\chapter{That $u$ is a Viscosity Solution}

Our goal this chapter is to use Feynman-Kac to see that the solutions to certain versions of the heat equations can be expressed as the expectation of a stocahstic integral. Parts of this work is heavily inspired from \cite{crandall_lions} and esp. \cite{Beck_2021}.
%\subsection{The case without $f$}
%\subsection{Linear Algebra Preliminaries}
%\begin{lemma}
%    For a matrix $A \in \R^{m\times n}$, $A[A]^*$ is symmetric.
%\end{lemma}
%\begin{proof}
%    Note that $[A]_{i,j} = [A^*]_{j,i}$, and hence:
%    \begin{align}
%    [AA^*]_{i,j} = \sum^n_{k=1} [A]_{i,k}[A^*]_{k,j} = [AA^*]_{j,i}
%    \end{align}
%\end{proof}
%\begin{lemma}
%    A symmetric matrix $A\in \R^{m\times n}$:
%    \begin{enumerate}[label = (\roman*)]
%    	\item Has only real eigenvalues.
%    	\item Is always diagonalizable.
%    \end{enumerate} 
%\end{lemma}
%\begin{proof}
%	\textit{(i)} Assume $\lambda \in \mathbb{C}$ is an eigenvalue for the symmetric matrix A. This indicates that $Av = \lambda v$, where $v\neq 0$ for some $v$ in our vector space and further that:
%	\begin{align}
%		v^*Av = \lambda v^*v
%	\end{align}
%	where $v^* = \overline{v}^T$. 
%	Note that since $A$ is symmetric we have:
%	\begin{align}
%		\lambda v^* v = v^*Av = \lp v^*Av\rp^* = \overline{\lambda}v^*v
%	\end{align}
%	This indicates that $\lambda = \overline{\lambda}$ whence $\lambda$ is real.
%	\medskip
%	
%	\textit{(ii)} Assume $A$ is not diagonalizable. Then there must exist some eigenvalue $\lambda_i$ of order 2 or more. This would indicate that there is some repeated eigenvalue $\lambda$, and $v\neq 0$ such that:
%	\begin{align}
%		(A-\lambda I)^2v = 0 & \text{ and }(A-\lambda I)v \neq 0
%	\end{align}
%	
%	But note that:
%	\begin{align}
%		0 = v^*(A-\lambda I)^2v = v^*(A-\lambda I)(A - \lambda I) \neq 0
%	\end{align}
%	Leading to a contradiction. Thus there are no generalized eigenvectors of order 2 or higher, and so $A$ must be diagonalizable.
%\end{proof}    
\section{Some Preliminaries}
	We take work previously pioneered by \cite{Ito1942a} and \cite{Ito1946}, and then seek to re-apply concepts applied in \cite{Beck_2021} and \cite{BHJ21}. 
	\begin{lemma}\label{lemma:2.7}
	Let $d,m \in \N$, $T \in (0,\infty)$. Let $\mu \in C^{1,2}([0,T] \times \R^d, \R^d)$ and $\sigma \in C^{1,2}([0,T] \times \R^d, \R^{d\times m})$ satisfying that they have non-empty compact supports and let $\mathfrak{S}= \supp(\mu)\cup \supp(\sigma) \subseteq [0,T] \times \R^d$. Let $( \Omega, \mathcal{F}, \mathbb{P}, ( \mathbb{F}_t )_{t \in [0,T]})$ be a filtered probability space satisfying usual conditions. Let $W:[0,T ]\times \Omega \rightarrow \R^m$ be a standard $(\mathbb{F}_t)_{t\in [0,T]}$ -Brownian motion, and let $\mathcal{X}:[0,T] \times \Omega \rightarrow \R^d$ be an $(\mathbb{F}_t)_{t\in [0,T]}$-adapted stochastic process with continuous sample paths satisfying for all $t \in [0,T]$ with $\mathbb{P}$-a.s. that:
		\begin{align}
			\mathcal{X}_t = \mathcal{X}_0 + \int^t_0 \mu(s, \mathcal{X}_s) ds + \int^t_0 \sigma(s, \mathcal{X}_s)dW_s
		\end{align}
		It then holds that:
		\begin{enumerate}[label = (\roman*)]
			\item $\lb \lp \mathbb{P} \lp \mathcal{X}_0 \not \in \mathfrak{S} \rp = 1 \rp \implies \lp \mathbb{P} \lp \forall t \in [0,T]: \mathcal{X}_t = \mathcal{X}_0 \rp =1 \rp \rb $
			\item $\lb \lp \mathbb{P} \lp \mathcal{X}_0 \in \mathfrak{S} \rp = 1 \rp \implies \lp \mathbb{P} \lp \forall t \in [0,T]: \mathcal{X}_t \in \mathfrak{S} \rp = 1 \rp \rb$
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		Assume that $\mathbb{P}(\mathcal{X}_0 \not \in \mathfrak{S} ) =1$, meaning that the particle almost surely starts outside $\mathfrak{S}$. It is then the case that $\mathbb{P}( \forall t \in [0, T]: \|\mu(t,\mathcal{X}_0)\|_E + \|\sigma(t,\mathcal{X}_0)\|_F =0)=1$ as the $\mu$ and $\sigma$ are outside their supports, and we integrate over zero over time.
		\medskip
		
		It is then the case that:
		\begin{align}
			\mathcal{Y}:= \lp \lb 0,T \rb \times \Omega \ni \lp t, \omega \rp \mapsto \mathcal{X}_0(\omega)\in \R^d \rp
		\end{align}
		is an $(\mathbb{F}_t)_{t \in [0,T]}$ adapted stochastic process with continuous sample paths satisfying that for all $t \in [0,T]$ with $\mathbb{P}$-almost surety that:
		\begin{align}
			\mathcal{Y}_t = \mathcal{X}_0 + \int^t_0 0 ds + \int^t_0 0 dW_s &= \mathcal{X}_0 + \int^t_0 \mu(s,\mathcal{X}_0)ds + \int^t_0 \sigma(s, \mathcal{X}_0) dW_s \nonumber\\
			&= \mathcal{X}_0 + \int^t_0 \mu(s,\mathcal{Y}_s) ds + \int^t_0 \sigma(s,\mathcal{Y}_s) dW_s
		\end{align}
		Note that since $\mu \in C^{1,2}([0, T] \times \R^d, \R^d)$ and $\sigma \in C^{1,2}([0, T] \times \R^d,  \R^{d \times m})$, and since continuous functions are locally Lipschitz, and since  this is especially true in the space variable for $\mu$ and $\sigma$, the fact that $\mathfrak{S}$ is compact and continuous functions over compact sets are Lipschitz and bounded, and  \cite[Theorem~5.2.5]{karatzas1991brownian} allows us to conclude that strong uniqueness holds, that is to say:
		\begin{align}
			\mathbb{P}\lp \forall t \in [0,T]: \mathcal{X}_t = \mathcal{X}_0 \rp  = \mathbb{P} \lp \forall t \in [0,T]: \mathcal{X}_t = \mathcal{Y}_t \rp=1 
		\end{align}
		establishing the case (i).
		
		Assume now that $\mathbb{P}(\mathcal{X}_0 \in \mathfrak{S})=1$ that is to say that the particle almost surely starts inside $\mathfrak{S}$. We define $\tau: \Omega \rightarrow [0,T]$ as $\tau= \inf\{t \in [0,T]: \mathcal{X}_t \not \in \overline{\mathfrak{S}}\}$. $\tau$ is an $(\mathbb{F}_t)_{t\in [0,T]}$-adapted stopping time. On top of $\tau$ we can define $\mathcal{Y}:[0,T] \times \Omega \rightarrow\R^d$, for all $t\in [0,T]$, $\omega \in \Omega$ as $\mathcal{Y}_t(\omega) =\mathcal{X}_{\min \{t,\tau\}}(\omega)$. $\mathcal{Y}$ is thus an $(\mathbb{F}_t)_{t \in [0,T]}$-adapted stochastic process with continuous sample paths. Note however that for $t > \tau$ it is the case $\|\mu(t, \mathcal{Y}_t)+\sigma(t, \mathcal{Y}_t) \|_E=0$ as we are outside their supports. For $t < \tau$ it is also the case that $\mathcal{Y}_t = \mathcal{X}_t$. This yields with $\mathbb{P}$-a.s. that:
		\begin{align}
			\mathcal{Y}_t = \mathcal{X}_{\min\{t,\tau \}} &= \mathcal{X}_0 + \int^{\min\{t,\tau\}}_0 \mu(s, \mathcal{X}_s) ds + \int^{\min\{t,\tau\}}_0 \sigma(s,\mathcal{X}_s)dW_s \nonumber \\
			&=\mathcal{X}_0 + \int^t_0 \mathbbm{1}_{\{0 < s \leqslant \tau \}} \mu (s, \mathcal{X}_s) ds+\int^t_0 \mathbbm{1}_{\{0<s\leqslant \tau \}}\sigma (s,\mathcal{X}_s) dW_s \nonumber \\
			&= \mathcal{X}_0 + \int^t_0 \mu(s,\mathcal{Y}_s)ds + \int^t_0 \sigma(s, \mathcal{Y}_s) dW_s 
		\end{align}
		Thus another application of \cite[Theorem~5.2.5]{karatzas1991brownian} and the fact that within our compact support $\mathfrak{S}$, the continuous functions $\mu$ and $\sigma$ are Lipschitz and hence locally Lipschitz, and also bounded gives us:
		\begin{align}
			\mathbb{P}(\forall t \in [0,T]: \mathcal{X}_t = \mathcal{Y}_t) =1
		\end{align}
		 Proving case (ii).
	\end{proof}
\begin{lemma}\label{lem:3.4}
	Let $d,m \in \N$, $T\in (0,\infty)$. Let $g \in C^2(\R^d, \R)$. Let $\mu \in C^{1,3}([0,T] \times \R^d,\R^d)$ and $\sigma \in C^{1,3}([0,T] \times \R^d,\R^{d \times m})$ have non-empty compact supports and let $\mathfrak{S} = \supp(\mu) \cup \supp(\sigma)$. Let $(\Omega, \mathcal{F}, \mathbb{P}, (\mathbb{F}_t)_{t \in [0,T]})$ be a stochaastic basis and let $W: [0,T] \times \Omega \rightarrow \R^m$ be a standard $(\mathbb{F}_t)_{t\in [0,T]}$-Brownian motion. For every $t\in [0,T]$ , $x\in \R^d$, let $\mathcal{X}^{t,x} = (\mathcal{X}^{t,x}_s)_{s\in [t,T]}:  [t,T] \times \Omega \rightarrow \R^d$ be an $(\mathbb{F}_s)_{s\in[t,T]}$-adapted stochastic process with continuous sample paths satisfying for all $s\in [t,T]$ with $\mathbb{P}$-almost surety that:
	\begin{align}\label{2.13}
		\mathcal{X}^{t,x}_s = x + \int^s_t \mu(r,\mathcal{X}^{t,x}_r) dr + \int^s_t \sigma(r,\mathcal{X}^{t,x}_s)dW_r
	\end{align}
	also let $u:\R^d \rightarrow \R$ satisfy for all $t \in [0,T]$, $x \in \R^d$ that:
	\begin{align}\label{2.14}
		u(t,x) = \E \lb g(\mathcal{X}^{t,x}_T) \rb 
	\end{align}
	then it is the case that we have:
	\begin{enumerate}[label = (\roman*)]
		\item $u \in C^{1,2}([0,T] \times \R^d, \R)$ and
		\item for all $t \in [0,T]$, $x \in \R^d$ that $u(T,x) = g(x)$ and:
		\begin{align}
			\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + \frac{1}{2}\Trace \lp \sigma \lp t,x \rp \lb \sigma \lp t,x \rp \rb^* \lp \Hess_x u \rp \lp t,x \rp \rp + \la \mu \lp t,x \rp, \lp \nabla_x u \rp\lp t,x \rp \ra = 0 
		\end{align}
	\end{enumerate}
	
\end{lemma}
\begin{proof}
We break the proof down into two cases, inside the support $\mathfrak{S} = \supp(\mu) \cup \supp(\sigma)$ and outside the support: $[0, T] \times (\mathbb{R}^d \setminus \mathfrak{S})$. 

For the case inside $\mathfrak{S}$. Note that we may deduce from Item $(i)$ of Lemma \ref{lemma:2.7} that for all $t \in [0,T]$, $x \in \R^d \setminus \mathfrak{S}$ it is the case that $\mathbb{P}(\forall s \in [t,T]: \mathcal{X}^{t,x}_s =x) =1$. Thus for all $t \in [0,T]$, $x \in \R^d \setminus \mathfrak{S}$ we have, deriving from (\ref{2.14}):
\begin{align}\label{(2.19)}
	u(t,x) = \E \lb g \lp \mathcal{X}^{t,x}_T \rp \rb = g(x)
\end{align}
Note that $g(x)$ only has a space parameter and so derivatives w.r.t. $t$ is $0$. Inhereting from the regularity properties of $g$ and (\ref{(2.19)}), we may assume for all $t \in [0,T]$, $x \in \R^d \setminus \mathfrak{S}$, that $u |_{[0,T] \times (\R^d \setminus \mathfrak{S})} \in C^{1,2}([0,T] \times (\R^d \setminus \mathfrak{S}))$. Note that the hypotheses that $\mu \in C^{1,3}([0,T] \times \R^d, \R^d)$ and $\sigma \in C^{1,3}([0,T] \times \R^d, \R^{d \times m})$ allow us to apply Theorem~7.4.3, Theorem~7.4.5 and Theorem~7.5.1 from \cite{da_prato_zabczyk_2002} for $t \in [0,T]$, $x \in \R^d \setminus \mathfrak{S}$, to give us:
\begin{enumerate}[label = (\roman*)]
\item $u \in C^{1,2} ([0,T] \times \R^d,\R)$.
\item \begin{align}
	0 &= \lp \frac{\partial}{\partial t} u \rp \lp t,x \rp \nonumber\\
	&= \lp \frac{\partial}{\partial t}u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x)\lb \sigma(t,x) \rb^* \lp \Hess_x u\rp \lp t,x \rp \rp + \la\mu(t,x), \lp \nabla_x u\rp \lp t,x \rp \ra 
\end{align} 
\end{enumerate}
\medskip


Now consider the case within support $\mathfrak{S}$. Note that by hypothesis $\mu$ and $\sigma$ must at least be locally Lipschitz. Thus \cite[Theorem~5.2.5]{karatzas1991brownian} allows us to conclude that within $\mathfrak{S}$ the pair $(\mu,\sigma)$ for our our stochastic process $\mathcal{X}^{t,x}_s$ defined in (\ref{2.13}) must exhibit a strong uniqueness property. 
\medskip

	Further note that Item $(ii)$ from  Lemma \ref{lemma:2.7} tells us that:
	\begin{align}
	 \mathbb{P}(\forall t \in [0,T]: \mathcal{X}^{t,x}_s \in \mathfrak{S}) = 1. 
	\end{align}
	Note that again the hypotheses that $\mu \in C^{1,3}([0,T] \times \R^d, \R^d)$ and $\sigma \in C^{1,3}([0,T] \times \R^d, \R^{d \times m})$, and $g \in C^2(\R^d)$ allow us to apply Theorem~7.4.3, Theorem~7.4.5 and Theorem~7.5.1 from \cite{da_prato_zabczyk_2002} for $t \in [0,T]$, $x \in \mathfrak{S}$, to give us:
\begin{enumerate}[label = (\roman*)]
\item $u \in C^{1,2} ([0,T] \times \R^d,\R)$.
\item \begin{align}
	 \lp \frac{\partial}{\partial t}u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x)\lb \sigma(t,x) \rb^* \lp \Hess_x u\rp \lp t,x \rp \rp + \la\mu(t,x), \lp \nabla_x u\rp \lp t,x \rp \ra =0
\end{align} 
\end{enumerate}
Note that (\ref{2.13}) and (\ref{2.14}) together prove that $u(T,x) = g(x)$. This completes the proof.
\end{proof}
\section{Viscosity Solutions} 
    
\begin{definition}[Symmetric Matrices] Let $d \in \N$. The set of symmetric matrices is denoted $\mathbb{S}_d$ given by $\mathbb{S}_d = \{ A \in \mathbb{S}_d : A^* = A \}$.
\end{definition}

\begin{definition}[Upper semi-continuity]
	A function $f: U \rightarrow \R$ is upper semi-continuous at $x_0$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that:
	\begin{align}
		f(x) < f(x_0) + \ve \text{ for all } x \in B\lp x_0, \delta \rp \cap U
	\end{align}
\end{definition}
\begin{definition}[Lower semi-continuity]
	A function $f: U \rightarrow \R$ is lower semi-continuous at $x_0$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that:
	\begin{align}
		f(x) > f(x_0) - \varepsilon \text{ for all } x \in B\lp x_0, \delta \rp \cap U
	\end{align}
\end{definition}
\begin{corollary}\label{sumofusc}
	Given two upper semi-continuous functions $f,g: \R^d \rightarrow \R$, their sum $(f+g): \R^d \rightarrow \R$ is also upper semi-continuous.
\end{corollary}
\begin{proof}
	From definitions, at any given $x_0 \in \R^d$, for any $\ve \in (0, \infty)$ there exist neighborhoods $U$ and $V$ around $x_0$ such that:
	\begin{align}
		\lp \forall x \in U \rp \lp f(x) \leqslant f(x_0) + \varepsilon \rp  \\
		\lp \forall x \in V \rp \lp g(x) \leqslant g(x_0) + \varepsilon \rp
	\end{align}
	and hence:
	\begin{align}
		\lp \forall x \in U \cap V \rp \lp f(x) + g(x) \leqslant f(x_0)+g(x_0) + 2\varepsilon \rp
	\end{align}
\end{proof}
\begin{corollary}\label{neglsc}
	Given an upper semi-continuous function $f:\R^d \rightarrow \R$, it is the case that $(-f):\R^d \rightarrow \R$ is lower semi-continuous.
\end{corollary}

\begin{proof}
		Let $f:\R^d \rightarrow \R$ be upper semi-continuous. At any given $x_0 \in \R^d$, for any $\ve \in (0, \infty)$ there exists a neighborhood $U$ around $x_0$ such that:
		\begin{align}
			\lp \forall x \in U \rp \lp f(x) \leqslant f(x_0) + \ve \rp
		\end{align}
		This also means that:
		\begin{align}
			&\lp \forall x \in U \rp \lp -f(x) \geqslant -f(x_0) - \ve \rp \nonumber\\
		\end{align}
		This completes the proof.
\end{proof}

\begin{definition}[Degenerate Elliptic Functions] Let $d \in \N$, $T \in \lp 0, \infty \rp$, let $\mathcal{O} \subseteq \R^d$ be a non-empty open set, and let $\la \cdot, \cdot \ra: \R^d \times \R^d \rightarrow \R$ be the standard Euclidean inner product on $\R^d$. $G$ is degenerate elliptic on $\lp 0,T \rp \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ if and only if:
\begin{enumerate}[label = (\roman*)]
	\item $G: \lp 0,T \rp \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ is a function, and
	\item for all $t \in \lp 0 ,T \rp$, $x \in \mathcal{O}$, $r \in \R$, $p \in \R^d$, $A,B \in \mathbb{S}_d$, with $\forall y \in \R^d$: $\la Ay,y \ra \leqslant \la By, y \ra$ that $G(t,x,r,p,A) \leqslant G(t,x,r,p,B)$.
\end{enumerate}
\end{definition}
\begin{remark}
	Let $t \in (0,T)$, $x \in \R^d$, $r\in \R$, $p\in \R^d$, $A \in \mathbb{S}_d$. Let $u \in C^{1,2} ([0,T] \times \R^d, \R)$, and let $\sigma: \R^d \rightarrow \R^{d\times d}$ and $\mu: \R^d \rightarrow \R^d$ be infinitely often differentiable.  The function $G:(0,T) \times \R^d \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ given by:
	\begin{align}
	G(t,x,r,p,A) = \frac{1}{2} \Trace \lp \sigma(x) \lb \sigma(x)\rb^* \lp \Hess_x u \rp \lp t,x \rp \rp + \la \mu(t,x), \nabla_xu\lp t,x\rp \ra 
	\end{align}
	where $\lp t,x,u(t,x),\mu(x), \sigma(x) \lb \sigma(x) \rb^* \rp \in \lp 0,T\rp \times \R^d \times \R \times \R^d \times \mathbb{S}_d$, is degenerate elliptic.
\end{remark}

\begin{lemma}\label{negdegel}
	Given a function $G: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ that is degerate elliptic on $(0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ it is also the case that $H: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ given by $H(t,x,r,p,A)=-G(t,x,-r, -p,-A)$ is degenerate elliptic on $(0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$.
\end{lemma}

\begin{proof}
	Note that $H$ is a function. Assume for $y\in \R^d$ it is the case that $\la Ay, y \ra \leqslant \la By,y \ra$ then it is also the case by (\ref{bigsum}) that $\la -Ay,y\ra \geqslant \la -By, y \ra$ for $y\in \R^d$. However since $G$ is monotoically increasing over the subset of $(0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ where $\la Ay,y \ra \leqslant \la By,y \ra$ then it is also the case that $H(t,x,r,p,A) =-G(t,x,-r,-p, -A) \geqslant -G(t,x,-r,-p,-B)=H(t,x,r,p,B)$. 
	
	
	
\end{proof}

\begin{definition}[Viscosity subsolutions]\label{def:viscsubsolution}
	Let $d \in \N$, $T \in \lp 0, \infty \rp$, let $\mathcal{O} \subseteq \R^d$ be a non-empty open set, and let $G: \lp 0, T \rp \times \mathcal{O} \times \R \times \R^d \times\mathbb{S}_d \rightarrow \R$ be degenrate elliptic. Then we say that $u$ is a viscosity solution of $\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + G \lp t,x,u(t,x), \lp \nabla_x u \rp \lp t,x \rp, \lp \Hess_x u \rp \lp t,x \rp \rp \geqslant 0$ for $\lp t,x, \rp \in \lp 0,T \rp \times \mathcal{O}$ if and only if there exists a set $A$ such that:
	\begin{enumerate}[label = (\roman*)]
		\item we have that $\lp 0,T \rp \times \mathcal{O} \subseteq A$.
		\item we have that $u: A \rightarrow \R$ is an upper semi-continuous function from $A$ to $\R$, and
		\item we have that for all $t\in \lp 0, T \rp$, $x \in \mathcal{O}$, $\phi \in C^{1,2} \lp \lp 0,T \rp \times \mathcal{O}, \R \rp$ with $\phi (t,x) = u (t,x)$ and $\phi \geqslant u$ that:
			\begin{align}
				\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + G \lp t,x, \phi(t,x), \lp \nabla_x \phi \rp \lp t,x \rp, \lp \Hess_x \phi \rp \lp t,x \rp \rp \geqslant 0
			\end{align}
	\end{enumerate}
\end{definition}

\begin{definition}[Viscosity supersolutions]\label{def:viscsupsolution}
	Let $d \in \N$, $T \in \lp 0, \infty \rp$, let $\mathcal{O} \subseteq \R^d$ be a non-empty open set, and let $G: \lp 0, T \rp \times \mathcal{O} \times \R \times \R^d \times\mathbb{S}_d \rightarrow \R$ be degenrate elliptic. Then we say that $u$ is a viscosity solution of $\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + G \lp t,x,u(t,x), \lp \nabla_x u \rp \lp t,x \rp, \lp \Hess_x u \rp \lp t,x \rp \rp \leqslant 0$ for $\lp t,x, \rp \in \lp 0,T \rp \times \mathcal{O}$ if and only if there exists a set $A$ such that:
	\begin{enumerate}[label = (\roman*)]
		\item we have that $\lp 0,T \rp \times \mathcal{O} \subseteq A$.
		\item we have that $u: A \rightarrow \R$ is an upper semi-continuous function from $A$ to $\R$, and
		\item we have that for all $t\in \lp 0, T \rp$, $x \in \mathcal{O}$, $\phi \in C^{1,2} \lp \lp 0,T \rp \times \mathcal{O}, \R \rp$ with $\phi (t,x) = u (t,x)$ and $\phi \leqslant u$ that:
			\begin{align}
				\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + G \lp t,x, \phi(t,x), \lp \nabla_x \phi \rp \lp t,x \rp, \lp \Hess_x \phi \rp \lp t,x \rp \rp \leqslant 0
			\end{align}
	\end{enumerate}
\end{definition}
\begin{definition}[Viscosity solution]\label{def:viscsolution}
	Let $d \in \N$, $T \in \lp 0, \infty \rp$, $\mathcal{O} \subseteq \R^d$ be a non-empty open set and let $G: \lp 0, T \rp \times \mathcal{O} \times  R \times \R^d \times \mathbb{S}_d \rightarrow \R$ be degenerate elliptic. Then we say that $u_d$ is a viscosity solution to $\lp \frac{\partial }{\partial t} u_d \rp (t,x) + G(t,x,u(t,x), \nabla_x(x,t), (\Hess_x u_d)(t,x))$ if and only if:
	\begin{enumerate}[label = (\roman*)]
		\item $u$ is a viscosity subsolution of $\lp \frac{\partial }{\partial t} u_d \rp (t,x) + G(t,x,u(t,x), \nabla_x(x,t), (\Hess_x u_d)(t,x))=0$ for $(t,x) \in (0,T) \times \mathcal{O}$
		\item $u$ is a viscosity supersolution of $\lp \frac{\partial }{\partial t} u_d \rp (t,x) + G(t,x,u(t,x), \nabla_x(x,t), (\Hess_x u_d)(t,x))=0$ for $(t,x) \in (0,T) \times \mathcal{O}$
	\end{enumerate}
	
\end{definition}

%
%\begin{lemma}
%    Let $T\in (0, \infty)$, let $\left( \Omega, \mathcal{F}, \mathbb{P} \right)$ be a probability space, let $\sigma_d: \R^d \rightarrow \R^{d\times d}$, $d\in \N$, be infinitely often differentiable functions, let $u_d \in C^{1,2} \left( \left[ 0,T \right] \times \R^d, \R \right)$, $d\in \N$, satisfy for all $d\in \N$, $t \in \left[ 0,T \right]$, $x \in \R^d$ that:
%    \begin{align}
%        \left( \frac{\partial}{\partial t} u_d \right)\left( t,x \right) + \text{Trace} \left( \sigma(x) \left[ \sigma(x) \right]* \left( \text{Hess}_x u_d \right)(t,x) \right) =0
%    \end{align}
%
%Let $W^d: [0,T] \times \Omega \rightarrow \R^d$, $d\in \N$, be standard Brownian motions, and let $\mathcal{X}^{d,t,x}: [t,T] \times \Omega \rightarrow \R^d$, $d \in \N$, $t \in [0,T]$, $x \in \R^d$, be a stochastic process with continuous sample paths satisfying that for all $d\in \N$, $t\in [0,T]$, $s \in [t,T]$, $x \in \R^d$, we have $\mathbb{P}$-a.s. that:
%\begin{align}
%    \mathcal{X}^{d,t,x}_s = x + \int^t_s \sqrt{2} \sigma \left( \mathcal{X}^{d,t,x} \right) dW^d_r
%\end{align}
%Then for all $d\in \N$, $t \in [0,T]$, $x \in \R^d$ it holds that:
%\begin{align}
%    u_d \left( t,x \right) = \E \left[ u_d \left( T, \mathcal{X}^{d,t,x}_T \right) \right]
%\end{align}
%\end{lemma}
%
%First we consider the case where $\sigma: \R^d \rightarrow \R^{d\times d}$ is the constant function:
%\begin{align*}
%\sigma: \begin{pmatrix}
%    x_1 \\
%    x_2 \\
%    \vdots\\
%    x_d
%\end{pmatrix} \rightarrow \begin{pmatrix}
%    c_{1,1} & c_{2,2} & \hdots & c_{1,d}\\
%    c_{2,1}  & & & \vdots \\
%    \vdots & & \ddots\\
%    c_{d,1} & \hdots & & c_{d,d}
%\end{pmatrix}
%\end{align*} 
%
%and hence $\sigma \sigma^*$ is a $d \times d$ matrix where $[\sigma \sigma^*]_{i,j} = \sum^d_{k=1}\left( c_{i,k} \times c_{j,k} \right)$. Observe also that:
%\begin{align*}
%    \text{Hess}_x u_d &= \begin{pmatrix}
%        \frac{\partial^2 u_d(t,x)}{\partial x_1 \partial x_1} & \cdots & \frac{\partial^2 u_d(t,x)}{\partial x_1 \partial x_d} \\  
%        \vdots & & \vdots \\
%        \frac{\partial^2 u_d(t,x)}{\partial x_d \partial x_1} & \cdots & \frac{\partial^2 u_d(t,x )}{\partial x_d^2}
%    \end{pmatrix}
%\end{align*}
%
%Their product is the matrix 
%\begin{align}
%[(\sigma [\sigma^*])(\text{Hess}_x u)(t,x)]_{i,j} = \sum^d_{l=1}\left( \sum^d_{k=1} \left( c_{i,k} \times c_{l,k} \right) \left(\frac{\partial u_d(t,x)}{\partial x_l \partial x_j} \right) \right)
%\end{align}
%And hence:
%\begin{align}
%    \text{Trace}[(\sigma [\sigma^*])(\text{Hess}_x u)(t,x)] 
%    &=  \sum^d_{m=1} \left(\sum^d_{l=1}\left( \sum^d_{k=1} \left( c_{m,k} \times c_{l,k} \right) \left(\frac{\partial u_d(t,x)}{\partial x_l \partial x_m} \right) \right) \right) \nonumber\\
%    &= \sum^d_{m=1} \left(\sum^d_{l=1}\left( \sum^d_{k=1} \left( c_{m,k} \times c_{l,k} \right) \right)\left(\frac{\partial u_d(t,x)}{\partial x_l \partial x_m} \right) \right) \nonumber\\
%    &= \sum^d_{m=1} \sum^d_{l=1}\mathfrak{C}^{m,d,l} \frac{\partial u_d(t,x)}{\partial x_l \partial x_m} 
%\end{align}
%\medskip
%
%Where $\mathfrak{C}^{m,d,l} = \sum^d_{k=1} \left( c_{m,k}\times c_{l,k} \right)$
%\medskip
%
%Note that for each row it is the case that:
%
%
%
%This also renders (2.12) as:
%\begin{align}
%\left( \frac{\partial}{\partial t} u_d \right)\left( t,x \right) + \sum^d_{m=1} \sum^d_{l=1}\mathfrak{C}^{m,d,l} \frac{\partial u_d(t,x)}{\partial x_l \partial x_m} = 0
%\end{align}
%
%This renders (2.13) as:
%
%\begin{align}
%    \mathcal{X}^{d,t,x}_s &= \begin{pmatrix}
%        x_1 \\
%        x_2 \\
%        \vdots\\
%        x_d
%    \end{pmatrix} + \int^t_s \sqrt{2} \begin{pmatrix}
%        c_{1,1} & c_{1,2} & \hdots & c_{1,d}\\
%        c_{2,1}  & & & \vdots \\
%        \vdots & & \ddots\\
%        c_{d,1} & \hdots & & c_{d,d}
%    \end{pmatrix} dW^d_r \nonumber\\
%    &=  \begin{pmatrix}
%        x_1 \\
%        x_2 \\
%        \vdots\\
%        x_d
%    \end{pmatrix} + \sqrt{2} \begin{pmatrix}
%        c_{1,1} & c_{1,2} & \hdots & c_{1,d}\\
%        c_{2,1}  & & & \vdots \\
%        \vdots & & \ddots\\
%        c_{d,1} & \hdots & & c_{d,d}
%    \end{pmatrix} W^d_{t-s}
%\end{align}
%
%\medskip
%
%Note that $W^d \in \R^d$, and specifically let $W^d_{t-s}$ be the $d$-dimensional vector:
%\begin{align}
%    W^d_{t-s} = \begin{pmatrix}
%        W_{1,t-s} \\
%        W_{2,t-s} \\
%        \vdots \\
%        W_{d,t-s}
%    \end{pmatrix}
%\end{align}
%\medskip
%
%This then renders (2.16) as:
%\begin{align}
%    \mathcal{X}^{d,t,x}_s &=
%    \begin{pmatrix}
%        x_1 \\
%        x_2 \\
%        \vdots\\
%        x_d
%    \end{pmatrix} + \sqrt{2} \begin{pmatrix}
%        c_{1,1} & c_{1,2} & \hdots & c_{1,d}\\
%        c_{2,1}  & & & \vdots \\
%        \vdots & & \ddots\\
%        c_{d,1} & \hdots & & c_{d,d}
%    \end{pmatrix} \begin{pmatrix}
%        W_{1,t-s} \\
%        W_{2,t-s} \\
%        \vdots \\
%        W_{d,t-s}
%    \end{pmatrix} \nonumber \\
%    &= \begin{pmatrix}
%        x_1 \\
%        x_2 \\
%        \vdots\\
%        x_d
%    \end{pmatrix} + \sqrt{2} \begin{pmatrix}
%        \sum^d_{i=1}\left(c_{1,i} W_{i,t-s}\right) \\
%        \sum^d_{i=1}\left(c_{2,i} W_{i,t-s}\right) \\
%        \vdots \\
%        \sum^d_{i=1}\left(c_{d,i} W_{i,t-s}\right) \\
%    \end{pmatrix} \nonumber \\
%    &= \begin{pmatrix}
%        x_1 + \sqrt{2}\sum^d_{i=1}\left(c_{1,i} W_{i,t-s}\right) \\
%        x_2 + \sqrt{2}\sum^d_{i=1}\left(c_{2,i} W_{i,t-s}\right) \\
%        \vdots \\
%        x_d + \sqrt{2}\sum^d_{i=1}\left(c_{d,i} W_{i,t-s}\right) \\
%    \end{pmatrix}
%\end{align}
%
%Let $\mathfrak{W}^{i,j,d}_{t-s} = \sum^d_{i=1} \left( c_{j, i} W_{i,t-s}\right)$, given that the product of a Brownian motion with a constant is a Brownian motion and the sum of Brownian motion is also a Brownian motion, we have that: $\mathfrak{W}^{i,j,d}_{t-s} = \sum^d_{i=1} \left( c_{j, i} W_{i,t-s}\right)$ is also a Brownian motion.
%\medskip
%
%For each row $j$ we therefore have $x_k + \sqrt{2} \mathfrak{W}^{i,j,d}_{t-s}$ 
%
\begin{lemma} \label{maxviscosity}
	Let $d\in \N$, $T \in \lp 0,\infty \rp$, $\mathfrak{t} \in \lp 0,T \rp$, let $\mathcal{O} \subseteq \R^d$ be an open set, let $\mathfrak{r} \in \mathcal{O}$, $\phi \in C^{1,2}\lp \lp 0,T \rp \times \mathcal{O},\R \rp$, let $G: \lp 0,T \rp \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ be degenerate elliptic and let $u_d: (0,T) \times \mathcal{O} \rightarrow \R$ be a viscosity solution of \\
	$\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + G \lp t,x,u(t,x), \lp \nabla_x u_D \rp \lp t,x \rp, \lp \Hess_x u_d \rp \lp t,x \rp \rp \geqslant 0$ for $(t,x) \in (0,T) \times \mathcal{O}$, and assume that $u-\phi$ has a local maximum at $(\mathfrak{t}, \mathfrak{r}) \in (0,T) \times \mathcal{O}$, then:
		\begin{align}
			\lp \frac{\partial}{\partial t} \phi \rp \lp \mathfrak{t},\mathfrak{r}\rp + G \lp \mathfrak{t}, \mathfrak{r}, u(\mathfrak{t}, \mathfrak{r}), \lp \nabla _x \phi \rp \lp \mathfrak{t}, \mathfrak{r} \rp, \lp \Hess_x \phi\rp\lp \mathfrak{t}, \mathfrak{r} \rp \rp \geqslant 0
		\end{align}
\end{lemma}
\begin{proof}
	That $u$ is upper semi-continuous ensures that there exists as a neighborhood $U$ around $(\mathfrak{t}, \mathfrak{r})$ and $\psi \in C^{1,2} ((0,T) \times \mathcal{O},\R)$ where:
	\begin{enumerate}[label = (\roman*)]
		\item for all $(t,x) \in (0,T) \times \mathcal{O}$ that $u(\mathfrak{t}, \mathfrak{r}) - \psi(\mathfrak{t}, \mathfrak{r}) \geqslant u(t,x) - \psi(t,x)$
		\item for all $(t,x) \in U$ that $\phi(t,x) = \phi(t,x)$.
	\end{enumerate}
	We therefore obtain that:
	\begin{align}
		&\lp \frac{\partial}{\partial t} \phi \rp \lp \mathfrak{t}, \mathfrak{r} \rp + G \lp \mathfrak{t}, \mathfrak{r}, u(\mathfrak{t},\mathfrak{r}), (\nabla_x) (\mathfrak{t}, \mathfrak{r}), (\Hess_x \phi)(\mathfrak{t}, \mathfrak{r}) \rp \nonumber \\
		&= \lp \frac{\partial}{\partial t} \psi \rp \lp \mathfrak{t}, \mathfrak{r} \rp + G \lp \mathfrak{t}, \mathfrak{r}, u(\mathfrak{t},\mathfrak{r}), (\nabla_x) (\mathfrak{t}, \mathfrak{r}), (\Hess_x \psi)(\mathfrak{t}, \mathfrak{r}) \rp \geqslant 0
	\end{align}
\end{proof}
\begin{lemma}\label{ungeq0}
	Let $d \in \N$, $T \in (0, \infty)$, let $\mathcal{O} \subseteq \R^d$ be a non-empty open set, let $u_n:(0,T) \times \mathcal{O} \rightarrow \R$, $n \in \N_0$ be functions, let $G_n: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$, $n \in \N$ be degenerate elliptic, assume that $G_\infty$ is upper semi-continuous for all non-empty compact $\mathcal{K} \subseteq (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ that:
	\begin{align}
		\limsup_{n \rightarrow \infty} \lb\sup_{(t,x,r,p,A) \in \mathcal{K}} \lp \lv u_n(t,x) -u_0(t,x) \rv + \lv G_n(t,x,r,p,A) - G_0(t,x,r,p,A) \rv \rp \rb=0 \label{limitofun}
	\end{align}
	and assume for all $n\in \N$ that $u_n$ is a viscosity solution of:
	\begin{align}
		\lp \frac{\partial}{\partial t} u_n \rp \lp t,x \rp + G_n \lp t,x,u_n(t,x),(\nabla_x u_n)(t,x), (\Hess_x u_n)(t,x) \rp \geqslant 0
	\end{align}
	then $u_0$ is a viscosity solution of:
	\begin{align}\label{hessungeq0}
		\lp \frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + G_n \lp t,x,u_0(t,x),(\nabla_x u_0)(t,x), (\Hess_x u_0)(t,x) \rp \geqslant 0
	\end{align}
\end{lemma}
\begin{proof}
	Let $(t_0, x_0) \in (0,T) \times \mathcal{O}$. Let $\phi_\epsilon \in C^{1,2}((0,T) \times \mathcal{O}, \R)$ satisfy for all $\epsilon \in (0, \infty)$, $s \in (0,T)$, $y \in \mathcal{O}$ that $\phi_0(t_0,x_0) = u_0(t_0,x_0)$, $\phi_0(t_0,x_0) \geqslant u_0(t_0,x_0)$, and:
	\begin{align}\label{phieps}
		\phi_\varepsilon(s,y) = \phi_o(s,y) + \varepsilon\lp \lv s - t_0 \rv + \| y - x_0 \|_E \rp  
	\end{align}
	Let $\delta \in (0,\infty)$ be such that $\{(s,y) \in \R^d \times \R: \max \lp|s-t_0|^2, \|y-x_0\|_E^2 \rp\leqslant \delta \}$. Note that this and (\ref{limitofun}) 
	then imply for all $\varepsilon \in (0,\infty)$ there exists an $\nu_\varepsilon \in \N$ such that for all $n \geqslant \nu_\varepsilon$, and $\max \lp|s-t_0|, \|y-x_0\|_E \rp\leqslant \delta$, it is the case that:
	\begin{align}
		\sup \lp \lv u_n(s,y) - u_0(s,y) \rv \rp \leqslant \frac{ \varepsilon \delta}{2}
	\end{align}
	Note that this combined with (\ref{phieps}) tells us that for all $\varepsilon \in (0,\infty)$, $n \in \N \cap [\nu_\epsilon, \infty)$, $s\in (0,T)$, $y\in \mathcal{O}$, with $|s-t_0| < \delta$, $\|y-x_0\|_E \leqslant \delta$, $|s-t_0| + \|y-x_0\|_E > \delta$ that:
	\begin{align}
		u_n(t_0,x_0) - \phi_\varepsilon(t_0,x_0) &= u_n(t_0,x_0) - \phi_0(t_0,x_0) \\
		&= u_n(t_0,x_0) - u_0(t_0,x_0) \nonumber\\
		&\geqslant \frac{-\varepsilon \delta}{2} \nonumber \\
		&\geqslant u_n(s,y) - u_0(s,y)-\varepsilon \lp |s-t_0|+\|y-x_0\|_E\rp \nonumber \\
		&\geqslant u_n(s,y) - \phi_0(s,y)-\varepsilon \lp |s-t_0|+\|y-x_0\|_E\rp \nonumber \\
		&= u_n(s,y) - \phi_\varepsilon (s,y)
	\end{align}
	\medskip
	
	Note that Corollary \ref{sumofusc} implies that for all $\epsilon \in (0,\infty)$ and $n\in \N$ that $u_n - \phi_\varepsilon$ is upper semi-continuous. There therefore exists for all $\epsilon \in (0,\infty)$ and $n \in \N$, a $\tau^\varepsilon_n \in (t_0 - \delta, t_0+\delta)$ and a $\rho^\varepsilon_n$, where $\|\rho_n^\varepsilon - x_0 \| \leqslant \delta$ such that:
	\begin{align}\label{maxphiu}
		u_n(\tau^\varepsilon_n, \rho_n^\varepsilon) - \phi_\epsilon(\tau_n^\varepsilon, \rho^\ve_n) \geqslant u_n(s,y) - \phi_\varepsilon(s,y)
	\end{align} 
	By Lemma \ref{maxviscosity}, it must be the case that for all $\varepsilon \in (0,\infty)$ and $n \in \N \cap [\nu_\varepsilon, \infty)$:
	\begin{align}
		\lp \frac{\partial}{\partial t} \phi_\varepsilon \rp \lp \tau^\varepsilon_n, \rho^\varepsilon_n \rp + G_n \lp \tau_n^\ve, \rho_n^\ve, u_n \lp \tau_n^\ve, \rho_n^\ve\rp, \lp \nabla_x \phi_\ve \rp \lp \tau^\ve_n, \rho^\ve_n \rp, \lp \Hess_x \phi_\ve \rp \lp \tau^\ve_n, \rho^\ve_n \rp \rp \geqslant 0
	\end{align}
	Note however that (\ref{maxphiu}) along with (\ref{phieps}) and (\ref{limitofun})  yields that for all $\ve \in (0,\infty)$ that:
	\begin{align}
		&\limsup_{n\rightarrow \infty}\lb u_n(\tau_n^\varepsilon, \rho_n^\varepsilon) - \phi_\epsilon(\tau_n^\varepsilon, \rho^\ve_n) \rb \nonumber\\
		&\geqslant \limsup_{n\rightarrow \infty} \lb u_n(\tau^\ve_n,\rho^\ve_n) - \lp\phi_0(\tau^\ve_n,\rho^\ve_n) + \ve \lp |\tau^\ve_n-t_0| + \|\rho^\ve_n- x_0\|_E \rp \rp \rb \nonumber \\
		&\geqslant \limsup_{n\rightarrow \infty} \lb u_n(\tau^\ve_n,\rho^\ve_n) -  u_0(\tau^\ve_n,\rho^\ve_n) - \ve \lp |\tau^\ve_n-t_0| + \|\rho^\ve_n- x_0\|_E \rp \rb \nonumber\\
		&= \limsup_{n\rightarrow \infty} \lb - \ve \lp |\tau^\ve_n-t_0| + \|\rho^\ve_n- x_0\|_E \rp \rb \leqslant 0
	\end{align}
	
	However note also that since $G_0$ is upper semi-continuous, further the fact that, $\phi_0 \in \lp \lp0,T\rp \times \mathcal{O}, \R \rp$, and then $(\ref{limitofun})$, and $(\ref{phieps})$, imply for all $\ve \in (0,\infty)$ we have that:
	 $\limsup_{n\rightarrow \infty} \lv \lp \frac{\partial}{\partial t}\phi_\ve \rp\lp \tau^\ve_n, \rho^\ve_n \rp - \\ 
	 \lp\frac{\partial}{\partial t} \phi_0\rp \lp t_0,x_0 \rp\rv = 0$ and: 
	\begin{align}
		& G_0 \lp t_0, x_0, \phi_0 \lp t_0, x_0 \rp, \lp \nabla_x \phi_0 \rp \lp t_0,x_0 \rp, \lp \Hess_x \phi_0 \rp \lp t_0,x_0 \rp+ \text{Id}_{\R^d} \rp \nonumber \\
		&= G_0 \lp t_0, x_0, u_0 \lp t_0, x_0 \rp, \lp \nabla_x \phi_\ve \rp \lp t_0,x_0 \rp, \lp \Hess_x \phi_\ve \rp \lp t_0,x_0\rp \rp \nonumber \\
		& \geqslant \limsup_{n \rightarrow \infty} \lb G_0 \lp \tau^\ve_n, \rho^\ve_n, u_n \lp \tau^\ve_n, \rho^\ve_n \rp, \lp \nabla_x \phi_\ve \rp \lp \tau^\ve_n,\rho^\ve_n \rp, \lp \Hess_x \phi_\ve \rp \lp \tau^\ve_n,\rho^\ve_n\rp \rp \rb \\
		&\geqslant \limsup_{n \rightarrow \infty} \lb G_n \lp \tau^\ve_n, \rho^\ve_n, u_n \lp \tau^\ve_n, \rho^\ve_n \rp, \lp \nabla_x \phi_\ve \rp \lp \tau^\ve_n,\rho^\ve_n \rp, \lp \Hess_x \phi_\ve \rp \lp \tau^\ve_n,\rho^\ve_n\rp \rp \rb
	\end{align}
	
	This with (\ref{maxphiu}) assures for all $\epsilon \in (0,\infty)$ that:
	\begin{align}
		\lp \frac{\partial}{\partial t} \phi_0 \rp \lp t_0,x_0 \rp + G_0 \lp t_0, x_0, \phi_0 \lp t_0, x_0 \rp, \lp \nabla_x \phi_0 \rp \lp t_0, x_0 \rp, \lp \Hess_x \phi_0 \rp \lp t_0, x_0 \rp + \ve \text{Id}_{\R^d} \rp \geqslant 0
	\end{align}
	That $G_0$ is upper semi-continuous then yields that:
	
	\begin{align}
		\lp \frac{\partial}{\partial t} \phi_0 \rp \lp t_0,x_0 \rp + G_0 \lp t_0, x_0, \phi_0\lp t_0, x_0 \rp, \lp \nabla_x \phi_0 \rp \lp t_0, x_0 \rp, \lp \Hess_x \phi_0 \rp \lp t_0, x_0 \rp + \ve \text{Id}_{\R^d} \rp \geqslant 0
	\end{align}	
	 This establishes $(\ref{hessungeq0})$ which establishes the lemma.
	
	
	
\end{proof} 

\begin{corollary}\label{unleq0}
	Let $d \in \N$, $T \in (0, \infty)$, let $\mathcal{O} \subseteq \R^d$ be a non-empty open set, let $u_n:(0,T) \times \mathcal{O} \rightarrow \R$, $n \in \N_0$ be functions, let $G_n: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$, $n \in \N_0$ be degenerate elliptic, assume that $G_0$ is lower semi-continuous for all non-empty compact $\mathcal{K} \subseteq (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ that:
	\begin{align}
		\limsup_{n \rightarrow \infty} \lb\sup_{(t,x,r,p,A) \in \mathcal{K}} \lp \lv u_n(t,x) -u_0(t,x) \rv + \lv G_n(t,x,r,p,A) - G_0(t,x,r,p,A) \rv \rp \rb = 0 \label{limitofun}
	\end{align}
	and assume for all $n\in \N$ that $u_n$ is a viscosity solution of:
	\begin{align}\label{viscsolutionvn}
		\lp \frac{\partial}{\partial t} u_n \rp \lp t,x \rp + G_n \lp t,x,u_n(t,x),(\nabla_x u_n)(t,x), (\Hess_x u_n)(t,x) \rp \leqslant 0
	\end{align}
	then $u_0$ is a viscosity solution of:
	\begin{align}\label{hessungeq0}
		\lp \frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + G_n \lp t,x,u_0(t,x),(\nabla_x u_0)(t,x), (\Hess_x u_0)(t,x) \rp \leqslant 0
	\end{align}
\end{corollary}

\begin{proof}
	Let $v_n:(0,T) \times \mathcal{O} \rightarrow \R$, $n \in \N_0$ and $H_n:(0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ satisfy for all $n\in \N_0$, $t\in (0,T)$, $x \in \mathcal{O}$, $r\in \R$, $p \in \R^d$, $A \in \mathbb{S}_d$ that $v_n(t,x) = -u_n(t,x)$ and that $H_n(t,x) = -G_n(t,x,-r,-p,-A)$. 
	\medskip
	
	Note that Corollary \ref{neglsc} gives us that $H_0$ is upper semi-continuous. Note also that since it is the case that for all $n\in \N_0$, $G_n$ is degenerate elliptic then it is also the case by Lemma \ref{negdegel} that  $H_n$ is degenerate elliptic for all $n\in \N_0$. These together with (\ref{viscsolutionvn}) ensure that for all $n\in \N$, $v_n$ is a viscosity solution of:
	\begin{align}\label{hgeq0}
		\lp \frac{\partial}{\partial t} v_n \rp \lp t,x \rp + H_n \lp t,x,v_n \lp t,x \rp, \lp \nabla_x v_n \rp \lp t,x \rp, \lp \Hess_x v_n \rp \lp t,x \rp \rp \geqslant 0
	\end{align}
	This together with (\ref{limitofun}) establish that:
	\begin{align}
		\limsup_{n \rightarrow \infty} \lb\sup_{(t,x,r,p,A) \in \mathcal{K}} \lp \lv u_n(t,x) -u_0(t,x) \rv + \lv H_n(t,x,r,p,A) - H_0(t,x,r,p,A) \rv \rp \rb = 0 
	\end{align}
	
	This (\ref{hgeq0}) and the fact that $H_0$ is upper semi-continuous then establish that:
	\begin{align}
		\lp \frac{\partial}{\partial t} v_0 \rp \lp t,x \rp + H_0 \lp t,x,v_0(t,x),(\nabla_x v_0)(t,x), (\Hess_x v_0)(t,x) \rp \geqslant 0
	\end{align}
	for $(t,x) \in (0,T) \times \mathcal{O}$. Hence $v_0$ is a viscosity solution of:
	\begin{align}
		\lp \frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + H_0 \lp t,x,u_0(t,x),(\nabla_x u_0)(t,x), (\Hess_x u_0)(t,x) \rp \leqslant 0
	\end{align}
	This completes the proof.	
\end{proof}

\begin{corollary}\label{unneq0}
	Let $d \in \N$, $T \in (0,\infty)$, let $\mathcal{O} \subseteq \R^d$ be a non-empty set, let $u_n: (0,T) \times \mathcal{O} \rightarrow \R$, $n \in \N_0$, be functions, let $G_n: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$, $n \in \N_0$ be degenerate elliptic, assume also that $G_0: (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ be consinuous and assume for all non-empty compact $\mathcal{K} \subseteq (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ it is the case that:
	\begin{align}
		\limsup_{n\rightarrow \infty} \lb \sup_{(t,x,r,p,A) \in \mathcal{K}} \lp \lv G_n \lp t,x,r,p,A \rp - G_0 \lp t,x,r,p,A \rp \rv + \lv u_n \lp t,x \rp - u_0 \lp t,x \rp \rv \rp \rb = 0
	\end{align} 
	and further assume for all $n \in \N$, that $u_n$ is a viscosity solution of:
	\begin{align}
		\lp \frac{\partial}{\partial t} u_n \rp \lp t,x \rp + G_n \lp t,x ,u_n \lp t,x \rp, \lp \nabla_x u_n \rp \lp t,x \rp, \lp \Hess_x u_n \rp \lp t,x \rp \rp = 0
	\end{align}
	for $(t,x) \in (0,T) \times \mathcal{O}$, then we have that $u_0$ is a viscosity solution of:
	\begin{align}
		\lp\frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + G_0 \lp t,x,u_0 \lp t,x \rp, \lp \nabla_x u_0 \rp \lp t,x \rp, \lp \Hess_x u_0 \rp \lp t,x \rp \rp= 0
	\end{align}
\end{corollary}
\begin{proof}
 	Note that Lemma \ref{ungeq0} gives us that $u_0$ is a viscosity solution of:
 	\begin{align}
		\lp \frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + G_n \lp t,x,u_0(t,x),(\nabla_x u_0)(t,x), (\Hess_x u_0)(t,x) \rp \geqslant 0
	\end{align}

for $(t,x) \in (0,T) \times \mathcal{O}$. Also note that Corollary \ref{unleq0} ensures that $u_0$ is a viscosity solution of:
	\begin{align}
		\lp \frac{\partial}{\partial t} u_0 \rp \lp t,x \rp + G_n \lp t,x,u_0(t,x),(\nabla_x u_0)(t,x), (\Hess_x u_0)(t,x) \rp \leqslant 0
	\end{align}
Taken together these prove the corollary.
\end{proof}
\begin{lemma}\label{absq}
	For all $a,b \in \R$ it is the case that $(a+b)^2 \leqslant 2a^2+2b^2$.
\end{lemma}
\begin{proof}
		Since for all $a,b\in\R$ it is the case that $(a-b)^2 \geqslant 0$ we then have that:
		\begin{align}
			(a+b)^2 &\leqslant (a+b)^2 + (a-b)^2 \nonumber \\
			&\leqslant a^2 + 2ab + b^2 + a^2 -2ab+b^2 \nonumber\\
			&=2a^2+2b^2 \nonumber
		\end{align}
		This completes the proof.
\end{proof}


\begin{lemma}\label{ugoesto0}
	Let $d,m \in \N$, $T \in (0,\infty)$. Let $\mathcal{O} \subseteq \R^d$ be a non-empty compact set, and for all $n\in \N_0$, $\mu_n \in C([0,T] \times \mathcal{O}, \R)$, $\sigma_n \in C([0,T] \times \mathcal{O}, \R^{d \times m})$
%	\begin{align}
%		\sup_{t\in [0,T]} \sup_{x\in \R^d} \sup_{y\in \R^d \setminus\{x\}} \lb \frac{\|\mu_n(t,x)-\mu_n(t,y)\|_E+\|\sigma_n(t,x) - \sigma_n(t,y)\|_F}{\|x-y\|_E} \rb < \infty
%	\end{align}
	assume also:
	\begin{align}\label{limsupis0}
		\limsup_{n \rightarrow \infty} \lb \sup_{t\in[0,T]} \sup_{x\in \mathcal{O}} \lp \right\|\mu_n(t,x) - \mu_0(t,x)\left\|_E + \left\|\sigma_n(t,x)-\sigma_0(t,x)\right\|_F \rp \rb = 0
	\end{align}
	Let $\lp \Omega, \mathcal{F}, \mathbb{R} \rp$ be a stochastic basis and let $W: [0,T] \times \Omega \rightarrow \R^m$ be a standard\\ $(\mathbb{F}_t)_{t\in [0,T]}$-Brownian motion for every $t\in [0,T]$, $x \in \mathcal{O}$, let $\mathcal{X}^{t,x} = (\mathcal{X}^{t,x}_s)_{s\in [t,T]}: [t,T] \times \Omega \rightarrow \R^d$ be an $(\mathbb{F}_s)_{s\in [t,T]}$ adapted stochastic process with continuous sample paths, satisfying for all $s \in [t, T]$ we have $\mathbb{P}$-a.s.
	\begin{align}\label{xnasintuvxn}
		\mathcal{X}^{n,t,x}_s = x + \int^s_t \mu_n(r,\mathcal{X}^{n,t,x}_s) dr + \int^s_t \sigma_n(r,\mathcal{X}^{n,t,x}_r) dW_r
	\end{align}
	then it is the case that:
	\begin{align}
		\limsup_{n\rightarrow \infty} \lb \sup_{t\in[0,T]}\sup_{s \in[t,T]}\sup_{x \in \mathcal{O}} \lp \E \lb \left\|\mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s\right\|_E^2\rb \rp \rb = 0
	\end{align}
	for $(t,x) \in (0,T) \times \R^d$.
\end{lemma}

\begin{proof}
	Since $\mathcal{O}$ is compact, let $L \in \R$ be such that for all $t \in [0,T]$, $x,y\in\mathcal{O}$ it is the case that:
	\begin{align}\label{lipformun}
		\|\mu_0(t,x) - \mu_0(t,y)\|_E - \|\sigma_0(t,x) + \sigma_0(t,y) \|_F \leqslant L\|x-y\|_E
	\end{align}
	
	Furthermore \cite[Theorem~5.2.9]{karatzas1991brownian} tells us that:
	\begin{align} \label{expofxisbounded}
		\sup_{s\in[t,T]}\E \lb \|\mathcal{X}^{n,t,x}_s\|_E\rb < \infty
	\end{align}
	Note now that (\ref{xnasintuvxn}) tells us that:
	\begin{align}\label{mathcalxn-mathclx0}
		\mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s = \int^s_t \mu_n(r,\mathcal{X}^{n,t,x}_s) - \mu_0(r,\mathcal{X}^{0,t,x}_s) dr + \int^s_t \sigma_n(r,\mathcal{X}^{n,t,x}_r) - \sigma_0(r,\mathcal{X}^{0,t,x}_r) dW_r
	\end{align} 
	Minkowski's Inequality applied to (\ref{mathcalxn-mathclx0}) then tells us for all $n\in \N$, $t\in [0,T]$, $s \in [t,T]$, and $x\in \mathcal{O}$ that:
	\begin{align}
		\lp \E \lb \left\|\mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s\right\|_E\rb \rp^{\frac{1}{2}} &\leqslant \int^s_t\lp \E \lb \left\|\mu_n(r, \mathcal{X}^{n,t,x}_r) - \mu_0(r,\mathcal{X}^{0,t,x}_r) \right\|_E^2 \rb\rp^{\frac{1}{2}} dr + \nonumber\\
		&\lp \E \lb \left\|\int^s_t (\sigma_n(r,\mathcal{X}^{n,t,x}_r)-\sigma_0(r,\mathcal{X}^{0,t,x}_r)) dW_r \right\|_E^2 \rb \rp^{\frac{1}{2}}
	\end{align}
	It\^o's isometry applied to the second summand yields:
	\begin{align}
		\lp \E \lb \left\|\mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s \right\|_E\rb \rp^{\frac{1}{2}} &\leqslant \int^s_t\lp \E \lb \left\|\mu_n(r, \mathcal{X}^{n,t,x}_r) - \mu_0(r,\mathcal{X}^{0,t,x}_r)\right\|_E^2 \rb\rp^{\frac{1}{2}} dr + \nonumber\\
		&\lp \int^s_t \E \lb \left\|\sigma_n(r,\mathcal{X}^{n,t,x}_r) - \sigma_0(r,\mathcal{X}^{0,t,x})\right\|_F^2 \rb dr \rp^\frac{1}{2}
	\end{align}
	Applying Lemma \ref{absq} followed by the Cauchy-Schwarz Inequality then gives us for all $n \in \N$, $t\in[0,T]$, $s \in [t,T]$, and $x\in \mathcal{O}$ that:
	\begin{align}
		\E \lb \|\mathcal{X}^{n,t,x}_s-\mathcal{X}^{n,t,x}_s	\|^2_E\rb &\leqslant 2 \lb \int^s_t \lp \E \lb \left\| \mu_n(r,\mathcal{X}^{n,t,x}_r) - \mu_0(r,\mathcal{X}^{0,t,x}_r)\right\|^2_E \rb \rp^{\frac{1}{2}} dr \rb^2 \nonumber\\
		&+2\int^s_t \E \lb \left\|\sigma_n(r,\mathcal{X}^{nt,x}_r) - \sigma_0(r,\mathcal{X}^{0,t,x}_r)\right\|_F^2 \rb dr \nonumber\\
		&\leqslant 2T \int^s_t \E \lb \left\|\mu_n(r,\mathcal{X}^{n,t,x}_r) -\mu_0(r,\mathcal{X}^{0,t,x}_r) \right\|_E^2\rb dr \nonumber\\
		&+ 2 \int^s_t \E \lb \left\|\sigma_n(r,\mathcal{X}^{n,t,x}_r) - \sigma_0(r,\mathcal{X}^{0,t,x}_r) \right\|_F^2\rb dr 
	\end{align}
	Applying Lemma \ref{absq} again to each summand then yields for all $n\in \N$, $t\in [0,T]$ $s\in [t,T]$, and $x\in \mathcal{O}$ it is the case that:
	\begin{align}
		&\E \lb \left\|\mathcal{X}^{n,t,x}_s-\mathcal{X}^{0,t,x}_s \right\|^2 \rb \nonumber \\
		&\leqslant 2T \int^s_t \lp 2\E \lb \left\|\mu_n(r,\mathcal{X}^{n,t,x}_r)-\mu_0(r,\mathcal{X}^{n,t,x}_r)\right\|^2_E\rb + 2\E \lb \left\|\mu_0(r,\mathcal{X}^{n,t,x}_r)-\mu_0(r,\mathcal{X}^{0,t,x}_r)\right\|^2_E\rb \rp dr \nonumber \\
		&+2\int^2_t \lp 2\E \lb \left\| \sigma_n(r,\mathcal{X}^{n,t,x}_r) - \sigma_0(r,\mathcal{X}^{n,t,x}_r) \right\|_F^2 \rb  + 2\E \lb\left\|\sigma_0(r,\mathcal{X}^{n,t,x}_r)-\sigma_0(r,\mathcal{X}^{0,t,x}_r)\right\|_F\rb \rp dr
	\end{align}
	However assumption (\ref{lipformun}) then gives us that for all $n\in \N$, $t \in [0,T]$, $s \in [t,T]$, and $x \in \mathcal{O}$ that:
	\begin{align}
		&\E \lb \left\|\mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s \right\|_E^2 \rb \leqslant 4L^2(T+1) \int^s_t\E\lb \left\|\mathcal{X}^{n,t,x}_r-\mathcal{X}^{0,t,x}_r \right\|_E^2 \rb dr \nonumber \\
		&+4T(T+1) \lb \sup_{r\in [0,T]}\sup_{y\in \R^d} \lp \left\| \mu_n(r,y) - \mu_0(r,y) \right\|_E^2 + \left\| \sigma_n(r,y) - \sigma_0(r,y) \right\|_F^2 \rp  \rb \nonumber 
	\end{align}
	
	Finally Gronwall's Inequality with assumption (\ref{expofxisbounded}) gives us for all $n\in \N$, $t\in [0,T]$, $s\in [t,T]$, $x \in \mathcal{O}$ that:
	\begin{align}
		&\E \lb \left\| \mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s \right\|_E^2 \rb \nonumber \\
		&\leqslant 4T(T+1) \lb \sup_{r\in [0,T]}\sup_{y\in \R^d} \lp \| \mu_n(r,y)-\mu_0(r,y) \|_E^2 + \|\sigma_n(r,y) - \sigma_)(r,y) \|_F^2 \rp \rb e^{4L^2T(T+1)} 
	\end{align}
	Applying $\limsup_{n\rightarrow \infty}$ to both sides and applying (\ref{limsupis0}) gives us for all $n \in \N$, $t \in [0,T]$, $s\in [t,T]$, $x \in \mathcal{O}$ that:
	\begin{align*}
		&\limsup_{n\rightarrow \infty} \E \lb \left\| \mathcal{X}^{n,t,x}_s - \mathcal{X}^{0,t,x}_s \right\|_E^2 \rb \nonumber
		\leqslant \\ &\limsup_{n\rightarrow \infty} \lb 4T(T+1) \lb \sup_{r\in [0,T]}\sup_{y\in \R^d} \lp \left\| \mu_n(r,y)-\mu_0(r,y) \right\|_E^2 + \left\|\sigma_n(r,y) - \sigma_0(r,y) \right\|_F^2 \rp \rb e^{4L^2T(T+1)} \rb \nonumber 
		\nonumber\\ &\leqslant \\ &4T(T+1) \lb \limsup_{n\rightarrow \infty} \lb\sup_{r\in [0,T]}\sup_{y\in \R^d} \lp \left\| \mu_n(r,y)-\mu_0(r,y) \right\|_E^2 + \left\|\sigma_n(r,y) - \sigma_0(r,y) \right\|_F^2 \rp \rb\rb e^{4L^2T(T+1)} \nonumber \\
		&\leqslant 0 \nonumber
	\end{align*} 
	This completes the proof.
\end{proof}

\begin{lemma}\label{2.19}
	Let $d,m \in \N$, $T \in (0,\infty)$, let $\mathcal{O} \subseteq [0,T] \times \R^d$, let $\mu \in C([0,T] \times \mathcal{O},\R^d)$ and $\sigma \in C([0,T] \times \mathcal{O}, \R^{d\times m})$ have compact supports such that $\supp(\mu) \cup \supp(\sigma) \subseteq [0,T] \times \mathcal{O}$
	let $g\in C(\R^d,\R)$. Let $\lp \Omega, \mathcal{F}, \mathbb{P}, (\mathbb{F}_t)_{t\in [0,T]} \rp$ be a stochastic basis, let $W:[0,T] \times \Omega \rightarrow \R^m$ be a standard $(\mathbb{F}_t)_{t\in [0,T]}$ Brownian motion, for every $t\in [0,T]$, $x\in \R^d$, let $\mathcal{X}^{t,x} = (\mathcal{X}^{t,x}_s)_{s\in [t,T]}: [t,T] \times \Omega \rightarrow \R^d$ be an $(\mathbb{F}_s)_{s\in [t,T]}$ adapted stochastic process with continuous sample paths satisfying for all $s\in [t,T]$ with $\mathbb{F}$-a.s. that:
	\begin{align}\label{2.59}
		\mathcal{X}^{t,x}_s = x + \int^s_t \mu \lp r, \mathcal{X}^{t,x}_r \rp dr + \int^s_t \sigma \lp r, \mathcal{X}^{t,x}_r \rp dW_r
	\end{align}
	and further let $u:[0,T] \times \R^d \rightarrow \R$ satisfy for all $t\in [0,T]$, $x\in \R^d$ that:
	\begin{align}\label{2.60}
		u(t,x) = \E \lb g \lp \mathcal{X}^{t,x}_T \rp \rb 
	\end{align}
	Then $u$ is a viscosity solution of:
	\begin{align}
		\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x)\lb \sigma(t,x) \rb^* \lp \Hess_xu \rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x u \rp \lp t,x \rp \ra = 0
	\end{align}
	and where $u(T,x) = g(x)$ for $(t,x) \in (0,T) \times \mathcal{O}$. 
\end{lemma}

\begin{proof}
	Let $\mathcal{S} = \supp(\mu) \cup \supp(\sigma) \subseteq [0,T] \times \mathcal{O}$ be bounded in space by $\rho \in (0,\infty)$, as $\mathcal{S} \subseteq [0,T] \times (-\rho, \rho)^d$. This exists as the supports are compact and thus by Hiene-B\"orel is closed and bounded. Let $\mathfrak{s}_n, \mathfrak{m}_n \in C^\infty([0,T] \times \R^d, \R^{d\times n})$ where $\bigcup_{n\in \N} \lb \supp(\mathfrak{s}_n) \cup \supp(\mathfrak{m}_n) \rb \subseteq [0,T] \times (-\rho, \rho)^d$ satisfy for $n\in \N$ that:
	\begin{align}\label{2.62}
		\limsup_{n\rightarrow \infty} \lb \sup_{t\in [0,T]} \sup_{x\in \R} \lp \left\| \mathfrak{m}_n(t,x) - \mu(t,x) \right\|_E + \left\|\mathfrak{s}_n - \sigma(t,x)\right\|_F \rp \rb = 0
	\end{align}
	We construct a set of degenerate elliptic functions, $G^{n}:(0,T) \times \R^d \times \R \times \R^d \times \mathbb{S}_d \rightarrow \R$ with $n \in \N_0$ such that:
	\begin{align} 		
		&G^{0}(t,x,r,p,A) = \frac{1}{2} \Trace\lp \sigma (t,x)[\sigma(t,x)]^*A \rp + \la \mu(t,x),p \ra \label{2.63}\\
		&\text{and} \nonumber\\
		&G^{n}(t,x,r,p,A) = \frac{1}{2} \Trace \lp \mathfrak{s}_n (t,x)[\mathfrak{s}_n(t,x)]^*A \rp + \la \mu(t,x),p \label{2.64}\ra 
	\end{align}
	Also let $\mathfrak{g}_n \in C^\infty(\R^d,\R)$ for $n\in \N$ satisfy for all $n\in \N$ that:
	\begin{align}\label{2.65}
		\limsup_{n\rightarrow \infty} \sup_{t\in [0,T]} \sup_{x\in \R^d} \lp \| \mathfrak{g}_n(x) - g(x) \|_E \rp = 0	
	\end{align}
	
	Further let $\mathfrak{X}^{n,t,x} = (\mathfrak{X}^{n,t,x}_s)_{s\in [t,T]}:[t,T] \times \Omega \rightarrow \R^d$ be an $(\mathbb{F}_s)_{s\in [t,T]}$-adapted stochastic process with continuous sample paths that satisfy:
	\begin{align}\label{2.66}
		\mathcal{X}^{n,t,x}_s = x + \int^s_t \mathfrak{m}_n(r, \mathcal{X}^{n,t,x}_r)dr + \int^s_t \mathfrak{s}_n(r,\mathcal{X}^{n,t,x}_r) dW_r
	\end{align} 
	% TODO Need to talk about karatzas and how the  proof really isnt there the one it's referencing
%	Note that then \cite{karatzas1991brownian} tells us that 
	Finally let $\mathfrak{u}^n: [0,T] \times \R^d \rightarrow \R$ for $n\in \N$ be:
	\begin{align}\label{ungn}
		\mathfrak{u}^n = \E \lb \mathfrak{g}_n \lp \mathfrak{X}^{n,t,x}_T \rp \rb
	\end{align}
	and:
	\begin{align}\label{u0gn}
		\mathfrak{u}^0 = \E \lb \mathfrak{g}_n \lp \mathcal{X}^{t,x}_T \rp \rb
	\end{align}
	
	Note that \cite[Lemma~2.2]{BHJ21} with $g \curvearrowleft \mathfrak{g}_k$, $\mu \curvearrowleft \mathfrak{m}_n$, $\sigma \curvearrowleft \mathfrak{s}_n$, $\mathcal{X}^{t,x} \curvearrowleft \mathcal{X}^{n,t,x}$ gives us $\mathfrak{u}^n \in C^{1,2}([0,T] \times \R^d, \R)$, and $\mathfrak{u}^n(t,x) = \mathfrak{g}_k(x)$ where:
	\begin{align}
		\lp \frac{\partial}{\partial t} \mathfrak{u}^n \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \mathfrak{s}_n(t,x)\lb \mathfrak{s}_n(t,x) \rb^* \lp \Hess_x \mathfrak{u}^n \rp \lp t,x \rp \rp + \la \mathfrak{m}_n(t,x), \lp \nabla_x \mathfrak{u}^n \rp \lp t,x \rp \ra = 0
	\end{align}
	And by Definitions \ref{def:viscsubsolution}, \ref{def:viscsupsolution}, and \ref{def:viscsolution} we have that $\mathfrak{u}^n$ is a viscosity solution of
	\begin{align}\label{2.70}
		\lp \frac{\partial}{\partial t} \mathfrak{u}^n \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \mathfrak{s}_n(t,x)\lb \mathfrak{s}_n(t,x) \rb^* \lp \Hess_x \mathfrak{u}^n \rp \lp t,x \rp \rp + \la \mathfrak{m}_n(t,x), \lp \nabla_x \mathfrak{u}^n \rp \lp t,x \rp \ra = 0
	\end{align}
	for $(t,x) \in (0,T) \times \R^d$.
\medskip

Since for all $n\in \N$, it is the case that $\mathcal{S} = \lp \supp(\mathfrak{m}_n)\cup \supp(\mathfrak{s}_n) \cup \supp(\mu) \cup \supp(\sigma) \rp \subseteq [0,T] \times (-\rho, \rho)^d$ and because of (\ref{2.59}) of (\ref{2.66}) we have that \cite[Lemma~3.2, Item~(ii)]{Beck_2021} which yields that for all $n\in \N$, $t\in [0,T]$, $x \in \R^d\setminus(-\rho, \rho)^d$ that $\mathbb{P}(\forall s \in [t,T]: \mathfrak{X}^{n,t,x}_s =x=\mathcal{X}^{t,x}_s)=1$. This in turn shows that for all $n\in \N$, $x \in \R^d \setminus(-\rho,\rho)^d$ that $\mathfrak{u}^n(t,x) = \mathfrak{u}^0(t,x)$ which along with (\ref{ungn}) and (\ref{u0gn}) yields that:
	\begin{align}\label{unu0}
	\sup_{t\in[0,T]}\sup_{x \in \R^d} \lb \lv \mathfrak{u}^n (t,x) - \mathfrak{u}^0(t,x) \rv \rb &= \sup_{t\in [0,T]} \sup_{x \in (-\rho,\rho)^d} \lb \lv \mathfrak{u}^n(t,x) - \mathfrak{u}^0(t,x) \rv \rb \nonumber\\
	&\leqslant \sup_{t\in [0,T]}\sup_{x \in (-\rho,\rho)^d} \lp \E \lb \lv \mathfrak{g}_k \lp \mathfrak{X}^{n,t,x}_T \rp - \mathfrak{g} \lp \mathcal{X}^{t,x}_T \rp \rv \rb \rp 
	\end{align}
	Note that Lemma \ref{ugoesto0} allows us to conclude  that:
	
	\begin{align}
		\limsup_{n\rightarrow \infty} \lb \sup_{t\in[0,T]} \sup_{x\in (-\rho, \rho)^d} \lp \E \lb \|\mathfrak{X}^{n,t,x}_T-\mathcal{X}^{t,x}_s \| \rb \rp \rb = 0
	\end{align} 
	But then we have that (\ref{unu0}) which yields that:
	\begin{align}\label{2.73}
		\limsup_{n\rightarrow 0} \lb \sup_{t\in [0,T]} \sup_{x \in \R^d} \lp \lv \mathfrak{u}^n(t,x) - \mathfrak{u}^0(t,x) \rv\rp \rb =0
	\end{align}
	However now note that (\ref{2.64}) and (\ref{2.70}) thus yield that for $n \in \N_0$, $\mathfrak{u}^n$ is a viscosity solution to:
	\begin{align}\label{2.74}
		\lp \frac{\partial}{\partial t} \mathfrak{u}^n \rp \lp t,x \rp + G^n \lp t,x,\mathfrak{u}^n \lp t,x \rp, \lp \nabla_x \mathfrak{u}^n \rp \lp t,x \rp, \lp \Hess_x \mathfrak{u}^n \rp \lp t,x \rp \rp =0
	\end{align}
	But since we've established (\ref{2.62}) we have that for a non-empty compact set $\mathcal{C} \subseteq (0,T) \times \mathcal{O} \times \R \times \R^d \times \mathbb{S}_d$ that:
	\begin{align}
	&\limsup_{n \rightarrow \infty} \lb \sup_{(t,x,r,p,A) \in \mathcal{C}} \lv G^n \lp t,x,r,p,A \rp - G^0 \lp t,x,r,p,A \rp \rv \rb \nonumber \\
	&\leqslant \limsup_{n\rightarrow \infty} \lb \sup_{(t,x,r,p,A) \in \mathcal{C}} \left\|\mu(t,x) - \mathfrak{m}_n(t,x) \|_E \right\| p \|_E \rb \nonumber \\
	&+ \limsup_{n\rightarrow \infty} \lb \sup_{(t,x,r,p,A) \in \mathcal{C}} \left\| \sigma(t,x)\lb \sigma(t,x) \rb^* - \mathfrak{s}_n(t,x) \lb \mathfrak{s}_n(t,x) \rb^* \right\|_F \left\|A\right\|_F \rb = 0	
	\end{align}
	
	This, together with (\ref{2.73}), (\ref{2.74}) and Corollary \ref{unneq0} yields that $\mathfrak{u}^0$ is also a viscosity solution to:
	\begin{align}\label{2.76}
		\lp \frac{\partial}{\partial t} \mathfrak{u}^0 \rp \lp t,x \rp + G^0 \lp t,x,\mathfrak{u}^0(t,x), \lp \nabla_x \mathfrak{u}^0\rp \lp t,x \rp, \lp \Hess_x \rp \lp t,x \rp \rp =0 
	\end{align}
	Finally note that (\ref{2.62}), (\ref{2.66}), (\ref{u0gn}), and 	(\ref{2.76}) yield that $u$ is a viscosity solution of::
	\begin{align}
		\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x) \lb \sigma(t,x) \rb^* \lp \Hess_x u \rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x \rp \lp t,x \rp \ra = 0 
	\end{align}
	for $(t,x) \in [0,T] \times \R^d$. Finally (\ref{2.59}) and (\ref{2.60}) allows us to conclude that for all $x \in \R^d$ it is the case that $u(T,x) = g(x)$. This concludes the proof.	
\end{proof}

\begin{lemma}
	Let $d,m \in \N$, $T \in (0,\infty)$, further let $\mathcal{O} \subseteq \R^d$ be a non, empty compact set. Let every $r \in (0,\infty)$ satisfy the condition that $O_r \subseteq \mathcal{O}$, where $O_r = \{x \in \mathcal{O}: \lp \|x\|_E \leqslant r \text{ and }\{y \in \R^d: \| y-x\|_E < \frac{1}{r} \} \subseteq \mathcal{O} \rp \}$ let $g \in C(\mathcal{O},\R)$, $\mu \in C([0,T] \times \mathcal{O},\R)$, $V \in C^{1,2}([0,T] \times \mathcal{O},(0,\infty))$, assume that for all $t \in [0,T]$, $x \in \mathcal{O}$ that:
	\begin{align}
		\sup \lp \left\{ \frac{\| \mu(t,x)-\mu(t,y)\|_E+\|\sigma(t,x)-\sigma(t,y)\|_F}{\|x-y\|_E}:t\in [0,T], x,y\in O_r, x \neq y \right\} \cup \{0\}\rp < \infty
	\end{align}
	\begin{align}
		\lp \frac{\partial}{\partial t} V \rp \lp t,x \rp + \frac{1}{2}\Trace \lp \sigma(t,x)\lb \sigma(t,x) \rb^* \lp \Hess_x V\rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x V \rp (t,x) \ra \leqslant 0 
	\end{align}	
	assume that $\sup_{r \in (0,\infty)} \lb \inf_{x \in \mathcal{O}\setminus O_r} V(t,x)\rb = \infty$ and $\inf_{r \in (0, \infty)} \lb \sup_{t \in [0,T]} \sup _{x \in \mathcal{O} \setminus O_r} \lp \frac{g(x)}{V(T,x)} \rp \rb = 0$.
	Let $\lp \Omega, \mathcal{F}, \mathbb{P}, (\mathbb{F}_t)_{t\in [0,T]} \rp$ be a stochastic basis and let $W: [0,T] \times \Omega \rightarrow \R^m$ be a standard $(\mathbb{F}_t)_{t \in [0,T]}$-Brownian motion, for every $t \in [0,T]$, $x \in \mathcal{O}$ let $\mathcal{X}^{t,x} = (\mathcal{X}^{t,x}_s)_{s\in [t,T]}: [t,T] \times \Omega \rightarrow \mathcal{O}$ be an $(\mathbb{F}_s)_{s\in [t,T]}$-adapted stochastic process with continuous sample paths satisfying that for all $s\in [t, T]$, we have $\mathbb{P}$-a.s. that:
	\begin{align}\label{2.79}
		\mathcal{X}^{t,x}_s = x+\int^s_t \mu(r,\mathcal{X}^{t,x}_r) dr + \int^s_t \sigma(r, \mathcal{X}^{t,x}_n) dW_r
	\end{align}
	also let $u:[0,T] \times \R^d \rightarrow \R$ satisfy for all $t \in [0,T]$, $x \in \R^d$ that:
	\begin{align}
		u(t,x) = \E \lb u(T,\mathcal{X}^{t,x}_T) \rb
	\end{align}
	It is then the case that $u$ is a viscosity solution to:
	\begin{align}\label{2.81}
		\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x) \lb \sigma(t,x) \rb^* \lp \Hess_x u \rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x \rp \lp t,x \rp \ra = 0 
	\end{align}
	for $(t,x) \in (0,T) \times \mathcal{O}$ with $u(T,x) = g(x)$.
	\end{lemma}
\begin{proof}
	Let it be the case, that throughout the proof, for $n\in \N$, we have that $\mathfrak{g}_n \in C(\R^d, \R)$, compactly supported and that $\lb \bigcup_{n\in\N}  \supp(\mathfrak{g}_m) \rb \subseteq [0, T] \times \mathcal{O}$ and further that:
	\begin{align}\label{2.82}
		\limsup_{n\rightarrow \infty} \lb \sup_{t\in [0,T]}\sup_{x\in \mathcal{O}} \lp \frac{|\mathfrak{g}_n(x)-g(x)|}{V(T,x)} \rp \rb = 0  
	\end{align}
	Let is also be the case that for $n\in \N$, $\mathfrak{m}_n \in C([0,T] \times \R^d, \R^d)$ and $\mathfrak{s}_n \in C([0,T] \times \R^d, \R^{d \times m})$ satisfy:
	\begin{enumerate}[label = (\roman*)]
	\item for all $n\in \N$:
	\begin{align}
		\sup_{t \in [0,T]} \sup_{x,y \in \R^d,x\neq y} \lb \frac{\|\mathfrak{m}_n(t,y)-\mathfrak{m}_n(t,y)\|_E + \|\mathfrak{s}_n(t,x)-\mathfrak{s}_n(t,y)\|_E}{\| x-y\|_E} \rb = 0
	\end{align}
	\item for all all $n\in \N$, $t \in [0,T]$, $x \in \mathcal{O}$:
	\begin{align}\label{2.84}
		\mathbbm{1}_{\{V \leqslant n \}}(t,x)\lb \| \mathfrak{m}_n(t,x) - \mu(t,x)\|_E + \| \mathfrak{s}_n(t,x) - \sigma(t,x) \|_F \rb = 0
	\end{align}
	and
	\item for all $n\in \N$, $t \in [0,T]$, $x \in \R^d \setminus \{V \leqslant n+1\}$ that:
	\begin{align}
		\| \mathfrak{m}_n(t,x)\|_E + \| \mathfrak{s}_n (t,x) \|_F = 0
	\end{align}
	\end{enumerate}
	Next for every $n\in \N$, $t\in [0,T]$ and $x\in \R^d$ let it be the case that $\mathfrak{X}^{n,t,x}_s = (\mathfrak{X}^{n,t,x}_s)_{s\in [t,T]}: [t,t] \times \Omega. \rightarrow \R^d$ be a stochastic process with continuous sample paths satisfying:
	\begin{align}\label{2.86}
		\mathfrak{X}^{n,t,x}_s = x + \int^s_t \mathfrak{m}_n(r, \mathfrak{X}^{n,t,x}_s) dr + \int^s_t \mathfrak{s}_n(r, \mathfrak{X}^{n,t,x}_s) dW_r
	\end{align} 
	Let $\mathfrak{u}^n: [0,T] \times \R^d \rightarrow \R$, $k \in \N$, $n \in \N_0$, satisfy for all $n\in \N$, $t \in [0,T]$, $x \in \R^d$ that:
	\begin{align}
		\mathfrak{u}^{n,k}(t,x) = \E \lb \mathfrak{g}_k (\mathfrak{X}^{n,t,x}_T) \rb 
	\end{align}
	and
	\begin{align}\label{2.88}
		\mathfrak{u}^{0,k}(t,x) = \E \lb \mathfrak{g}_k \lp \mathcal{X}^{t,x}_T \rp \rb 
	\end{align}
	and finally let, for every $n\in \N$, $t \in [0,T]$, $x \in \mathcal{O}$, there be $\mathfrak{t}^{t,x}_n: \Omega \rightarrow [t,T]$ which satisfy $\mathfrak{t}^{t,x}_n = \inf \lp \{ s \in [t,T], \max \{V(s,\mathfrak{X}^{t,x}_s),V(s,\mathcal{X}^{t,x}_s)\} \geqslant n \} \cup \{T\} \rp$. We may apply Lemma \ref{2.19} with $\mu \curvearrowleft \mathfrak{m}_n$, $\sigma \curvearrowleft \mathfrak{s}_n$, $g \curvearrowleft \mathfrak{g}_k$ to show that for all $n,k \in \N$ we have that $\mathfrak{u}^{n,k}$ is a viscosity solution to:
	\begin{align}\label{2.89}
		&\lp \frac{\partial}{\partial t} \mathfrak{u}^{n,k} \rp (t,x) + \frac{1}{2} \Trace \lp \mathfrak{s}_n(t,x) \lb \mathfrak{s}_n(t,x) \rb^* \lp \Hess_x \mathfrak{u}^{n,k} \rp (t,x) \rp + \la \mathfrak{m}_n(t,x), \lp \nabla_x(\mathfrak{u}^{n,k} \rp(t,x) \ra \nonumber\\& = 0 \nonumber
	\end{align}  
	for $(t,x) \in (0,T) \times \R^d$. But note that items (i)-(iii) and \ref{2.86} give us that, in line with \cite[Lemma~3.5]{Beck_2021}:
	\begin{align}
		\mathbb{P}\lp \forall s \in [t,T]: \mathbbm{1}_{\{s \leqslant \mathfrak{t}^{t,x}_n\}} \mathfrak{X}^{n,t,x}_s = \mathbbm{1}_{\{s\leqslant \mathfrak{t}^{t,x}_n\}} \mathcal{X}^{t,x}_s \rp =1
	\end{align}
	Further this implies that for all $n,k \in \N$, $t \in [0,T]$, $x \in \mathcal{O}$ that:
	\begin{align}
		\E \lb \lv \mathfrak{g}_k \lp \mathfrak{X}^{n,t,x}_T) - \mathfrak{g}_k(\mathcal{X}^{t,x}_T \rp \rv \rb &= \E \lb \mathbbm{1}_{\{\mathfrak{t}^{t,x}_n < T\}} \lv \mathfrak{g}_k(\mathfrak{X}^{n,t,x}_T) - \mathfrak{g}_k(\mathcal{X}^{t,x}_T) \rv \rb \nonumber\\
		&\leqslant 2 \lb \sup_{y \in \mathcal{O}} \lv \mathfrak{g}_k(y) \rv \rb \mathbb{P} \lp \mathfrak{t}^{t,x}_n < T \rp \nonumber
	\end{align}
	Note that this combined with \cite[Lemma~3.1]{Beck_2021} implies for all $t \in [0,T]$, $x \in \mathcal{O}$, $n \in \N$ we have that $\E \lb V \lp \mathfrak{t}^{t,x}_n, \mathcal{X}^{t,x}_{\mathfrak{t}^{t,x}_n}\rp \rb \leqslant V(t,x)$, which then further proves that:
	\begin{align}
		\lv \mathfrak{u}^{n,k}(t,x) - \mathfrak{u}^{0,k}(t,x) \rv &\leqslant 2 \lb \sup_{y\in \mathcal{O}} \lv \mathfrak{g}_k(y) \rv \rb \mathbb{P}\lp \mathfrak{t}^{t,x}_n < T \rp \nonumber\\
		&\leqslant 2 \lb\sup _{y \in \mathcal{O}} \lv \mathfrak{g}_k(y) \rv \rb \mathbb{P} \lp V \lp \mathfrak{t}^{t,x}_n, \mathcal{X}^{t,x}_{\mathfrak{t}^{t,x}_n}\rp \geqslant n\rp \nonumber \\
	%TODO: Ask Dr. Padgett how this came about	
		&\leqslant \frac{2}{n} \lb \sup_{y\in \mathcal{O}} \lv \mathfrak{g}_k(y) \rv \rb \E \lb V \lp \mathfrak{t}^{t,x}_n, \mathcal{X}^{t,x}_{\mathfrak{t}^{t,x}_n} \rp \rb \nonumber \\
		&\leqslant \frac{2}{n} \lb \sup_{y \in \mathcal{O}} \lv \mathfrak{g}_k(y) \rv \rb V \lp t,x, \rp \nonumber
	\end{align} 
	Together these imply that for all $k\in\N$ and compact $\mathcal{K} \subseteq [0,T] \times \mathcal{O}$:
	\begin{align}\label{2.90}
		\limsup_{k \rightarrow \infty}\lb \sup_{(t,x) \in \mathcal{K}} \lp \lv \mathfrak{u}^{n,k}(t,x) - \mathfrak{u}^{0,k}(t,x) \rv \rp \rb = 0
	\end{align}
	But again note that since have that $\sup_{r\in (0,\infty)}\lb \inf_{t \in [0,T], x \in \R^d \setminus O_r} V(t,x) \rb = \infty$ and (\ref{2.84}) tell us that for all compact $\mathcal{K} \subseteq [0,T] \times \mathcal{O}$ we have that:
	\begin{align}
		\limsup_{n \rightarrow \infty} \lb \sup_{(t,x) \in \mathcal{K}} \lp \|\mathfrak{m}_n(t,x) - \mu(t,x) \|_E + \| \mathfrak{s}_n(t,x)-\sigma(t,x) \|_F \rp \rb = 0  
	\end{align}
	Note that (\ref{2.89}), (\ref{2.90}) and Corollary \ref{unneq0} tell us that for all $k \in \N$ we have that $\mathfrak{u}^{0,k}$ is a viscosity solution to:
	\begin{align}\label{2.93}
		\lp \frac{\partial}{\partial t} \mathfrak{u}^{0,k} \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x) \lb \sigma(t,x) \rb^* \lp \Hess_x \mathfrak{u}^{0,k} \rp (t,x) \rp + \la \mu(t,x), \lp \nabla_x \mathfrak{u}^{0,k} \rp (t,x) \ra = 0
	\end{align}
	for $(t,x) \in (0,T) \times \mathcal{O}$. However note that (\ref{2.79}),(\ref{2.82}), (\ref{2.88}) prove that for all compact $\mathcal{K} \subseteq [0,T] \times \mathcal{O}$ we have:
	\begin{align}
		\limsup_{k \rightarrow \infty} \lb \sup_{(t,x) \in \mathcal{K}} \lv \mathfrak{u}^{0,k}(t,x) - u(t,x) \rv \rb = 0
	\end{align}
	This together with (\ref{2.93}), (\ref{2.82}), Corollary \ref{unneq0} shows that $u_0$ is a viscosity solution to:
	\begin{align}\label{2.95}
		\lp \frac{\partial}{\partial t} u \rp \lp t,x \rp + \frac{1}{2} \Trace \lp \sigma(t,x) \lb \sigma(t,x) \rb^*\lp \Hess_x u \rp (t,x) \rp + \la \mu(t,x), \lp \nabla_x u \rp \ra = 0
	\end{align}
	for $(t,x) \in (0,T) \times \mathcal{O}$. By (\ref{2.81}) we are ensured that for all $x\in \R^d$ we have that $u(T,x) = g(x)$ which together with proves the proposition.

\end{proof}


\section{Solutions, Characterization, and Computational \\ Bounds}
% \begin{proof}
% From Feynman-Kac, especially from \cite[(1.5)]{hutzenthaler_strong_2021} and setting $f=0$ in the notation of \cite[(1.5)]{hutzenthaler_strong_2021} we have that:
% \begin{align}
%     u_d(t,x) = \E \left[ u_d \left( 0,x+ \sqrt{2}W^{d}_t \right) \right]
% \end{align}

% Substituting (2.2) and applying the time reversal property of Brownian motions then gives us: 
% \begin{align}
%     u_d(t,x) &= \E \left[ u_d \left( 0,x+ \sqrt{2}W^{d}_t \right) \right] \nonumber \\
%     &= \E \left[ u_d \left( 0,x+ \sqrt{2}W^{d}_{-t} \right) \right] \nonumber \\
%     &= \E \left[ u_d \left( 0,x+ \sqrt{2}W^{d}_{0-t} \right) \right] \nonumber \\
%     &= \E \left[ u_d \left( 0, \mathcal{X}^{d,t,x}_{0} \right) \right]
% \end{align}
% Looking closely at (2.3), where according to the notation of (2.2), the time reversal property of Brownian motions, (2.5), and the shift-invariance of Brownian motions we have that:

% \begin{align}
%     u_d(t,x) = \E \left[ u_d \left( T,\mathcal{X}_T^{d,t,x} \right) \right] &= \E \left[ u_d\left( T,x+\sqrt{2}W^{d,\theta}_{T-t}\right) \right] \nonumber \\
%     &= \E \left[ u_d\left( T,x+\sqrt{2}W^{d,\theta}_{t-0}\right) \right] \nonumber \\
%     &= \E \left[ u_d\left( T,x+\sqrt{2}W^{d,\theta}_{0-t}\right) \right] \nonumber \\
%     &= \E \left[ u_d\left(T, \mathcal{X}^{d,t,x}_0\right) \right]
% \end{align}
% The independence of the Brownian motions then indicates that (2.5) and (2.6) are equal.

% This completes the proof of Lemma 2.1.
% \end{proof}



\begin{theorem}[Existence and characterization of $u_d$]\label{thm:3.21}
Let $T \in (0,\infty)$. Let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. Let $\sigma_d \in C \lp \R^d,\R^{d\times d} \rp $ and $\mu_d\in C \lp \R^d, \R^d \rp$ for $d \in \N$, let $u_d \in C^{1,2} \lp \lb 0,T \rb \times \R^d, \R \rp$ satisfy for all $d \in \N$, $t \in \lb 0,T \rb$ , $x \in \R^d$ that:
	\begin{align}\label{3.3.1}
		\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \frac{1}{2}\Trace\lp \sigma_d(x) \lb \sigma_d(x) \rb^* \lp \Hess_x u_d \rp (t,x) \rp + \la \mu_d(x), \lp \nabla_x u_d \rp(t,x) \ra = 0
	\end{align}
	let $\mathcal{W}^d:[0,T] \times \Omega \rightarrow \R^d$, $d \in \N$ be a standard Brownian motions and let $\mathcal{X}^{d,t,x}: \lb t, T\rb \times \Omega \rightarrow \R^d$, $d \in \N$, $ t\in \lb 0,T \rb$, be a stochastic process with continuous sample paths satisfying for all $d \in \N$, $t \in \lb 0,T \rb$, $s \in \lb t,T \rb$, $x \in \R^d$, we have $\mathbb{P}$-a.s. that:
	\begin{align}\label{3.102}
		\mathcal{X}^{d,t,x} = x + \int^t_s \mu_d \lp \mathcal{X}^{d,t,x}_r \rp dr + \int^t_s \sigma \lp \mathcal{X}^{d,t,x}_r \rp d\mathcal{W}^d_r
	\end{align} 
	Then for all $d \in N$, $t \in \lb 0,T \rb$, $x \in \R$, it holds that:
	\begin{align}
		u_d(t,x) = \E \lb u_d \lp T,\mathcal{X}_t^{d,t,x} \rp \rb
	\end{align}
	Furthermore, $u_d$ is a viscosity solution to (\ref{3.3.1}).
\end{theorem}


\begin{proof}
	This is a consequence of Lemma \ref{lem:3.4} and \ref{2.19}. 
\end{proof}
\begin{corollary}\label{lem:3.19} Let $T \in (0,\infty)$,\\ let $\left( \Omega, \mathcal{F}, \mathbb{P} \right)$ be a probability space, let $u_d \in C^{1,2} \left( \left[ 0,T \right] \times \R^d, \R \right)$, $d \in \N$ satisfy for all $d \in \N$, $t \in [0,T]$, $x \in \R^d$ that:
\begin{align}
    \left( \frac{\partial}{\partial t} u_d \right) \left(t,x\right) + \frac{1}{2}\left(\nabla^2_x u_d\right) \left(t,x\right) = 0
\end{align}
Let $\mathcal{W}^d: [0,T] \times \Omega \rightarrow \R^d$, $d \in \N$ be standard Brownian motions, and let $\mathcal{X}^{d,t,x}: [t,T] \times \Omega \rightarrow \R^d$, $d\in \N$, $t \in [0,T]$, $x \in \R^d$, be a stochastic process with continuous sample paths satisfying that for all $d \in N$, $t \in [0,T]$, $s \in [t,T]$, $x \in \R^d$ we have $\mathbb{P}$-a.s. that: 
\begin{align}
    \mathcal{X}^{d,t,x}_s = x + \int^s_t d\mathcal{W}^d_r = x + \mathcal{W}^d_{t-s}
\end{align}
Then for all $d\in \N$, $t \in [0,T]$, $x \in \R^d$ it holds that:
\begin{align}
    u_d(t,x) = \E \left[u_d \left(T,\mathcal{X}^{d,T,x}_t\right)\right]
\end{align}
\end{corollary}
\begin{proof}
	This is a special case of Theorem \ref{thm:3.21}. It is the case where $\sigma_d(x) = \mathbb{I}_d$, the uniform identity function where $\mathbb{I}_d$ is the identity matrix in dimension $d$ for $d \in \N$, and $\mu_d(x) = \mymathbb{0}_{d}$ where $\mymathbb{0}_d$ is the zero vector in dimension $d$ for $d \in \N$.
\end{proof}
 


\begin{lemma}\label{3.3.2}
	Let $T \in \lp 0,\infty \rp$, let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$, be a probability space, let $\alpha_d\in C^2_b \lp \R^d,\R \rp$, and $\alpha \in \mathcal{O}\lp x^2 \rp $ for $d \in N$, be infinitey often differentiable function, let $u_d \in C^{1,2} \lp \lb 0,T \rb \times \R^d,\R \rp$, $d \in \N$, satisfy for all $d \in \N$, $t \in \lb 0,T \rb$, $x \in \R^d$, that: 
	\begin{align}\label{3.3.7}
		\lp \frac{\partial}{\partial t} u_d\rp \lp t,x \rp + \frac{1}{2}\lp \nabla^2_x u_d \rp \lp t,x \rp +\alpha_d \lp x \rp u_d \lp t,x \rp = 0 
	\end{align}
	Let $\mathcal{W}^d: \lb 0,T \rb \times \Omega \rightarrow \R^d$ be standard Brownian motions and let $\mathcal{X}^{d,t,x}: \lb t,T \rb \times \Omega \rightarrow \R^d$, $d \in \N$, $t \in \lb 0,T \rb$, $x \in \R^d$ be a stochastic process with continuous sample paths satisfying that for all $d \in \N$, $ t\in \lb 0,T \rb$, $s \in \lp t,T \rb$, $x \in \R^d$, we have $\mathbb{P}$-a.s. that:
	\begin{align}
		\mathcal{X}^{d,t,x}_s = x + \int^t_s \frac{1}{2} d \mathcal{W}^d_r =\frac{1}{2} \mathcal{W}^d_{t-r}
	\end{align}
	Then for all $d \in \N$, $t \in \lb 0,T \rb$, $x \in \R^d$ it holds that:
	\begin{align}
		u_d \lp t,x \rp = \E \lb \exp \lp \int^T_t \alpha_d \lp \mathcal{X}^{d,t,x}_r \rp dr \rp u_d \lp T, \mathcal{X}^{d,t,x}_T \rp \rb 
	\end{align}
\end{lemma}
\begin{proof} Let $v_d: \R^d \rightarrow \R$ be continuous. Throughout the proof let $u_d \lp t,x\rp = e^{-t\alpha_d(x)}v_d(t,x)$ for all $d \in \N$, $t \in [0,T]$, $x \in \R^d$. For notational simplicity, we will drop the $d,t,x$ wherever it is obvious.  Therefore the derivatives become:
%\begin{align}
%	\lp \frac{\partial}{\partial t}u_d \rp \lp t,x \rp &= -\alpha_d(x)e^{-t\alpha_d(x)}v_d(t,x)+e^{-t\alpha_d(x)} v_d^{(1,0)}(t,x) \nonumber\\
%	\lp \nabla_x^2 u_d \rp \lp t,x \rp &= e^{-t\alpha_d(x)} \lb -2t\alpha_d'(x) (\nabla_x v_d) (t,x) + tv(t,x) \lb t\lp \alpha_d'(x) \rp^2 -\alpha_d''(x)\rb  + \lp \nabla_x^2v_d \rp (t,x)\rb
%\end{align}
%	Substituting this into (\ref{3.3.7}) renders it as:
%	\begin{align}
%		0 &= - \cancel{\alpha_d(x)e^{-t\alpha_d(x)}v_d(t,x)}+e^{-t\alpha_d(x)} v_d^{(1,0)} (t,x) \nonumber\\
%		&+ e^{-t\alpha_d(x)} \lb -2t\alpha_d'(x) (\nabla_x v_d) (t,x) + tv(t,x) \lb t\lp \alpha_d'(x) \rp^2 -\alpha_d''(x)\rb  + v_d^{(0,2)}(t,x)\rb \nonumber\\
%		&+ \cancel{\alpha_d(x)e^{-t\alpha_d(x)}v_d(t,x)} \nonumber \\
%		&= e^{-t\alpha_d(x)} v_d^{(1,0)}(t,x) \nonumber\\
%		&+ e^{-t\alpha_d(x)} \lb -2t\alpha_d'(x) v_d^{(0,1)} (t,x) + tv(t,x) \lb t\lp \alpha_d'(x) \rp^2 -\alpha_d''(x)\rb  + v_d^{(0,2)}(t,x)\rb \nonumber
%	\end{align}
%	Multiplying both sides by $e^{-t\alpha_d(x)}$ then gives us:
%	\begin{align}
%		v_d^{(1,0)}(t,x) -2t\alpha'_d(x)v_d^{(0,1)}(t,x)+tv(t,x)\lb t \lp \alpha_d'(x) \rp^2 - \alpha''_d(x) \rb + v_d^{(0,2)}(t,x) &= 0 \nonumber\\
%		v_d^{(1,0)}(t,x) = 2t\alpha'_d(x)v_d^{(0,1)}(t,x)-v(t,x)\lb t^2 \lp \alpha_d'(x) \rp^2 - t\alpha''_d(x) \rb - v_d^{(0,2)}(t,x) \nonumber\\
%	\end{align}
%	From the variable substitution observe that:
%	\begin{align}
%		v_d(t,x) = e^{t \alpha_d(x)}u_d(t,x) \nonumber
%	\end{align}
%	And hence:
%	\begin{align}
%		v_d(0,x) = e^{0}u_d(0,x) \nonumber\\
%		v_d(0,x) = u_d(0,x) \nonumber
%	\end{align}
\begin{align}
	u_t  &= -\alpha e^{-t\alpha}v+e^{-t\alpha} v_t \\
	\frac{1}{2}\nabla^2_x u &= \frac{1}{2} \lb e^{-t\alpha} \nabla^2_xv+2\la \nabla_x v, \nabla_x e^{-t\alpha} \ra + v\nabla^2_x e^{-t\alpha} \rb
%	u_{xx} &= \frac{1}{2}e^{-t\alpha} \lb -2t\alpha_x v_x + tv \lb t(\alpha_x)^2-\alpha_{xx}\rb + v_{xx}\rb 
\end{align}
This then renders (\ref{3.3.7}) as:
\begin{align}
	\cancel{-\alpha e^{-t\alpha}v} + e^{-t \alpha}v_t + \frac{1}{2} \lb e^{-t\alpha }\nabla^2_xv + 2\la \nabla_xv,\nabla_xe^{-t\alpha}\ra +v\nabla_x^2e^{-t\alpha} \rb + \cancel{\alpha e^{-t\alpha}v} &= 0 \nonumber\\
	e^{-t \alpha}v_t + \frac{1}{2} \lb  e^{-t\alpha }\nabla^2_xv - 2te^{-t\alpha}\la \nabla_x v, \nabla_x \alpha  \ra +v\nabla_x^2e^{-t\alpha} \rb &= 0 \nonumber\\
	e^{-t \alpha}v_t + \frac{1}{2} \lb e^{-t\alpha }\nabla^2_xv - 2te^{-t\alpha}\la \nabla_x v, \nabla_x \alpha  \ra -tve^{-t\alpha}\nabla^2_x\alpha \rb &= 0  \nonumber\\
	v_t + \frac{1}{2} \lb \nabla_x^2v-2t\la \nabla_xv,\nabla_x\alpha \ra -tv\nabla^2_x\alpha \rb &= 0 \nonumber \\
	v_t + \frac{1}{2} \lb \nabla_x^2v-2t\la \nabla_x \alpha,\nabla_x v \ra -tv\nabla^2_x\alpha \rb &= 0 \nonumber \\
	v_t + \frac{1}{2}\nabla_x^2v+\la -t\nabla_x \alpha,\nabla_x v \ra -\frac{1}{2}tv\nabla^2_x\alpha &= 0 \label{3.3.12}
\end{align}
%\begin{align}
%	\cancel{-\alpha e^{-t\alpha}v}+e^{-t\alpha} v_t + \frac{1}{2}e^{-t\alpha} \lb -2t\alpha_xv_x + tv \lb t(\alpha_x)^2-\alpha_{xx}\rb + v_{xx}\rb + \cancel{\alpha e^{-t\alpha}v} &= 0 \nonumber\\
%	e^{-t\alpha}v_t + \frac{1}{2} e^{-t\alpha} \lb -2t\alpha_x v_x + tv \lb t(\alpha_x)^2- \alpha_{xx} \rb + v_{xx} \rb &= 0 \nonumber \\
%	v_t -t\alpha_xv_x+\frac{1}{2}tv\lb t(\alpha_x )^2-\alpha_{xx} \rb + v_{xx} &= 0 \nonumber \\
%	v_t = 2t\alpha_xv_x - tv \lb t(\alpha_x)^2-\alpha_{xx} \rb &-v_{xx}
%\end{align} 
Let $\sigma(t,x) = \mathbb{I}_d$, i.e. the uniform identity function. Let $\mu(t,x) = -t\nabla_x \alpha$ for $t \in [0,T], x\in \R^d$, and for fixed $\alpha$. Let $f(t,x,v) = -\frac{1}{2}tv\nabla_x^2 \alpha$ for $t \in [0,T], x \in \R^d$.

\begin{claim}
	It is the case that for for all $x \in \R^d$ and $t \in [0,T]$ that $\la x, \mu(t,x) \ra \leqslant L \lp 1 + \|x\|_E \rp$ for some constant $L \in (0,\infty)$.  
\end{claim}
\begin{proof}
Since $\alpha$ has bounded first and second derivatives let:
\begin{align}
 \mathfrak{B} = \max \left\{\sup_{x \in \R^d} \| \nabla_x \alpha \|_E, \sup_{x\in \R^d} \lv \nabla_x^2\alpha \rv \right\} \label{3.3.13}
 \end{align}
 Note that we then have the Cauchy-Schwarz inequality:
\begin{align}
	\la x, \mu(t,x) \ra \leqslant \| \la x, -t\nabla_x \alpha \ra \|_E &\leqslant \| x\|_E\|t\nabla_x\alpha\|_E \nonumber\\
	&\leqslant T \lp \| x \|_E \mathfrak{B} \rp \nonumber \\
	&\leqslant  T \lp\mathfrak{B} +d \rp \|x\|_E \nonumber\\
	&= L\| x\|_E \leqslant L\lp 1 + \|x \|_E^2 \rp
\end{align} 
It also follows that $\|\sigma(t,x) \|_F =\sqrt{d}\leqslant L \leqslant L(1+\|x\|_E)$. 
\end{proof}
\begin{claim}
	It is the case that for all $x,y \in \R^d$, and $t \in [0,T]$ that: $\|\mu(t,x) - \mu(t,y) \|_E + \| \sigma(t,x) - \sigma(t,y)\|_E \leqslant \mathfrak{C} \lp \|x\|_E+\|y\|_E \rp \lp \|x-y\|_E\rp$ for some constant $\mathfrak{C} \in (0,\infty)$.
\end{claim}
\begin{proof}
The fact that for all $x,y \in \R^d$ and $t \in [0,T]$ it is the case that $\|\sigma(t,x)-\sigma(t,y)\|_F=0$, the fact that for all $x,y \in \R^d$ it is the case that $(\|x\|_E+\|y\|_E)(\|x-y\|_E) \geqslant 0$ and (\ref{3.3.13}) tells us that:
\begin{align}
	\| \mu(t,x) - \mu(t,y) \|_E+\|\sigma(t,x) - \sigma(t,y) \|_F  &= 	\| \mu(t,x) - \mu(t,y) \|_E+0  \nonumber\\
	&= \|t\nabla_x\alpha(x) - t\nabla_x \alpha(y) \|_E \nonumber\\
	&\leqslant T\|\nabla_x\alpha(x)-\nabla_x\alpha(y)\|_E \nonumber\\
	&\leqslant 2T\mathfrak{B} \label{3.3.15}
\end{align}
Now consider a function $\mathfrak{f} \in C \lp [0,T] \times \R^d,\R^d \rp$, where for all $x,y \in \R^d$ it is the case that $\mathfrak{f}(x) - \mathfrak{f}(y) \leqslant \mathscr{C}\lp \|x\|_E+\|y\|_E \rp \lp \|x+y\|_e \rp $. Note then that setting $y=x+h$ gives us:
\begin{align}
	\left|\frac{\mathfrak{f}(x+h)-\mathfrak{f}(x)}{h} \right| &\leqslant \mathscr{C} \lp \|x\|_E + \|x+h\|_E \rp \nonumber \\
	\lim_{h \rightarrow 0} \left|\frac{\mathfrak{f}(x+h)-\mathfrak{f}(x)}{h} \right| &\leqslant \lim_{h\rightarrow 0}\mathscr{C} \lp \|x\|_E + \|x+h\|_E \rp \nonumber \\
	\left| \nabla_x\mathfrak{f}\lp x \rp \right| &\leqslant 2\mathscr{C}\|x\|_E = \mathscr{K}\|x\|_E
\end{align} 
This suggests that $\nabla_x\mathfrak{f} \in O \lp x \rp$ and in particular that $\mathfrak{f} \in O \lp x^2 \rp$. However with $\mathfrak{f} \curvearrowleft \mu$ we first notice that because $\mu \leqslant 2T\mathfrak{B}$ in (\ref{3.3.15}) it must also be that case that $\mu \in O(1)$ by Corollary \ref{1.1.20.1}. However since $O(c) \subseteq O(x) \subseteq O \lp x^2 \rp$ by Corollary \ref{1.1.20.2} it is also the case that $\mu \in O \lp x^2 \rp$, and hence there exists a $\mathfrak{C}$ satisfying the claim. This proves the claim.
\end{proof}
\begin{claim}\label{3.3.5}
	It is the case that $\lv f(t,x,v) - f(t,x,w) \rv \leqslant L \lv v-w \rv$
\end{claim}
\begin{proof}
Note that by the absolute homogeneity property of norms, we have:
\begin{align}
	\left|f(t,x,v) -f(t,x,w) \right| &= \left|\frac{1}{2}tv\nabla_x^2 \alpha - \frac{1}{2} tw \nabla_x^2 \alpha \right| \nonumber\\
	&= \left| \frac{1}{2}t\nabla_x^2\alpha \right| \left|v-w \right| \nonumber\\
	&\leqslant \frac{1}{2}T \left| \nabla_x^2 \alpha \right|\left| v-w \right| \nonumber\\
	&\leqslant \frac{1}{2}T \mathfrak{B} \lv v-w \rv \nonumber\\
	&\leqslant T(\mathfrak{B}+d)\lv v-w \rv \nonumber\\
	&=L \lv v-w \rv 
\end{align}
\end{proof}

Note that we may rewrite (\ref{3.3.12}) as:
\begin{align}
	\lp \frac{\partial}{\partial t} v\rp \lp t,x \rp + \frac{1}{2}\Trace \lp \sigma \lp t,x \rp\lb \sigma \lp t,x \rp \rb^* \lp \Hess_x v\rp \lp t,x \rp \rp + \la \mu \lp t,x \rp , \lp \nabla_x v \rp \lp t,x \rp \ra \nonumber\\
	+ f \lp t,x, v \lp t,x \rp \rp = 0 \nonumber
\end{align}





We realize that (\ref{3.3.12}) is a case of \cite[Corollary~3.9]{bhj20} where it is the case that: $u(t,x) \curvearrowleft v(t,x)$, where  $\sigma_d(x) = \mathbb{I}_d$ for all $x \in \R^d$, $d \in \N$, where $\mu(t,x) = -t\nabla_x\alpha$ for fixed $\alpha$ and for all $t \in [0,T]$, $x \in \R^d$, and where $f \lp t,x,u \lp t,x \rp \rp = -\frac{1}{2}tu\nabla_x^2\alpha$ for fixed $\alpha$ and for all $t\in [0,T]$, $x \in \R^d$.

We thus have that there exists a unique, at most polynomially growing viscosity solution $v \in C\lp \lb 0,T \rb \times \R^d, \R \rp$ given as:

\begin{align}
	v(t,x) &= \E \lb v \lp T, \mathcal{Y}^{t,x}_T \rp+ \int^T_t f \lp s,\mathcal{Y}^{t,x}_s, v \lp s,\mathcal{Y}^{t,x}_s \rp \rp ds \rb \label{3.2.21}
\end{align}
Let $\mathcal{V}: \lb 0,T \rb \times \Omega \rightarrow \R^m$ be a standard $\lp \mathbb{F}_t\rp _{t \in \lb 0,T \rb }$-Brownian motion. Note that this also implies that the $\mathcal{Y}$ in (\ref{3.2.21}) is characterized as:
\begin{align}
	\mathcal{Y}^{t,x}_s = x + \int^s_t \mu \lp r,\mathcal{Y}^{t,x}_r \rp dr + \int^s_t \sigma\lp s, \mathcal{X}^{t,x}_r \rp d\mathcal{V}_r 
\end{align}
With substitution, this is then:
\begin{align}
	\mathcal{Y}^{t,x}_s &= x + \int^s_t -r\nabla_x\alpha \lp \mathcal{Y}^{t,x}_r \rp dr + \int^s_t \mathbb{I} d\mathcal{V}_r \nonumber\\
	\mathcal{Y}^{t,x}_s &=x - \int^s_t r\nabla_x\alpha \lp \mathcal{Y}^{t,x}_s \rp   dr + \mathcal{V}_{s-t} \nonumber
\end{align}
Note that our initial substitution tells us: $v(t,x) = e^{t\alpha(x)}u(t,x)$. And so we have that:
\begin{align}
	v(t,x) &= \E \lb v\lp T, \mathcal{X}_T^{t,x} \rp + \int ^T_t f \lp s, \mathcal{X}^{t,x}_s, v \lp s, \mathcal{X}^{t,x}_s \rp  \rp ds\rb \label{3.3.20}\\
	v(t,x)&=\E \lb v\lp T, \mathcal{X}^{t,x}_T \rp -\frac{1}{2} \int^T_ttv\lp s,\mathcal{X}^{t,x}_s \rp \nabla_x^2 \alpha \lp \mathcal{X}^{t,x}_s \rp ds\rb \nonumber\\
	e^{t\alpha(x)} u(t,x)&= \E \lb \exp \lb T\alpha \lp \mathcal{X}^{t,x}_T \rp \rb  u\lp T, \mathcal{X}^{t,x}_T \rp -\frac{1}{2} \int^T_t t  \exp \lb t\alpha \lp \mathcal{X}^{t,x}_s \rp \rb u\lp t, \mathcal{X}^{t,x}_s \rp \nabla_x^2\alpha \lp \mathcal{X}^{t,x}_s \rp  ds\rb \nonumber \\
	u(t,x) &= \E \lb \exp \lb T \alpha\lp \mathcal{X}^{t,x}_T \rp -t\alpha (x)\rb u \lp T, \mathcal{X}^{t,x}_T \rp \rb  \nonumber\\
	&- \E \lb\frac{1}{2e^{t\alpha(x)}} \int^T_t t \exp \lb t\alpha \lp \mathcal{X}^{t,x}_s \rp \rb u\lp  t,\mathcal{X}^{t,x}_s \rp \nabla_x^2 \alpha \lp \mathcal{X}^{t,x}_s \rp ds \rb  \nonumber
\end{align}
\end{proof}
