathcal K$ $\subset \mathbf R ^{n},$ if we define

$$\begin{aligned} B_{f,m}=\sup \left\{ c\ge 0\left|{ f=\chi _{_{B( \mathbf R ^{n})}}^{m}\circ f \forall \left( t,x,u\right) \in [0,T]\times cB( \mathbf R ^{n})\times U}\right. \right\} \end{aligned}$$

(124)

and

$$\begin{aligned} B_{\ell ,m}=\sup \left\{ c\ge 0\left|{ \ell =\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell \forall \left( t,x,u\right) \in [0,T]\times cB( \mathbf R ^{n})\times U}\right. \right\} \end{aligned}$$

(125)

and

$$\begin{aligned} B_{G_{1},m}=\sup \left\{ c\ge 0\left|{G_{1}=\chi _{_{B( \mathbf R ^{n})}}^{m}\circ G_{1} \forall \left( t,x,u\right) \in [0,T]\times cB( \mathbf R ^{n})\times U}\right. \right\} \nonumber \\ \end{aligned}$$

(126)

if we consider the following function

$$\begin{aligned} D_{m,X,u}^{s,x}= \mathbf E _{s,x}\exp \left\{ \varphi _{ \mathbf X _{t}^{m},u}(t,\omega )\right\} - \mathbf E _{s,x}\exp \left\{ \varphi _{ \mathbf X ,u}(t,\omega )\right\} . \end{aligned}$$

(127)

and if we define

$$\begin{aligned} \varOmega _{m}=\left\{ \omega \in \varOmega \left|{\left|{ \mathbf X _{t}(\omega )}\right|\le \min (B_{f,m},B_{\ell ,m},B_{G_{1},m})\quad \forall t\in [0,T]}\right. \right\} \end{aligned}$$

(128)

then, since $\mathbf X _{t}(\omega )= \mathbf X _{t}^{m}(\omega )$ on $\varOmega _{m},$ the following inequality

$$\begin{aligned} \left|{D_{m,X,u}^{s,x}}\right|\le \left|{ \mathbf E _{s,x} \mathbf 1 _{\varOmega _{m}^{c}}\exp \left\{ \varphi _{ \mathbf X _{t}^{m},u}(t,\omega )\right\} + \mathbf E _{s,x} \mathbf 1 _{\varOmega _{m}^{c}}\exp \left\{ \varphi _{ \mathbf X ,u}(t,\omega )\right\} } \right|\end{aligned}$$

(129)

must hold. Therefore, by the definition of the functions $\varphi _{ \mathbf X ^{m},x,u}(s,\omega )$ and $\varphi _{ \mathbf X ,x,u}(s,\omega )$ and the properties of the functions $\ell$ and $G_{1}$, we have

$$\begin{aligned} \left|{D_{m,X,u}^{s,x}}\right|&\le \mathbf E _{s,x} \mathbf 1 _{\varOmega _{m}^{c}}\exp \left\{ \int _{0}^{T}K_{1}\left( 1+\left|{ \mathbf X _{t}^{m}}\right|^{2}\right) \text{ d}t+K_{2}\left( 1+\left|{ \mathbf X _{T}^{m}}\right|^{2}\right) \right\} \\&+\, \mathbf E _{s,x} \mathbf 1 _{\varOmega _{m}^{c}}\exp \left\{ \int _{0}^{T}K_{1}\left( 1+\left|{ \mathbf X _{t}}\right|^{2}\right) \text{ d}t+K_{2}\left( 1+\left|{ \mathbf X _{T}}\right|^{2}\right) \right\} . \end{aligned}$$

Since we proved in the first part of Lemma 2 that $\left|{ \mathbf X _{t}^{m}}\right|$ and $\left|{ \mathbf X _{t}}\right|$ are dominated by the process $Q(t)=Q^{x}(t)$ ⁵ which is the solution of the equation $Q^{x}(t)=\left\Vert{x}\right\Vert+K(t-s)+K\int _{s}^{t}Q^{x}(\tau )\text{ d}\tau +\sigma \left|{ \mathbf B _{t}}\right|$ (note that $\mathcal K$ compact implies $\max _{x\in \mathcal K }\left\Vert{x}\right\Vert$ is a finite positive number) then if we set

$$\begin{aligned} c_{1}=(1+T) \mathbf \max (K_{1},K_{2}) \end{aligned}$$

(130)

we can write that

$$\begin{aligned} \left|{D_{m,,X,u}^{s,x}}\right|\le 2\exp {c_{1}}\int _{\varOmega } \mathbf 1 _{\varOmega _{m}^{c}}(\omega )\exp \left\{ \frac{c_{1}}{1+T} \int _{0}^{T}Q^{x}(t)^{2}+Q^{x}(t)^{2}\right\} \text{ d} \mathbf P (\omega )\nonumber \\ \end{aligned}$$

(131)

Now we observe that

$$\begin{aligned} \mathbf P (\varOmega _{m}^{c})\le \mathbf P ({\max _{0\le t\le T}}\left|{Q_{t}}\right|>\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})) \end{aligned}$$

(132)

and $\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})\rightarrow \infty$ as $m\rightarrow \infty .$ To complete the proof we need to show that $\mathbf P (\varOmega _{m}^{c})\rightarrow 0$ therefore we note that $\left|{Q^{x}(t)}\right|\le \left\Vert{x}\right\Vert+KT+K\int _{0}^{t}\left|{Q^{x}(\tau )}\right|\text{ d}\tau +\sigma \max$ $_{0\le t\le T}\left|{ \mathbf B _{t}}\right|$ and the Gronwall’s Inequality implies

$$\begin{aligned} \left|{Q^{x}(t)}\right|&\le \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|{ \mathbf B _{t}}\right|\\&+K\int _{s}^{t}\exp {K(t-s)}\left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right) \\&\le \left( 1+KT\exp {KT}\right) \left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right) . \end{aligned}$$

The last estimate implies that

$$\begin{aligned} \mathbf P (\max _{0\le t\le T}\left|{Q_{t}}\right|>\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})) \end{aligned}$$

(133)

is smaller than

$$\begin{aligned} \mathbf P \left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|>\frac{\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})}{1+KT\exp {KT}}\right) \end{aligned}$$

(134)

therefore we can conclude that

$$\begin{aligned} \mathbf P (\varOmega _{m}^{c})< \mathbf P \left( \max _{0\le t\le T}\left|\mathbf B _{t}\right|>\frac{\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})}{\sigma \left( 1+KT\exp {KT}\right) }-\frac{1}{ \sigma }\left( \left\Vert{x}\right\Vert+KT\right) \right)\qquad \quad \end{aligned}$$

(135)

and, since

$$\begin{aligned} \lim _{m\rightarrow \infty }\frac{\min (B_{f,m},B_{\ell ,m},B_{G_{1},m})}{\sigma \left( 1+KT\exp {KT}\right) }=\infty \end{aligned}$$

(136)

and

$$\begin{aligned} \max _{x\in \mathcal K }\left\Vert{x}\right\Vert<\infty \end{aligned}$$

(137)

with $\mathcal K$ $\subset \mathbf R ^{n}$ compact, we can claim that this implies $\mathbf P (\varOmega _{m}^{c})\rightarrow 0$ uniformly$.$ We also observe that the following inequality

$$\begin{aligned} \int _{0}^{T}Q^{x}(t)^{2}+Q^{x}(T)^{2}&\le \int _{0}^{T}\left( 1+KT\exp {KT}\right)^{2}\left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right)^{2}\text{ d}\tau \\&+\left( 1+KT\exp {KT}\right)^{2}\left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right) ^{2} \\&\le \left( 1+T\right) \left( 1+KT\exp {KT}\right)^{2} \\&\times \left( \left\Vert{x}\right\Vert+KT+\sigma \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right)^{2} \\&\le 4\left( 1+T\right) \left( 1+KT\exp {KT}\right)^{2} \\&\times \left(\left\Vert{x}\right\Vert^{2}+K^{2}T^{2}+\sigma ^{2}\left( \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right) ^{2}\right) , \end{aligned}$$

and the Dominated Convergence Theorem and the Doob’s Maximal Inequality imply that

$$\begin{aligned}&\lim _{m\rightarrow \infty }2\exp {c_{1}}\int _{\varOmega } \mathbf 1 _{\varOmega _{m}^{c}}(\omega )\exp \left\{ 4c_{1}\left( 1+KT\exp {KT}\right)^{2}\right.\nonumber \\&\quad \times \left.\left( \left\Vert{x}\right\Vert^{2}+K^{2}T^{2}+\sigma \left( \max _{0\le t\le T}\left|\mathbf B _{t}\right|\right)^{2}\right) \right\} d \mathbf P (\omega )=0,\qquad \end{aligned}$$

(138)

therefore we can conclude that, uniformly for $x\in \mathcal K ,$

$$\begin{aligned} \lim \limits _{m\rightarrow \infty }2\exp {c_{1}}\int _{\varOmega } \mathbf 1 _{\varOmega _{m}^{c}}(\omega )\exp \left\{ \frac{c_{1}}{1+T} \int _{0}^{T}Q^{x}(t)^{2}+Q^{x}(T)^{2}\right\} \text{ d} \mathbf P (\omega )=0\qquad \quad \end{aligned}$$

(139)

and this completes the proof.

Theorem 10.

On compact sets we have $V_{m}\rightarrow V$ and $\varPsi _{m}\rightarrow \varPsi$ uniformly and the value function $V$ satisfies the following problem

$$\begin{aligned} 0=\frac{\partial V}{\partial t}+\frac{1}{2}\sigma ^{2}\Delta V+\frac{1}{2} \sigma ^{2}\left|\nabla V\right|^{2}+\min _{u\in \mathcal U } \left[ \left\langle f(t,x,v),\nabla V\right\rangle +\ell (t,x,v)\right]\qquad \quad \end{aligned}$$

(140)

Proof

If we denote $\log \mathcal J _{m}$by $\mathcal I _{m}$ and we remind the reader that we have just proved that

$$\begin{aligned} \lim _{m\rightarrow \infty }\left( \sup _{u,x} \left|\mathbf E _{s,x}\exp \left\{ \varphi _{X^{m},x,u}(s,\omega )\right\} - \mathbf E _{s,x}\exp \left\{ \varphi _{X,x,u}(s,\omega )\right\} \right|\right) =0\qquad \quad \end{aligned}$$

(141)

then we can claim that there is uniform convergence on compact sets for the following sequences

$$\begin{aligned} \mathcal J _{m}\rightarrow \mathcal J \end{aligned}$$

(142)

and

$$\begin{aligned} \mathcal I _{m}\rightarrow \mathcal I \end{aligned}$$

(143)

To prove that we also have uniform convergence on compact sets for the following sequences

$$\begin{aligned} V_{m}\rightarrow V \end{aligned}$$

(144)

and

$$\begin{aligned} \varPsi _{m}\rightarrow \varPsi \end{aligned}$$

(145)

it is enough to observe that if we fix a $\epsilon >0$ and we fix a compact set $\mathcal K$ then, if $m$ is big enough, we have $\left|\mathcal I _{m}- \mathcal I \right|<\epsilon$ therefore we can find a control $u$ such that

$$\begin{aligned} V_{m}(s,x)-V(s,x)< \mathcal I _{m}(s,x,u)- \mathcal I (s,x,u)+\epsilon \end{aligned}$$

(146)

as a consequence we have, for any $x\in K,$ the following

$$\begin{aligned} V_{m}(s,x)-V(s,x)<2\epsilon \end{aligned}$$

(147)

and this implies that $V_{m}\rightarrow V$ uniformly on compact sets. In the same way, we can prove that $\varPsi _{m}\rightarrow \varPsi$ uniformly on compact sets.

Since we have

$$\begin{aligned} 0&= \frac{\partial V_{m}}{\partial t}+\frac{1}{2}\sigma ^{2}\Delta V_{m}+\frac{1}{2}\sigma ^{2}\left|\nabla V_{m }\right|^{2} \\&+\min _{u\in \mathcal U } \left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ f(t,x,v),\nabla V_{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (t,x,v)\right] \end{aligned}$$

and

$$\begin{aligned} 0=\frac{\partial V}{\partial t}+\frac{1}{2}\sigma ^{2}\Delta V+\frac{1}{2} \sigma ^{2}\left|\nabla V\right|^{2}+\min _{u\in \mathcal U } \left[ \left\langle f(t,x,v),\nabla V\right\rangle +\ell (t,x,v)\right]\qquad \end{aligned}$$

(148)

if we set

$$\begin{aligned} D_{m}=\left[ \left\langle f(t,x,v),\nabla V\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \ell (t,x,v)\right] \end{aligned}$$

(149)

and

$$\begin{aligned} D=\left[ \left\langle f(t,x,v),\nabla V\right\rangle +\ell (t,x,v)\right] \end{aligned}$$

(150)

then we can claim $D_{m}\rightarrow D$ uniformly on any compact set $\mathcal K .$ Therefore, by Lemma 2.6.2 in Fleming and Soner 1992, we can claim that $V$ is the viscosity solution and since we can apply the local existence theorems to the equation with viscosity data we can conclude that the solution of the equation is actually a classical solution. Clearly, the result goes for the function $\varPsi$ since this is just a differentiable transform of the function $V.$

We are now in the position to prove the main result. In fact, we have the following

Theorem 11.

Given the model $\mathcal (PA)$ described at the beginning, if there exists an open set $\mathcal O$ such that $\mathbf X _{t}$ $\in \mathcal O$ and for all $y\in \mathcal O$ $\sigma$ is not singular with the inverse matrix satisfying the following condition

$$\begin{aligned} \exists \mathcal G \in \mathcal C ^{2}( \mathcal O , \mathbf R ^{n}) \qquad \mathcal G _{y}(y)=\sigma ^{-1} \quad \forall y\in \mathcal O \end{aligned}$$

(151)

and $F\circ \mathcal G ^{-1},$ $\widetilde{ \mathbf c }_{f,c,\sigma }(t, \mathcal G ^{-1},u_{t})$ and the drift term of $\mathcal G ( \mathbf X _{t})$ satisfy the assumptions ($\mathbf F$), ($\mathbf C$), ($\mathbf G$) and ($\mathbf H$) then the solution of the m-truncated problem $\mathcal (PA) _{m}$ converges to the solution of $\mathcal (PA)$.

Proof

If we start with

$$\begin{aligned} \ \text{ d} \mathbf X _{t}=\psi ( \mathbf X _{t},u_{t})\text{ d}t+\sigma ( \mathbf X _{t})\text{ d} \mathbf B _{t} \end{aligned}$$

(152)

then, by assumption, we can find $\mathcal G \in \mathcal C ^{2}( \mathcal O , \mathbf R ^{n})$ such that $D \mathcal G (y)=\sigma ^{-1}\quad \forall y\in \mathcal O$. Hence, if we apply the Ito Formula to

$$\begin{aligned} \mathbf Y _{t}=\widetilde{\sigma } \mathcal G ( \mathbf X _{t}) \end{aligned}$$

(153)

($\widetilde{\sigma }$ is an arbitrary positive constant) we obtain

$$\begin{aligned} \text{ d}Y_{k,t}&= \widetilde{\sigma }\frac{\partial \mathcal G _{k}}{\partial t}(t, \mathbf X _{t})+\sum _{j}\widetilde{\sigma }\frac{\partial \mathcal G _{k}}{ \partial x_{j}}(t, \mathbf X _{t})\text{ d}X_{j,t}+\frac{1}{2}\sum _{j,i}\widetilde{ \sigma }\frac{\partial \mathcal G _{k}}{\partial x_{j}\partial x_{i}}(t, \mathbf X _{t})\text{ d}X_{i,t}\text{ d}X_{j,t} \\&+\sum _{j}\widetilde{\sigma }\frac{\partial \mathcal G _{k}}{\partial x_{j}} (t, \mathbf X _{t})\text{ d}X_{j,t}+\frac{1}{2}\sum _{j,i}\widetilde{\sigma }^{2}\frac{ \partial \mathcal G _{k}}{\partial x_{j}\partial x_{i}}(t, \mathbf X _{t})\text{ d}X_{i,t}\text{ d}X_{j,t} \end{aligned}$$

where $\mathbf Y _{t}=\left( Y_{1,t},\ldots ,Y_{n,t}\right)$ and $\mathbf X _{t}=\left( X_{1,t},\ldots ,X_{n,t}\right) .$ Therefore, if we set $\widetilde{ \sigma } \mathcal G= \widetilde{ \mathcal G }$, and we observe that for any $t\in [0,T]$ we have $\mathbf X _{t}:\varOmega \rightarrow \mathcal O$ and $\widetilde{ \mathcal G }^{-1}: \mathcal O \rightarrow \mathbf R ^{n},$ $\widetilde{ \mathbf c }:$ $[0,T]\times \mathbf R ^{n}\times U\rightarrow \mathbf R$ and $F_{1}: \mathbf R ^{n}\rightarrow \mathbf R$ then, by using $\widetilde{ \mathcal G }^{-1},$ we can rewrite the problem $\mathcal (PA)$ and we obtain

$$\begin{aligned} \left\{ \begin{array}{l} \max _{u\in \mathcal U } \mathbf E \left[ -\exp -R\left\{ F_{1}\circ \widetilde{ \mathcal G }^{-1}( \mathbf Y _{T})+\int _{0}^{T}\widetilde{ \mathbf c }(t,\widetilde{ \mathcal G }^{-1}( \mathbf Y _{t}),u_{t})\text{ d}t\right\} \right] \\ \\ \text{ d} \mathbf Y _{t} = \widetilde{\psi }( \mathbf Y _{t},u_{t})\text{ d}t+\widetilde{ \sigma }\text{ d} \mathbf B _{t} \\ \mathbf Y _{0} = \widetilde{ \mathcal G }(x) \end{array} \right. \end{aligned}$$

(154)

and since this problem is equivalent to solving

$$\begin{aligned} \left\{ \begin{array}{l} \min _{u\in \mathcal U } \log \mathbf E \left[ \exp -R\left\{ F_{1}\circ \widetilde{ \mathcal G }^{-1}( \mathbf Y _{T})+\int _{0}^{T}\widetilde{ \mathbf c }(t,\widetilde{ \mathcal G }^{-1}( \mathbf Y _{t}),u_{t})\text{ d}t\right\} \right] \\ \\ \text{ d} \mathbf Y _{t} = \widetilde{\psi }( \mathbf Y _{t},u_{t})\text{ d}t+\widetilde{ \sigma }\text{ d} \mathbf B _{t} \\ \mathbf Y _{0} = \widetilde{ \mathcal G }(x) \end{array} \right. \end{aligned}$$

(155)

then the result follows from what we proved in this section $($with $s=0)$ with

$$\begin{aligned} \ell (t, \mathbf Y _{t},u_{t})&= -R\widetilde{ \mathbf c }(t,\widetilde{ \mathcal G }^{-1}( \mathbf Y _{t}),u_{t}) \\&= Rc(t,\widetilde{ \mathcal G }^{-1} \mathbf Y _{t},u_{t})-Ra(t,\widetilde{ \mathcal G }^{-1} \mathbf Y _{t}) \\&+\frac{Rr}{2}\left\Vert\nabla _{u}c(t,\widetilde{ \mathcal G }^{-1} \mathbf Y _{t},u_{t})D_{u}^{-1}f(t,\widetilde{ \mathcal G }^{-1} \mathbf Y _{t},u_{t})\sigma (t,\widetilde{ \mathcal G }^{-1} \mathbf Y _{t})\right\Vert^{2} \\ G_{1}( \mathbf Y _{T})&= -RF_{1}\circ \widetilde{ \mathcal G }^{-1}( \mathbf Y _{T}) \\&= -RF_{1}\left( \widetilde{ \mathcal G }^{-1}( \mathbf Y _{T})\right) . \end{aligned}$$

In fact, in Lemma 22, we proved that there exists a sequence of set $\{A_{m}\}$ such that $\mathbf P (A_{m})\rightarrow 1$ and for every $(\omega ,t)\in A_{m}\times [0,T]$ $\mathbf Y ^{m}(\omega ,t)= \mathbf Y (\omega ,t)$. Actually, it is enough to set $A_{m}=\varOmega _{m}$ where $\varOmega _{m}$ is defined exactly as in the cited Lemma i.e.⁶

$$\begin{aligned} \varOmega _{m}=\left\{ \omega \in \varOmega \left|\left|{Y_{t}(\omega )}\right|\le \min (B_{\widetilde{\psi },m}, B_{\widetilde{ \mathbf c }(\cdot ,\widetilde{ \mathcal G }^{-1}(\cdot ), \cdot ),m}, B_{F_{1}\circ \widetilde{ \mathcal G }^{-1},m})\forall t\in [0,T]\right. \right\} .\nonumber \\ \end{aligned}$$

(156)

Moreover, we observe that since in the last Theorem, we proved that $\mathbf H 1$ holds and since we know that if $g_{m}(x,t)$ is the Borel function (see Fleming and Rishel 1975) so defined

$$\begin{aligned} g_{m}(y,s)=\arg {\min }_{u\in U}\left[ \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ \widetilde{\psi }(s,y,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ )\widetilde{ \mathbf c }(s,\widetilde{ \mathcal G }^{-1}(y),u)\varPsi _{m}\right]\nonumber \\ \end{aligned}$$

(157)

then we have $u_{t}(t,\widetilde{Y}_{t})=g(t,\widetilde{Y}_{t})$ where $\widetilde{Y}_{t}$ is the unique strong solution of the equation $d \widetilde{Y}_{t}=\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \widetilde{\psi }(t, \widetilde{Y}_{t},g(t,\widetilde{Y}_{t}))\text{ d}t+\widetilde{\sigma }d \mathbf B _{t}$. If we define

$$\begin{aligned} \left\langle \chi _{_{B( \mathbf R ^{n})}}^{m}\circ \widetilde{\psi } (s,y,u),\nabla \varPsi _{m}\right\rangle +\chi _{_{B( \mathbf R ^{n})}}^{m}\circ \widetilde{ \mathbf c }(s,\widetilde{ \mathcal G }^{-1}(y),u)\varPsi _{m}=F_{m}(s,y,u) \end{aligned}$$

(158)

and

$$\begin{aligned} \left\langle \widetilde{\psi }(s,y,u),\nabla \varPsi _{m}\right\rangle + \widetilde{ \mathbf c }(s,\widetilde{ \mathcal G }^{-1}(y),u)\varPsi _{m}=F(s,y,u) \end{aligned}$$

(159)

then, for any $K>0$ and $\forall t\in [0,T]$, we have

$$\begin{aligned} \lim _{m\rightarrow \infty }\sup _{\left\Vert{y}\right\Vert_{ \mathbf R ^{n}}\le K}\left|{F_{m}(s,y,u)-F(s,y,u)}\right|=0 \end{aligned}$$

(160)

and this fact implies the convergence of controls.

Appendix II

Proof

(Theorem 5) We are going to present the proof for the diagonal case in detail; the general case is the same, but notationally very messy. Therefore, by assumption, if we define

$$\begin{aligned} \widehat{\sigma }\widehat{\sigma }^{\mathtt T }=\left[ a_{ij}\right] _{i,j=1,\ldots ,n} \end{aligned}$$

(161)

we are going to assume $a_{ij}=0$ if $i\ne j$. Since we are only going to work on a discrete space or lattice, we shall use the following standard approximations for the derivatives

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} \varphi _{t} \mapsto&\frac{1}{\delta }\left[ \varphi (x,t+\delta )-\varphi (x,t)\right] \\ \\ \varphi _{x_{j}}\quad \text{ if} \ b_{j}\ge 0 \mapsto&\frac{1}{h} \left[ \varphi (x+h \mathbf e _{j},t+\delta )-\varphi (x,t+\delta )\right] \\ \\ \varphi _{x_{i}}\quad \text{ if} b_{j}<0 \mapsto&\frac{1}{h}\left[ \varphi (x,t+\delta )-\varphi (x-h \mathbf e _{j},t+\delta )\right] \\ \\ \varphi _{x_{i}x_{i}}\quad {\mapsto }&\frac{1}{h^{2}}\left[ \varphi (x+h \mathbf e _{i},t+\delta )+\varphi (x-h \mathbf e _{i},t+\delta )-2\varphi (x)\right] \end{array} \right. \end{aligned}$$

(162)

where $\mathbf e _{j}=(0,\ldots ,1,\ldots ,0)\in \mathbf R ^{n}$ for $j=1,\ldots ,n$ (we do not need to compute $\varphi _{x_{i}x_{j}}$ if $i\ne j$ since $a_{ij}=0$ if $i\ne j).$ Therefore, if we define

$$\begin{aligned} W(x,t) \mathcal = \mathbf E _{x}\left[ -\exp -R\left\{ \widehat{F}( \mathbf X _{T})+\int _{t}^{T}\widehat{ \mathbf c }(t, \mathbf X _{t},u_{t})\text{ d}t\right\} \right] \end{aligned}$$

(163)

we have

$$\begin{aligned} \frac{\partial W}{\partial t}+\sum _{j=1}^{n}\widehat{\psi }(x,u)\frac{ \partial W}{\partial x_{j}}(x,t,u)+\frac{1}{2}\sum _{j=1}^{n}a_{j}(x,u)\frac{ \partial ^{2}W}{\partial x_{j}^{2}}(x,t,u)-\widehat{ \mathbf c }W(x,t,u) =0\nonumber \\ \end{aligned}$$

(164)

and, after we discretize the partial derivatives, if we define $\widehat{ \psi }^{+}= \mathbf \max (\widehat{\psi },0)$ and $\widehat{\psi }^{-}= \mathbf \max (-\widehat{\psi },0),$ we obtain the following

$$\begin{aligned} 0&= \frac{W^{h,\delta }(x,t+\delta ,u)-W^{h,\delta }(x,t,u)}{\delta } \\&+\sum _{j=1}^{n}\widehat{\psi }^{+}(x,u)\frac{\left( W^{h,\delta }(x+h \mathbf e _{j},t+\delta ,u)-W^{h,\delta }(x,t+\delta ,u)\right) }{h} \\&-\sum _{j=1}^{n}\widehat{\psi }^{-}(x,u)\frac{\left( W^{h,\delta }(x,t+\delta ,u)-W^{h,\delta }(x-h \mathbf e _{j},t+\delta ,u)\right) }{h} \\&+\frac{1}{2}\sum _{j=1}^{n}a_{jj}(x,u)\frac{W^{h,\delta }(x+h \mathbf e _{j},t+\delta ,u)+W^{h,\delta }(x-h \mathbf e _{j},t+\delta ,u)-2W^{h,\delta }(x,t+\delta ,u)}{2h^{2}} \\&-R\widehat{ \mathbf c }(x,n\delta ,u)W^{h,\delta }(x,n\delta ,u\left( x,n\delta \right) ) \end{aligned}$$

with the boundary condition on $W^{h,\delta }$. Hence, we have

$$\begin{aligned}&W^{h,\delta }(x,n\delta ,u) \\&\quad =\sum _{j=1}^{n}\left[ \frac{\delta }{2h^{2}}a_{jj}(x)+\widehat{\psi } ^{+}((x,u(x,n\delta ))\frac{\delta }{h}\right] W^{h,\delta }(x+h,n\delta +\delta ,u) \\&\qquad +\sum _{j=1}^{n}\left[ \frac{\delta }{2h^{2}}a_{jj}(x)+\widehat{\psi } ^{-}(x,u(x,n\delta ))\frac{\delta }{h}\right] W^{h,\delta }(x-h,n\delta +\delta ,u) \\&\qquad +W^{h,\delta }(x,n\delta +\delta ,u)\left[ 1-\sum _{j=1}^{n}\frac{\delta }{ h^{2}}a_{jj}(x)-\sum _{j=1}^{n}\left|\psi (x,u(x,n\delta ))\right|\frac{\delta }{h}\right] \\&\qquad -Rc(x,n\delta ,u)W^{h,\delta }(x,n\delta ,u\left( x,n\delta \right) )\delta \end{aligned}$$

it follows that, if we define the quantities

$$\begin{aligned} \widehat{p}^{h,\delta }(x,x+h,\left|{ u(x,n\delta )}\right. )&= \frac{ \delta }{2h^{2}}a_{jj}(x)+\widehat{\psi }^{+}((x,u(x,n\delta ))\frac{\delta }{h} \\ \widehat{p}^{h,\delta }(x,x-h,\left|{u(x,n\delta )}\right. )&= \frac{ \delta }{2h^{2}}a_{jj}(x)+\widehat{\psi }^{-}(x,u(x,n\delta ))\frac{\delta }{ h} \\ \widehat{p}^{h,\delta }(x,x,\left|{u(x,n\delta )}\right. )&= 1-\frac{ \delta }{h^{2}}a_{jj}(x)-\left|\widehat{\psi }(x,u(x,n\delta ))\right|\frac{\delta }{h} \end{aligned}$$

we can write

$$\begin{aligned}&W^{h,\delta }(x,n\delta ,u)(1+R\widehat{ \mathbf c }(x,n\delta ,u)\delta ) \\&\quad =\sum _{y}\widehat{p}^{h,\delta }(x,y,\left|{u(x,n\delta )}\right. )W^{h,\delta }(x,n\delta +\delta ,u) \\&\qquad +\widehat{p}^{h,\delta }(x;x\left|{u(x,n\delta )}\right. )W^{h,\delta }(x,n\delta +\delta ,u) \end{aligned}$$

and most importantly, given our boundedness assumptions, we have, for each element in the grid, positive numbers with

$$\begin{aligned} \sum _{y}\widehat{p}^{h,\delta }(x,n\delta ;y,n\delta \left|\gamma \right. )+\widehat{p}^{h,\delta }(x,n\delta ;x,n\delta +\delta \left|\gamma \right. )=1. \end{aligned}$$

(165)

Therefore, we can write the discretized optimal control as

$$\begin{aligned} \begin{array}{l} \max \limits _{u\in \mathcal U } \mathbf E _{x,t}^{u}\left[ -\exp {-R\left\{ -\sum \widehat{ \mathbf c }(x_{x,n}^{u,h,\delta },u(x_{x,n}^{u,h,\delta }))\Delta t^{h}(x_{x,n}^{u,h,\delta },u(x_{x,n}^{u,h,\delta }))+\widehat{F} (x_{x,n}^{u,h,\delta })\right\} }\right] \end{array}\nonumber \\ \end{aligned}$$

(166)

and we obtain a solution since we are optimizing a continuous function on a compact space. Let $u:G_{h,\delta }\rightarrow \mathbf R$ be the solution. So, if we consider the probabilities above as transition probabilities, we can write the Markov Chain $\left\{ \mathbf X ^{h,\delta }\right\}$ on the grid $G_{h,\delta }$. For $\left\{ \mathbf X ^{h,\delta }\right\}$ we have

$$\begin{aligned} \left\{ \begin{array}{l} \mathbf E _{x,n}^{u,h,\delta }\Delta \mathbf X ^{h,\delta }=\widehat{\psi } \delta +o(\delta ) \\ \\ cov_{x,n}^{u,h,\delta }\Delta \mathbf X ^{h,\delta }=\widehat{\sigma } \widehat{\sigma }^{\mathtt T }(x)\delta +o(\delta ) \end{array} \right. \end{aligned}$$

(167)

and this gives the claimed local consistency.

Remark 7

It is worth noticing that the equation we wrote above could actually allow us to compute the value $W^{h,\delta }(x,n\delta ,u)$ by using induction. In fact, on the right side, we have only values of the form $W^{h,\delta }(x,n\delta +\delta ,u)$ and this is due to the fact that the spatial derivatives are computed at time $n\delta +\delta$ in the approximation scheme.

Remark 8

It is worth noticing that in the scalar case, we have the following standard approximations for the derivatives

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _{t} \mapsto \varphi (x,t+\delta )-\varphi (x,t)/\delta \\ \\ \varphi _{x} \text{ if} b\ge 0 \mapsto \varphi (x+h,t+\delta )-\varphi (x,t+\delta )/h \\ \\ \varphi _{x} \text{ it} b<0 \mapsto \varphi (x,t+\delta )-\varphi (x-h,t+\delta )/h \\ \\ \varphi _{xx} \mapsto \left\{ \varphi (x+h,t+\delta )+\varphi (x-h,t+\delta )-2\varphi (x,t+\delta )\right\} /h^{2} \end{array} \right. \end{aligned}$$

(168)

and $n=1$ gives us

$$\begin{aligned} \frac{\partial W}{\partial t}+b(x,u)\frac{\partial W}{\partial x}(x,t)+\frac{ 1}{2}\sigma ^{2}(x)\frac{\partial ^{2}W}{\partial x^{2}}(x,t)-\widetilde{ \mathbf c }W(x,t) =0 \end{aligned}$$

(169)

and, after we discretize the partial derivatives, we obtain the following

$$\begin{aligned} W^{h,\delta }(x,n\delta ,u)&= \left[ \frac{\delta }{2h^{2}}\sigma ^{2}(x)+b^{+}(x,u(x,n\delta ))\frac{ \delta }{h}\right] W^{h,\delta }(x+h,n\delta +\delta ,u) \\&+\left[ \frac{\delta }{2h^{2}}\sigma ^{2}(x)+b^{-}(x,u(x,n\delta ))\frac{ \delta }{h}\right] W^{h,\delta }(x-h,n\delta +\delta ,u) \\&+W^{h,\delta }(x,n\delta +\delta ,u)\left[ 1-\frac{\delta }{h^{2}}\sigma ^{2}(x)+\left|{b(x,u(x,n\delta ))}\right|\frac{\delta }{h}\right] \\&-R\widetilde{ \mathbf c }(x,n\delta ,u)W^{h,\delta }(x,n\delta ,u)\delta \end{aligned}$$

So we can write

$$\begin{aligned}&W^{h,\delta }(x,n\delta ,u)(1+\widetilde{ \mathbf c }(x,n\delta ,u)\delta ) \\&\quad =\sum _{y}\widehat{p}^{h,\delta }(x,n\delta ;y,n\delta \left|{u(x,n\delta )}\right. )W^{h,\delta }(x,n\delta +\delta ,u) \\&\qquad +\widehat{p}^{h,\delta }(x,n\delta ;x,n\delta +\delta \left|{u(x,n\delta )}\right. )W^{h,\delta }(x,n\delta +\delta ,u) \end{aligned}$$

Remark 9

About the general case $\sigma \sigma ^{\mathtt T }=\left[ a_{ij}\right]_{i,j=1,\ldots ,n}$ we are going to use the following discretizations

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _{t} \mapsto \varphi (x,t+\delta )-\varphi (x,t)/\delta \\ \\ \varphi _{x_{j}} \quad \text{ if} \ b_{j}\ge 0 {\mapsto } \varphi (x+h \mathbf e _{j},t+\delta )-\varphi (x,t+\delta )/h \\ \\ \varphi _{x_{i}} \quad \text{ if} \ b_{j}<0 {\mapsto } \varphi (x,t+\delta )-\varphi (x-h \mathbf e _{j},t+\delta )/h \end{array} \right. \end{aligned}$$

(170)

and

$$\begin{aligned} \left\{ \begin{array}{ccc} \varphi _{x_{i}x_{j}}&\text{ if} \ a_{ij}\ge 0 \mapsto&\left( \begin{array}{l} 2\varphi (x)+\varphi (x+h\sum \limits _{k=i,j} \mathbf e _{k},t+\delta ) \\ -\varphi (x-h\sum \limits _{i,j} \mathbf e _{k},t+\delta ) \\ -[\varphi (x+h \mathbf e _{i},t+\delta )+\varphi (x-h \mathbf e _{i},t+\delta )]/2h^{2} \\ -[\varphi (x+h \mathbf e _{j},t+\delta )+\varphi (x-h \mathbf e _{j},t+\delta )/2h^{2} \end{array} \right) \cdot \frac{1}{2h^{2}} \\&&\\ \varphi _{x_{i}x_{j}}&\text{ if} \ a_{ij}<0 \mapsto&\left( \begin{array}{l} -\left[ \begin{array}{l} 2\varphi (x)+\varphi (x+h \mathbf e _{i}-h \mathbf e _{j},t+\delta ) \\ -\varphi (x-h \mathbf e _{i}+h \mathbf e _{j},t+\delta ) \end{array} \right] \\ +[\varphi (x+h \mathbf e _{i},t+\delta )+\varphi (x-h \mathbf e _{i},t+\delta )] \\ +[\varphi (x+h \mathbf e _{j},t+\delta )+\varphi (x-h \mathbf e _{j},t+\delta ) \end{array} \right) \cdot \frac{1}{2h^{2}} \\&&\end{array} \right. \end{aligned}$$

(171)

We now present the following

Proof

(Theorem 14) To simplify the notation, we define

$$\begin{aligned} F_{1}( \mathbf X _{T})+\int _{0}^{T}\widetilde{c}(t, \mathbf X _{t},u_{t})\text{ d}t=H( \mathbf X ,u) \end{aligned}$$

(172)

and

$$\begin{aligned} \chi _{_{B( \mathbf R ^{n})}}^{n}\circ F_{1}( \mathbf X _{T})+\int _{0}^{T}\chi _{_{B( \mathbf R ^{n})}}^{n}\circ \widetilde{c}(t, \mathbf X _{t},u_{t})\text{ d}t=H_{n}( \mathbf X ,u) \end{aligned}$$

(173)

we can write

$$\begin{aligned}&\left|\mathbf E \exp {-RH( \mathbf X ,u)}- \mathbf E \exp {-RH( \mathbf X ^{n},u^{n})}\right|\\&\quad \le \left|\mathbf E \exp {-RH( \mathbf X ,u)}- \mathbf E \exp {-RH_{n}( \mathbf X ^{n},u^{n})}\right|\\&\qquad +\left|\mathbf E \exp {-RH_{n}( \mathbf X ^{n},u^{n})}- \mathbf E \exp {-RH( \mathbf X ^{n},u^{n})}\right|\end{aligned}$$

Given our assumptions A1, A2 and A3, we need only to study the difference $\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c( \mathbf X _{t}^{n},u_{t}^{n})-c( \mathbf X _{t},u_{t}).$ If we fix an $\epsilon >0$ and we observe that for every $n$ we have the solution $( \mathbf X _{n,t},u_{n,t})$ of the $n$-truncated problem, then we can choose an $n$ big enough such that there is a set $A_{n}$ with $\mathbf P (A_{n}^{c})\le \epsilon$ and for every $(\omega ,t)\in A_{n}\times [0,T]$ $\ \mathbf X _{n,t}(\omega ,t)= \mathbf X (\omega ,t).$ We also observe that

$$\begin{aligned} \left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})-c(t, \mathbf X _{t},u_{t})\right|&= \left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t},u_{t}^{n})-c(t, \mathbf X _{t},u_{t})\right|\mathbf 1 _{A_{n}} \\&+\left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})-c(t, \mathbf X _{t},u_{t})\right|\mathbf 1 _{A_{n}^{c}} \end{aligned}$$

and

$$\begin{aligned} \left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})-c(t, \mathbf X _{t},u_{t})\right|\mathbf 1 _{A_{n}^{c}}&\le \left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})\right|\mathbf 1 _{A_{n}^{c}} \\&+\left|{c(t, \mathbf X _{t},u_{t})}\right|\mathbf 1 _{A_{n}^{c}} \\&\le 2K\left( 1+\left|\mathbf X _{t}^{n}\right|^{2}+\left|\mathbf X _{t}\right|^{2}\right) \mathbf 1 _{A_{n}^{c}} \\&\le 2K\left( 1+2\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}+3\left|\mathbf X _{t}\right|^{2}\right) \mathbf 1 _{A_{n}^{c}} \end{aligned}$$

if we define

$$\begin{aligned} Y_{n}=\sup _{[0,T]}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2} \end{aligned}$$

(174)

we know that (see Gard 1988)

$$\begin{aligned} Y_{n}\rightarrow 0 \end{aligned}$$

(175)

so we can conclude $Y_{n}\rightarrow 0$ in probability and $\mathbf 1 _{A_{n}^{c}}\int _{0}^{T}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}\le B$ a.e. on $\varOmega$, we also notice that there exists a constant $L>0$ such that, if we set

$$\begin{aligned} \left|\mathbf E \exp {\int _{0}^{T}\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})\text{ d}t}- \mathbf E \exp {\int _{0}^{T}c(t, \mathbf X _{t},u_{t})\text{ d}t}\right|= \mathcal I , \end{aligned}$$

(176)

we have, by A3,

$$\begin{aligned} \mathcal I&\le L \mathbf E \left|\exp {\int _{0}^{T}\left|\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})-c(t, \mathbf X _{t}^{n},u_{t}^{n})\right|\mathbf 1 _{A_{n}^{c}}\text{ d}t}-1\right|^{2} \\&\le L \mathbf E \left|\exp {\int _{0}^{T}2K\left( 1+2\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}+3\left|\mathbf X _{t}\right|^{2}\right) \mathbf 1 _{A_{n}^{c}}\text{ d}t}-1\right|^{2} \\&\le L \mathbf E \left|\exp { \mathbf 1 _{A_{n}^{c}}(T+4K\int _{0}^{T}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}\text{ d}t)+6K \mathbf 1 _{A_{n}^{c}}\int _{0}^{T}\left|\mathbf X _{t}\right|^{2}\text{ d}t}-1\right|^{2} \end{aligned}$$

We also notice that for every $\alpha \in \mathbf R$ we have, by Fubini ’s Theorem,

$$\begin{aligned} \mathbf E \exp {\alpha \mathbf 1 _{A_{n}^{c}}\int _{0}^{T}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}\text{ d}t}&= \mathbf E \sum _{\ell =0}^{\infty }\frac{1}{\ell !}\alpha ^{\ell }\left( \mathbf 1 _{A_{n}^{c}}\int _{0}^{T}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}\text{ d}t\right)^{\ell } \\&= \sum _{\ell =0}^{\infty }\frac{1}{\ell !}\alpha ^{\ell } \mathbf E \left( \mathbf 1 _{A_{n}^{c}}\int _{0}^{T}\left|\mathbf X _{t}^{n}- \mathbf X _{t}\right|^{2}\text{ d}t\right)^{\ell } \\&\le \sum _{\ell =0}^{\infty }\frac{1}{\ell !}\alpha ^{\ell }B^{\ell } \\&< \infty \end{aligned}$$

and this implies, by the Dominated Convergence Theorem, that

$$\begin{aligned} \left|\mathbf E \exp {\int _{0}^{T}\chi _{_{B( \mathbf R ^{n})}}^{n}\circ c(t, \mathbf X _{t}^{n},u_{t}^{n})\text{ d}t}- \mathbf E \exp {\int _{0}^{T}c(t, \mathbf X _{t},u_{t})\text{ d}t}\right|\rightarrow 0 \end{aligned}$$

(177)

since $\mathbf P (A_{n}^{c})\rightarrow 0$ and $\mathbf P (\varOmega )=1.$ Finally, we notice that for $( \mathbf X _{n}^{h_{n},\delta _{n}},u_{n}^{h_{n},\delta _{n}})$ and $\left( \mathbf X _{t}^{n},u_{t}^{n} \right)$ we can use the weak convergence and therefore we can claim that

$$\begin{aligned} \left|\mathbf E \left[ \exp {-R\left\{ F( \mathbf X _{n}^{h_{n},\delta _{n}})+\int _{0}^{T}c(t, \mathbf X _{n}^{h_{n},\delta _{n}},u_{n}^{h_{n},\delta _{n}})\text{ d}t\right\} }\right] - \mathbf E \exp {H( \mathbf X ^{n},u^{n})}\right|\rightarrow 0\nonumber \\ \end{aligned}$$

(178)

and this completes the proof.

Proof

(Theorem 16) We fix an $\epsilon >0$ and we observe that for every $m$ we have the solution $( \mathbf X _{t}^{m},u_{t}^{m})$ of the $m$-truncated problem. If we choose an $m$ big enough, then there exists an $A_{m}$ such that $\mathbf P (A_{m}^{c})\le \epsilon$ and for every $(\omega ,t)\in A_{m}\times [0,T]$ $\mathbf X _{t}^{m}(\omega ,t)= \mathbf X _{t}(\omega ,t).$ To prove our result, we must analyze the different parts of the sharing rule.

We start by proving that $\int _{0}^{T}c(t, \mathbf X _{t}^{m},u_{t}^{m})\text{ d}t \rightarrow$ $\int _{0}^{T}c(t, \mathbf X _{t},u_{t}^{*})\text{ d}t.$ We observe that we can write

$$\begin{aligned} \left|{c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{m})}\right|&\le \left|{c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))}\right|\\&+\left|{c(t, \mathbf X _{t}^{m},u_{t}^{*}))-c(t, \mathbf X _{t}^{m},u_{t}^{m})}\right|\end{aligned}$$

and

$$\begin{aligned} \left|{c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))}\right|\le (1+\left|\mathbf X _{t}\right|+\left|\mathbf X _{t}^{m})\right|)\left|\mathbf X _{t}- \mathbf X _{t}^{m}\right|\end{aligned}$$

(179)

therefore, by Jensen’s Inequality, we have

$$\begin{aligned} \mathbf E \left|\int _{0}^{T}\left( c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))\right) \text{ d}t\right|^{2}&\le \mathbf E \int _{0}^{T}\left|{c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))}\right|^{2}\text{ d}t \\&\le \mathbf E \int _{0}^{T}((1+\left|\mathbf X _{t}\right|+\left|\mathbf X _{t}^{m})\right|)^{2}\left|\mathbf X _{t}- \mathbf X _{t}^{m}\right|^{2}\text{ d}t \\&\le 4 \mathbf E \int _{0}^{T}(1+\left|\mathbf X _{t}\right|^{2}+\left|\mathbf X _{t}^{m}\right|^{2})\left|\mathbf X _{t}- \mathbf X _{t}^{m}\right|^{2}\text{ d}t \end{aligned}$$

and hence, by Cauchy–Schwarz’s Inequality,

$$\begin{aligned}&\mathbf E \left|\int _{0}^{T}\left( c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))\right) \text{ d}t\right|^{2} \\&\quad \le 4\left( \mathbf E \int _{0}^{T}(1+\left|\mathbf X _{t}\right|^{2}+\left|\mathbf X _{t}^{m}\right|^{2})^{2}\text{ d}t\right) ^{\frac{1}{2}} \left( \mathbf E \int _{0}^{T}\left|\mathbf X _{t}- \mathbf X _{t}^{m}\right|^{4}\text{ d}t\right)^{\frac{1}{2}} \\&\quad \le 16\left( \mathbf E \int _{0}^{T}(1+\left|\mathbf X _{t}\right|^{4}+\left|\mathbf X _{t}^{m}\right|^{4})\text{ d}t\right) ^{\frac{1}{2}} \left( \mathbf E \int _{0}^{T}\left|\mathbf X _{t}- \mathbf X _{t}^{m}\right|^{4}\text{ d}t\right) ^{\frac{1}{2}} \end{aligned}$$

and the last term goes to zero as $m\rightarrow \infty .$ We also notice that, by condition iv in Sect. 3, there exists $K_{1}$ such that

$$\begin{aligned} \left|{c(t, \mathbf X _{t}^{m},u_{t}^{*}))-c(t, \mathbf X _{t}^{m},u_{t}^{m})}\right|&\le \left|{c(t, \mathbf X _{t}^{m},u_{t}^{*}))-\chi _{_{B( \mathbf R ^{n})}}^{m}\circ c(t, \mathbf X _{t}^{m},u_{t}^{m})}\right|\\&+\left|\chi _{_{B( \mathbf R ^{n})}}^{m}\circ c(t, \mathbf X _{t}^{m},u_{t}^{*}))-c(t, \mathbf X _{t}^{m},u_{t}^{m})\right|\\&\le 2K_{1}(1+\left|\mathbf X _{t}^{m}\right|^{2}) \end{aligned}$$

and since there exists a constant $K_{2}$ such that (see Gard 1988 Theorem 3.8)

$$\begin{aligned} \mathbf E \int _{0}^{T}(1+\left|\mathbf X _{t}^{m}\right|^{2})^{2}&= \int _{0}^{T} \mathbf E (1+\left|\mathbf X _{t}^{m}\right|^{2})^{2} \\&\le 2\int _{0}^{T} \mathbf E (1+\left|\mathbf X _{t}^{m}\right|^{4}) \\&\le 2\int _{0}^{T}(1+(1+x)\exp {K_{2}t}) \\&< \infty \end{aligned}$$

therefore, by the Dominated Convergence Theorem, we obtain that

$$\begin{aligned} \mathbf E \int _{0}^{T}\left|{c(t, \mathbf X _{t},u_{t}^{*})-c(t, \mathbf X _{t}^{m},u_{t}^{*}))}\right|^{2}\text{ d}t\rightarrow 0 \text{ as} m\rightarrow \infty \end{aligned}$$

(180)

and this implies the convergence of both terms. Now if we use the fact that $\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) \cdot \sigma$ is bounded and the fact that, for a finite constant $C_{1},$

$$\begin{aligned} \mathbf E \left[ \int _{0}^{T}\left|\mathbf X _{t}^{m}(\omega ,t)- \mathbf X _{t}(\omega ,t)\right|^{2}\text{ d}t\right]&\le \int \limits _{\varOmega }\int _{0}^{T} \mathbf 1 _{A_{m}^{c}}\left|\mathbf X _{t}^{m}(\omega ,t)- \mathbf X _{t}(\omega ,t)\right|^{2}\text{ d}t\text{ d} \mathbf P \\&\le C_{1} \mathbf P (A_{m}^{c})T \end{aligned}$$

it follows that, in the $L^{2}$ sense,

$$\begin{aligned} \int _{0}^{T}\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) ( \mathbf X _{t}^{m},u_{t}^{m})\sigma ( \mathbf X _{t}^{m})\text{ d} \mathbf B _{t}\rightarrow \int _{0}^{T}\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) ( \mathbf X _{t},u_{t}^{*})\sigma ( \mathbf X _{t})\text{ d} \mathbf B _{t}.\nonumber \\ \end{aligned}$$

(181)

If we set

$$\begin{aligned} I =\left\Vert\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) ( \mathbf X _{t}^{m},u_{t}^{m})\sigma ( \mathbf X _{t}^{m})\right\Vert^{2}\text{ d}t \end{aligned}$$

(182)

then there exists a constant $M$ such that

$$\begin{aligned}&\mathbf E \left|\int _{0}^{T}I-\left[ \int _{0}^{T}\left|\nabla _{u}c(t, \mathbf X _{t},u_{t})D_{u}^{-1}f(t, \mathbf X _{t},u_{t})\sigma (t, \mathbf X _{t})\right|^{2}\text{ d}t\right] \right|^{2} \\&\quad \le \mathbf E \left|\int _{0}^{T}\left( I-\left|\nabla _{u}c(t, \mathbf X _{t},u_{t})D_{u}^{-1}f(t, \mathbf X _{t},u_{t})\sigma (t, \mathbf X _{t})\right|^{2}\right) \text{ d}t\right|^{2} \\&\quad \le \mathbf E \left|\int _{0}^{T}\left( I-\left|\nabla _{u}c(t, \mathbf X _{t},u_{t})D_{u}^{-1}f(t, \mathbf X _{t},u_{t})\sigma (t, \mathbf X _{t})\right|^{2}\right) \right|^{2}\text{ d}t \\&\quad \le 2M^{2}T \end{aligned}$$

and this implies that

$$\begin{aligned} \int _{0}^{T}\left\Vert\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{t}^{m},u_{t}^{m})\sigma (t, \mathbf X _{t}^{m})\right\Vert^{2} \end{aligned}$$

(183)

converges, in the $L^{2}$ norm, to

$$\begin{aligned} \int _{0}^{T}\left\Vert\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{t},u_{t}^{*})\sigma (t, \mathbf X _{t})\right\Vert^{2} \end{aligned}$$

(184)

Finally, for any $m$ we have $\left\{ \left( \mathbf X _{m}^{h_{n},\delta _{n}},u_{m}^{h_{n},\delta _{n}}\right) \right\} _{n}$ and, by using Skorokhod Theorem, we can assume that there is convergence with probability one. Therefore, by the boundedness of the functions $\nabla _{u}cD_{u}^{-1}f\sigma$ and $\chi _{_{B( \mathbf R ^{n})}}^{m}\circ c$, we have clearly the following

$$\begin{aligned} \int _{0}^{T}\chi _{_{B( \mathbf R ^{n})}}^{m}\circ c(t, \mathbf X _{m}^{h_{n},\delta _{n}},u_{m}^{h_{n},\delta _{n}})\text{ d}t\rightarrow \int _{0}^{T}\chi _{_{B( \mathbf R ^{n})}}^{m}\circ c(t, \mathbf X _{t}^{m},u_{t}^{m})\text{ d}t \end{aligned}$$

(185)

and

$$\begin{aligned} \int _{0}^{T}\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{m}^{h_{n},\delta _{n}},u_{m}^{h_{n},\delta _{n}})\sigma (t, \mathbf X _{m}^{h_{n},\delta _{n}})\text{ d} \mathbf B _{t} \end{aligned}$$

(186)

converges to

$$\begin{aligned} \int _{0}^{T}\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{t}^{m},u_{t}^{m})\sigma (t, \mathbf X _{t}^{m})d \mathbf B _{t} \end{aligned}$$

(187)

and

$$\begin{aligned} \int _{0}^{T}\left\Vert\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{m}^{h_{n},\delta _{n}},u_{m}^{h_{n},\delta _{n}})\sigma (t, \mathbf X _{m}^{h_{n},\delta _{n}})\right\Vert^{2}\text{ d}t \end{aligned}$$

(188)

converges

$$\begin{aligned} \int _{0}^{T}\left\Vert\left( \nabla _{u}c\cdot D_{u}^{-1}f\right) (t, \mathbf X _{t}^{m},u_{t}^{m})\sigma (t, \mathbf X _{t}^{m})\right\Vert^{2}\text{ d}t \end{aligned}$$

(189)

therefore we obtain

$$\begin{aligned} L^{2}-\lim _{n\rightarrow \infty }S( \mathbf X _{n})=S( \mathbf X ). \end{aligned}$$

(190)