dissertation_work/Scratch/Theorems with f = 0.tex

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\documentclass{article}
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\title{Reformulation without f}
\author{Shakil Rafi}
\begin{document}
\maketitle
\begin{abstract}
\noindent This document aims to reformulate, rather naively, the theorems and lemmas of HJKP21 with $f=0$. Things that can be outright dismissed with will be typed with a \st{strikethrough}. Those that are to be modified will be given with the color \textcolor{red}{red}, but only the first instance of that variable will be colored. Note that \textcolor{blue}{blue} represents confusion and doubt. Reformulated theorems and lemmas will be referred to by their corresponding number in HJKP21 but the with suffix R. For example Lemma 3.5 from HJKP21, once reformulated will be referred to as Lemma R3.5
\end{abstract}
\textbf{Theorem 1.1 Reformulation} Let $T,\kappa,\delta, p \in (0, \infty)$ and $\Theta = \bigcup_{n\in \mathbb{N}}\mathbb{Z}^n$ be Lipschitz continuous. Let $u_d\in C^{1,2}([0,T] \times \mathbb{R}^d,\mathbb{R}),d\in \mathbb{N}$, satisfy $\forall d\in \mathbb{N}, t\in[0,T], x = (x_1,x_2,...,x_d)\in \mathbb{R}^d$ that
$|u_d(t,x)|\leq \kappa d^{\kappa}(1+\sum^d_{k=1}|x_k|)^\kappa$ and:
\begin{align}
\left(\frac{\partial}{\partial t} u_d\right)(t,x) = (\Delta_x u_d)(t,x)
\end{align}
Let $(\Omega, \mathcal{F},\mathbb{P})$, represent a probability space and let $W^{d,\theta} : [0,T] \times \Omega \rightarrow \mathbb{R}^d, d\in \mathbb{N}, \theta \in \Theta$ be independent standard Brownian motions, assume that $(W^{d,\theta})_{(d,\theta) \in \mathbb{N}\times \Theta}$ are independent, let $\phi: \mathbb{N} \rightarrow \mathbb{N}$ and $U^{d,\theta}_{n,m}:[0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R},d,n,m\in \mathbb{Z}, \theta \in \Theta$, satisfy $\forall n \in \mathbb{N}_0$$d,m \in \mathbb{N}, \theta \in \Theta, t\in [0,T], x\in \mathbb{R}^d$ that $\phi(m) = \max\{k\in \mathbb{N}: k \leq \exp(|ln(m)|^{\frac{1}{2}})\}$ and:
\begin{align*}
U^{d,\theta}_{n,m}(t,x)=\frac{}{(\phi(m))^n}\left[\sum^{(\phi(m))^n}_{k=1} u_d(0,x+\sqrt{2} W_t^{d,(\theta,0,-k)})\right]
\end{align*}
\medskip
and for every $d,n,m\in \mathbb{N}$, let $\mathfrak{C}_{d,n,m} \in \mathbb{N}$ be the number of function evaluatons of $f$ and of $u_d(0,\cdot)$ the number of realizations of scalar random variables that are used to compute one realization of $U^{d,0}_{n,m}(T,0):\Omega \rightarrow \mathbb{R}$. Then there exists $c\in \mathbb{R}$ and $\mathcal{n}: \mathbb{N}\times (0,1] \rightarrow \mathbb{N}$ such that $\forall d \in \mathbb{N}, \epsilon \in (0,1]$ it holds:
\begin{align}
(\mathbb{E}[|u_d(T,0)-U^{d,0}_{\mathcal{n}(d,\epsilon),\mathcal{n}(d,\epsilon)}(T,0)|^p])^{\frac{1}{p}} \leq \epsilon \quad \text{ and } \quad \textcolor{red}{\mathfrak{C}_{d,n(d,\epsilon),n(d,\epsilon)} \leq cd^c \epsilon^{-(2+\delta)}}
\end{align}
\textbf{Reformulation of (1.5)}. Feynman-Kac proves that the solution functions $u_d:[0,T] \times \mathbb{R}^d \rightarrow \mathbb{R},d\in \mathbb{N}$ to (1) are the unique at most polynomially growing functions which satisfy $\forall d\in \mathbb{N}, \theta \in \Theta, t\in [0,T], x\in \mathbb{R}^d$ that:
\begin{align*}
u_d(t,x) = \mathbb{E}\left[u_d(0,x+\sqrt{2}W^{d,\theta}_t)\right]
\end{align*}
\textbf{Reformulation of (1.6)}{Since in our case we have that $f=0$, we can then say from Lemmas 3.3,3.5, and (3.24) in HJKP21 that:
\begin{align*}
\mathbb{E}[U^{d,\theta}_{n,m}(t,x)] - \mathbb{E}[u_d(0,x+\sqrt{2}W^{d,\theta}_t)] &= 0
\end{align*}
This follows from Lemmas R3.3, R3.5, and (R3.24)
\medskip
\section{Reformulation for the case without a forcing term}\label{sec:formulanof}
\textbf{Setting 1.1}\label{setting11} Let $d,m \in \mathbb{N}$ $T,L, \mathfrak{L},p \in [0,\infty)$, $\mathfrak{p} \in [2,\infty)$ $\mathfrak{m} = \mathfrak{k}_{\mathfrak{p}}\sqrt{\mathfrak{p}-1}$, $\Theta = \bigcup_{n\in \mathbb{N}}\mathbb{Z}^n$, $g \in C(\mathbb{R}^d,\mathbb{R})$, assume $\forall t \in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}
\max\{|g(x)|\} \leq \mathfrak{L}(1+||x||^p)
\end{align}
and let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space. Let $W^{\theta}: [0,T] \times \Omega \rightarrow \mathbb{R}^d$, $\theta \in \Theta$ be independent standard brownian motions, $(W^\theta)_{\theta \in \Theta}$ are independent, let $u \in C([0,T] \times \mathbb{R}^d,\mathbb{R})$ satisfy $\forall t \in [0,T]$,$x\in \mathbb{R}^d$, that $\mathbb{E}[|g(x+W^0_{T-t})] < \infty$ and:
\begin{align}
u(t,x) &= \mathbb{E}[g(x+W^0_{T-t})]
\end{align}
and let let $U^\theta_n:[0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R}$, $\theta \in \Theta$ satisfy $\forall n \in \mathbb{N}_0$,$\theta \in \Theta, t \in [0,T],x\in \mathbb{R}^d$, that:
\begin{align}
U^\theta_n(t,x) = \frac{1}{m^n}\left[\sum^{m^n}_{k=1}g(x+W^{(\theta,0,-k)})\right]
\end{align}
\textbf{Lemma R3.3:} Assume Setting \ref{seting11}. Then:
\begin{enumerate}[label = (\roman*)]
\item it holds $\forall n \in \mathbb{N}_0$ $\theta \in \Theta$ that $U^\theta_n:[0,T] \times \mathbb{R}^d\times \Omega \rightarrow \mathbb{R}$ is a consinuous random field.
\item it holds $\forall n \in \mathbb{N}_0$, $\theta \in \Theta$ that $\sigma(U^\theta_n) \subseteq \sigma ((W^{(\theta, \mathcal{V})})_{\mathcal{V} \in \Theta})$
\item it holds $\forall n \in \mathbb{N}_0$ that $(U^\theta_n)_{\theta \in\Theta}$,$(W^\theta)_{\theta \in \Theta}$, are independent
\item it holds $\forall n,m \in \mathbb{N}_0$, $i,k,\mathfrak{i},\mathfrak{k}\in \mathbb{Z}$, with $(i,k) \neq (\mathfrak{i},\mathfrak{k})$ that $(U_n^{(\theta,i,k)})_{\theta \in \Theta}$ and $(U_n^{(\theta,\mathfrak{i},\mathfrak{k})})_{\theta \in \Theta}$ are independent and,
\item it holds $\forall n \in \mathbb{N}_0$ that $(U^\theta_n)_{\theta \in \Theta}$ are identically distributed random variables
\end{enumerate}
\textit{Proof of Lemma R3.3:} For (i) Consider that $W^{(\theta,0,-k)}_{T-t}$ are continuous random fields and that $g\in C(\mathbb{R}^d,\mathbb{R})$, we have that $U^\theta_n(t,x)$ is the composition of continuous functions with $m^n > 0$ by hypothesis, ensuring no singularities. Thus $U^\theta_n: [0,T] \times \mathbb{R}^d\times \Omega \rightarrow \mathbb{R}$.
\medskip
For (ii) observe that $\forall \theta \in \Theta$ it holds that $W^\theta$ is $\mathcal{B}([0,T] \otimes \sigma(W^\theta))/\mathcal{B}(\mathbb{R}^d)$-measurable, this, and induction on $\mathbb{N}_0$ prove item \textit{ii}.
\medskip
Moreover observe that item (ii) and the fact that $\forall \theta \in \Theta$ it holds that $(W^{(\theta, \vartheta)}_{\vartheta \in \Theta})$, $W^\theta$ are independent establish item (iii).
\medskip
Furthermore, note that (ii) and the fact that $\forall i,k,\mathfrak{i},\mathfrak{k} \in \mathbb{Z}$,$\theta \in \Theta$, with $(i,k) \neq (\mathfrak{i},\mathfrak{k})$ it holds that $(W^{(\theta,i,k,\vartheta)})_{\vartheta \in \Theta}$ and $(W^{(\theta,\mathfrak{i},\mathfrak{k},\vartheta)})_{\vartheta \in \Theta}$ are independent establish item (iv).
\medskip
Hutzenhaler [30, Corollary 2.5] and induction on $\mathbb{N}_0$ establish item (v). This completes the proof of Lemma R3.3.
$\hfill \square$
\medskip
\textbf{Lemma R3.4} Assume setting R3.2. Then it holds $\forall n \in \mathbb{N}_0,\theta \in \Theta, s \in [0,T], t\in [s,T],x\in \mathbb{R}^d$ that:
\begin{align}
\mathbb{E}[|U^\theta_n(t,x+W^\theta_{t-s})|] +\mathbb{E}[|g(x+W^\theta_{t-s})|]+ \int^T_s \mathbb{E}[|U^\theta_n(r,x+W^\theta_{r-s})|]dr < \infty
\end{align}
\textit{Proof or Lemma R3.4:} Note that (3), the fact that $\forall r,a,b \in [0,\infty)$ it holds that $(a+b)^r \leq 2^{\max\{r-1,0\}}(a^r+b^r)$, and the fact that $\forall \theta \in \Theta$ it holds that $\mathbb{E}[||W^\theta_T||] < \infty$ assure that for all $s \in [0,T],t\in[s,T],\theta \in \Theta$ it holds that:
\begin{align}
\mathbb{E}[|g(x+W^\theta_{t-s})|] &\leq \mathbb{E}[\mathfrak{L}(1+||x+W^\theta_{t-s}||^p)] \\
&\leq \mathfrak{L}[1+2^{\max\{p-1,0\}}(||x||^p+\mathbb{E}[||W^\theta_T||^p])]<\infty
\end{align}
We next claim that $\forall n \in \mathbb{N}_0, s\in [0,T],t\in[s,T],\theta \in \Theta$ it holds that:
\begin{align}
\mathbb{E}[|U^\theta_n(t,x+W^\theta_{t-s})|]+ \int^T_s \mathbb{E}[|U^\theta_n(r,x+W^\theta_{r-s})|]dr < \infty \label{foo}
\end{align}
To prove this claim observe the triangle inequality and \textbf{Setting R3.2}, demonstrate that for all $s\in[0,T]$ $t\in[s,T]$ $\theta \in \Theta$ it holds that:
\begin{align}
\mathbb{E}[|U^\theta_n(t,x+W^\theta_{t-s})|] \leq \frac{}{m^n}\left[ \sum^{m^n}_{i=1}\mathbb{E}[|g(x+W^\theta_{t-s}+W^{(\theta,0,-i)}_{T-t})]\right]
\end{align}
Now observe that (8) and the fact that $(W^\theta)_{\theta \in \Theta}$ are independent imply that for all $s \in [0,T]$$t\in [s,T]$$\theta \in \Theta$ $i\in \mathbb{Z}$ it holds that:
\begin{align}
\mathbb{E}[|g(x+W^\theta_{t-s}+W^{(\theta,0,i)}_{T-t})|] = \mathbb{E}[|g(x+W^\theta_{(t-s)+(T-t)})|] = \mathbb{E}[|g(x+W^\theta_{T-s})|] <\infty
\end{align}
\medskip
Combining (10) and (11) demonstrate that for all $s \in [0,T]$ $t\in[s,T]$ $\theta \in \Theta$ it holds that:
\begin{align}
\mathbb{E}[|U^\theta_n(t,x+W^\theta_{t-s})|] < \infty
\end{align}
Finally observe that for all $s\in [0,T]$ $\theta \in \Theta$ it holds that:
\begin{align}
\int^T_s \mathbb{E}[|U^\theta_n(r,x+W^\theta_{r-s})|] &\leq (T-s) \sup_{r\in [s,T]} \mathbb{E} [|U^\theta_n(r,x+W^\theta_{r-s})|] \\
&\leq (T-s)(\mathbb{E}[|g(x+W^\theta_{T-s})]) <\infty
\end{align}
Combining (9), (12), and (13) prove Lemma R 3.4. This completes the proof of Lemma R3.4.
$\hfill \square$
\medskip
\textbf{Lemma R3.5} Assume reformulated setting 3.2, then:
\begin{enumerate}[label = (\roman*)]
\item it holds $\forall n \in \mathbb{N}_0, t \in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}
\mathbb{E}[|U^0_n(t,x)|]+ \mathbb{E}[|g(x+W^{(0,0,-1)}_{T-t})] < \infty
\end{align}
\item it holds $\forall n\in \mathbb{N}_0, t\in [0,T],x\in \mathbb{R}^d$ that:
\begin{align*}
\mathbb{E}[U^0_n(t,x)] = \mathbbm{1}_{\mathbb{N}}(n) \mathbb{E}[g(x+W^{(0,0,-1)}_{T-t})]
\end{align*}
\end{enumerate}
\textit{Proof of Lemma R3.5}. (i) is a restatement of Lemma R3.4 in that for all $n\in \mathbb{N}_0$ $t\in [0,T]$:
\begin{align}
\mathbb{E}[|(U^0_n)t,x)|] + \mathbb{E}[|g(x+W^{(0,0,-1)}_{T-t})|] &< \\
\mathbb{E}[|U^\theta_n(t,x+W^\theta_{t-s})|] +\mathbb{E}[|g(x+W^\theta_{t-s})|]+ \int^T_s \mathbb{E}[|U^\theta_n(r,x+W^\theta_{r-s})|]dr &< \infty
\end{align}
Furthermore (ii) is a restatement of (5) with $\theta = 0$
This completes the proof of Lemma R3.5 $\hfill \square$
\medskip
\textbf{Monte Carlo Approximations:} Let $p \in (2,\infty)$,$n\in \mathbb{N}$, let $(\Omega, \mathcal{F}, \mathbb{P})$, be a probability space and let $X_i: \Omega \rightarrow \mathbb{R}$, $i \in \{1,2,...,n\}$ be i.i.d. random variables with $\mathbb{E}[|X_1|]<\infty$. Then it holds that:
\begin{align}
(\E[|\E[X_1]-\frac{1}{n}(\sum^n_{i=1}X_i)|^p])^{\frac{1}{p}} \leq \left[\frac{p-1}{n}\right]^{\frac{1}{2}}\left(\E[|X_1-\E[X_1]|^p)]\right)^{\frac{1}{p}}
\end{align}
The hypothesis that for all $i \in \{1,2,...,n\}$ it holds that $X_i:\Omega \rightarrow \mathbb{R}$ ensures that:
\begin{align}
\E[|\E[X_1]-\frac{1}{n}\left(\sum^n_{i=1}X_i \right)|^p] = \E \left[\frac{1}{n} \left(\sum^n_{i=1}\left(\E[X_1]-X_i\right) \right)\right]
\end{align}
\medskip
\textbf{Lemma R3.9}: Assume reformualated Setting 3.2. Then it holds for all $n\in \mathbb{N}_0$,$t\in [0,T]$,$x\in \mathbb{R}^d$, that:
\begin{align}
&\left(\E\left[\left|U^0_n(t,x+W^0_t)-\E \left[U^0_n \left(t,x+W^0_t \right)\right]\right|^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \\
&\leq \frac{\mathfrak{m}}{m^{\frac{n}{2}}} \left[\left(\E\left[|g(x+W^0_T)|^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}}\right] \\
\end{align}
\textit{Proof of Lemma R3.9}: For notational simplicity, let $G_k: [0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R}$,$k\in \mathbb{Z}$, satisfy for all $k\in \mathbb{Z}$,$t\in[0,T]$,$x\in \mathbb{R}^d$ that:
\begin{align}
G_k(t,x) = g\left(x+W^{(0,0,-k)}_{T-t}\right)
\end{align}
\medskip
Observe that the hypothesis that $(W^\theta)_{\theta \in \Theta}$ are independent Brownian motions and the hypothesis that $g \in C(\mathbb{R}^d,\mathbb{R})$ assure that for all $t \in [0,T]$,$x\in \mathbb{R}^d$ it holds that $(G_k(t,x))_{k\in \mathbb{Z}}$ are i.i.d. random variables. This and Corollary 3.8 (applied for every $n\in \mathbb{N}$, $t\in [0,T]$, $x\in \mathbb{R}^d$ with $p \curvearrowleft \mathfrak{p}$, $n \curvearrowleft m^n$,$(X_k)_{k\in \{1,2,...,m^n\}} \curvearrowleft (G_k(t,x))_{k\in \{1,2,...,m^n\}}$), with the notation of Corollary 3.8 ensure that for all $n\in \mathbb{N}_0$, $t\in [0,T]$, $x \in \mathbb{R}^d$, it holds that:
\begin{align}
\left( \E \left[ \left| \frac{1}{m^n} \left[ \sum^{m^n}_{k=1} G_k(t,x) \right] - \E \left[ G_1(t,x) \right] \right| ^\mathfrak{p} \right] \right)^{\frac{1}{\mathfrak{p}}} \leq \frac{\mathfrak{m}}{m^{\frac{n}{2}}}\left(\E \left[|G_1(t,x)|^\mathfrak{p} \right] \right)^{\frac{1}{\mathfrak{p}}}
\end{align}
\medskip
Combining this, with (23), (24), and item (ii) of Lemma R3.5 yields that:
\begin{align}
&\left(\E\left[\left|U^0_n(t,x) - \E \left[U^0_n(t,x)\right]\right|^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} \\
&= \left(\E \left[\left|\frac{}{m^n}\left[\sum^{m^n}_{k=1}G_k(t,x)\right]- \E \left[G_1(t,x)\right]\right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}} \\
&\leq \frac{\mathbbm{1}(n)\mathfrak{m}}{m^{\frac{n}{2}}}\left(\E \left[\left| G_1(t,x)\right| ^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \\
&= \frac{\mathbbm{1}(n)\mathfrak{m}}{m^{\frac{n}{2}}} \left[\left(\E \left[\left|g\left(x+W^1_{T-t}\right)\right|^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}}\right]
\end{align}
This and the fact that $W^0$ has independent increments ensure that for all $n\in \mathbb{N}_0$, $t\in [0,T]$, $x\in \mathbb{R}^d$ it holds that:
\begin{align}
\left(\E \left[\left| U^0_n \left(t,x+W^0_t\right) - \E \left[U^0_n \left(t,x+W^0_t\right)\right]\right|^\mathfrak{p}\right]\right)^{\frac{1}{\mathfrak{p}}} \leq \frac{\mathfrak{m}}{m^{\frac{n}{2}}} \left[\left(\E \left[\left| g \left(x+W^0_T\right)\right|^\p\right]\right)^{\frac{1}{\mathfrak{p}}} \right]
\end{align}
This completes the proof of Lemma R3.9
$\hfill \square$
\medskip
\textbf{Lemma R3.10:} Assume setting R3.2. Then it holds for all $n\in \mathbb{N}_0$, $t\in [0,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
\left(\E \left[ \left| U^0_n \left(t,x+W^0_t\right) - u \left(t,x+W^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} \leq \left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right) \left( \E \left[\left| g \left(x+W^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
Observe that from Lemma R3.5 item (ii) we have:
\begin{align}
\E\left[U^0_n(t,x)\right] = \E \left[ g \left(x+W^{(0,0,-1)}_{T-t}\right) \right]
\end{align}
This and ($4$) ensure that:
\begin{equation}
u(t,x) - \E \left[U^0_n(t,x) \right] = \begin{cases}
u(t,x) & :n=0 \\
0 & :n\in \mathbb{N}
\end{cases}
\end{equation}
This, and the fact that $W^0$ has independent increments, assure that for all $n\in \mathbb{N}_0$, $t\in [0,T]$, $x\in \mathbb{R}^d$, it holds that:
\begin{align}
\left(\E \left[\left| \E \left[U^0_n(t,x+W^0_t)\right] - u(t,x+W^0_t)\right| ^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} \leq \mathbbm{1}_{\{0\}}(n) \left(\E \left[|u(t,x+W^0_t)|^\p\right]\right)
\end{align}
This along with (4) ensure that:
\begin{align}
\left(\E \left[\left| \E \left[U^0_n(t,x+W^0_t)\right] - u(t,x+W^0_t)\right| ^\mathfrak{p}\right]\right)^\frac{1}{\mathfrak{p}} \leq \mathbbm{1}_{\{0\}}(n) \left( \E \left[\left| g \left(x+W^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
Notice that the triangle inequality then gives us:
\begin{align}
\left(\E \left[ \left| U^0_n \left(t,x+W^0_t\right) - u \left(t,x+W^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} \leq \left(\E \left[\left| U^0_n(t,x+W^0_t) - \E \left[U^0(t,x+W^0_t)\right]\right|^\p\right]\right)^{\frac{1}{\p}} \\
+\left(\E \left[\left| \E \left[U^0_n(t,x+W^0_t)\right]-u(t,x+W^0_t)\right|^\p\right]\right)^{\frac{1}{\p}}
\end{align}
This, combined with (34) and (29) gives us:
\begin{align}
\left(\E \left[ \left| U^0_n \left(t,x+W^0_t\right) - u \left(t,x+W^0_t\right) \right|^\mathfrak{p} \right]\right)^\frac{1}{\mathfrak{p}} \leq \left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right) \left( \E \left[\left| g \left(x+W^0_T\right) \right|^\mathfrak{p} \right]\right)^{\frac{1}{\mathfrak{p}}}
\end{align}
This completes the proof of Lemma R3.10
$\hfill \square$
\medskip
\textbf{Lemma R3.14:} Assume setting R3.2 then it holds for all $n \in \mathbb{N}_0$, $t\in [0,T]$, $x\in \mathbb{R}^d$ that:
\begin{align}
\left( \E \left[ \left| U^0_n \left(t,x+W^0_t \right) - u \left( t, x+W^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
\medskip
\textit{Proof of Lemma R3.14}: Observe that Lemma R3.10 ensures that:
\begin{align}
\left( \E \left[ \left| U^0_n \left(t,x+W^0_t \right) - u \left( t, x+W^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leq \left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\E \left[\left| g \left(x+W^0_T\right)\right|^\p\right]\right)^\frac{1}{\p}
\end{align}
Observe next that (4) ensures that:
\begin{align}
\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\E \left[\left| g \left(x+W^0_T\right)\right|^\p\right]\right)^\frac{1}{\p} \leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\E \left[ \left( 1+ \left| \left| x+W^0_T \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Which in turn yields that:
\begin{align}
\mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\E \left[ \left( 1+ \left| \left| x+W^0_T \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p} \leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Combining (36),(37), and (38) yields that:
\begin{align}
\left( \E \left[ \left| U^0_n \left(t,x+W^0_t \right) - u \left( t, x+W^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leq \left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\E \left[\left| g \left(x+W^0_T\right)\right|^\p\right]\right)^\frac{1}{\p} \leq \\
\mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s\in[0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
This completes the proof of Lemma R3.14
$\hfill \square$
\medskip
\textbf{Corollary R3.15:} Assume setting R3.2. Then it holds for all $n\in \mathbb{N}_0$, $t\in[0,T]$, $x\in \R^d$ that:
\begin{align}
\left( \E \left[ \left| U^0_n(t,x) -u(t,x) \right| ^\p \right] \right) ^{\frac{1}{\p}} \leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
\textit{Proof of Corollary R3.15:} Observe that for all $t \in [0,T-\mft]$ and $\mft \in [0,T]$, and the fact that $W^0$ has independent increments it is the case that:
\begin{align}
u(t+\mft,x) = \E \left[g \left(x+W^0_{T-(t+\mft)}\right)\right] = \E \left[g \left(x+W^0_{(T-\mft)-t)}\right)\right]
\end{align}
And it is also the case that:
\begin{align*}
U^\theta_n(t+\mft,x) = \frac{}{m^n} \left[\sum^{m^n}_{k=1} g \left(x+W^{(\theta,0,-k)}_{T-(t+\mft)}\right)\right] = \frac{}{m^n} \left[\sum^{m^n}_{k=1} g \left(x+W^{(\theta,0,-k)}_{(T-\mft)-t}\right)\right]
\end{align*}
\medskip
Then, applying \textbf{Lemma 3.14}, applied for all $\mft \in [0,T]$, with $\mathfrak{L} \curvearrowleft \mathfrak{L}$, $p \curvearrowleft p$, $\mathfrak{p} \curvearrowleft \mathfrak{p}$, $T \curvearrowleft (T-\mft)$ is such that for all $\mft \in [0,T]$, $t \in [0,T-\mft]$, $x \in \R^d$, and $n\in \mathbb{N}_0$ we have:
\begin{align}
\left( \E \left[ \left| U^0_n \left(t+\mft,x+W^0_t \right) - u \left( t+\mft, x+W^0_t \right) \right|^\p \right] \right)^{\frac{1}{\p}} \leq \\
\mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T-\mft]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p} \\
\leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
Thus we get for all $\mft \in [0,T]$, $x\in \R^d$, $n \in \mathbb{N}_0$:
\begin{align}
\left( \E \left[ \left| U^0_n \left(\mft,x \right) - u \left(\mft, x \right) \right|^\p \right] \right)^{\frac{1}{\p}} &= \left( \E \left[ \left| U^0_n \left(\mft,x+W^0_0 \right) - u \left(\mft, x+W^0_0 \right) \right|^\p \right] \right)^{\frac{1}{\p}} \\
&\leq \mathfrak{L}\left( \frac{ \mathfrak{m}}{m^{\frac{n}{2}}}\right)\left(\sup_{s \in [0,T]}\E \left[ \left( 1+ \left| \left| x+W^0_s \right| \right|^p \right)^\p\right]\right)^\frac{1}{\p}
\end{align}
This completes the proof of Lemma R3.15
$\hfill \square$
\medskip
\textbf{Proposition 4.4:} Let $T,L,p,q \in [0,\infty), m \in \mathbb{N}$, $\Theta = \bigcup_{n\in \mathbb{N}} \Z^n$, let $g_d\in C(\R^d,\R)$, and assume that $d\in \N$, $t \in [0,T]$, $x = (x_1,x_2,...,x_d)\in \R^d$, $v,w \in \R$ and that $\max \{ |g_d(x)|\} \leq Ld^p \left(1+\Sigma^d_{k=1}\left|x_k \right|\right)$, let $\left(\Omega, \mathcal{F}, \mathbb{P}\right)$ be a probability space, let $W^{d,\theta}: [0,T] \times \Omega \rightarrow \R^d$, $d\in \N$, $\theta \in \Theta$, be independent standard Brownian motions, assume for every $d\in \N$ that $\left(W^{d,\theta}\right)_{\theta \in \Theta}$ are independent, let $u_d \in C([0,T] \times \R^d,\R)$, $d \in \N$, satisfy for all $d\in \N$, $t\in [0,T]$, $x \in \R^d$ that $\E \left[g_x \left(x+W^{d,0}_{T-t} \right)\right] < \infty$ and:
\begin{align}
u_d\left(t,x\right) = \E \left[g_d \left(x + W^{d,0}_{T-t}\right)\right]
\end{align}
Let $U^{d,\theta}_{n}: [0,T] \times \R^d \times \Omega \rightarrow \R$, $d$, $n\in \Z$, $\theta \in \Theta$, satisfy for all $n \in \N_0$, $d\in \N$, $\theta \in \Theta$, $t\in [0,T]$, $x\in \R^d$ that:
\begin{align}
U^{d,\theta}_{n}(t,x) = \frac{1}{m} \left[\sum^{m}_{k=1} g_d \left(x + W^{d,(\theta, 0,-k)}_{T-t}\right)\right]
\end{align}
and let $\mathfrak{C}_{d,n} \in \R$, $d$, $n$, $ \in \N_0$, satisfy for all $d \in \N$, $n\in \N_0$, that:
\begin{align}
\mathfrak{C}_{d} \leq \mathbbm{1}_\N(n) \alpha d^{\mathfrak{d}}m
\end{align}
Then there exists $\mathscr{M}_\epsilon \in \mathbb{Z}$ such that it holds that:
$\mathscr{N}: \N \times \R \rightarrow \N$ such that for all $d \in \N$, $\epsilon$, $\delta \in (0,\infty)$, $ \mathfrak{p} \in [2,\infty)$ with $\text{limsup}_{j \rightarrow \infty} \left[\frac{(m_j)^{\frac{p}{2}}}{j}\right] < \infty$ it holds that:
\begin{align}
(\n(d,\epsilon))^\beta \C_{d,\n(d,\epsilon),\n(d,\epsilon)} \leq \alpha d^{\mathfrak{d}}( \mathscr{N} (d,\epsilon))^\beta(1+ \sqrt{2})^{\n(d,\epsilon)} \\
\leq \alpha c_{\p,\delta} d^{\mathfrak{d}+(p+q)(2+\delta)}(\min\{1,\epsilon\})^{-(2+\delta)} \leq \epsilon
\end{align}
and:
\begin{align}
\sup_{t\in[0,T]} \sup_{x \in [-L,L]^d} \left(\E \left[\left| u_d(t,x) - U^{d,0}_{\n(d,\epsilon),\n(d,\epsilon)}\right|^\p\right]\right)^\frac{1}{\p} \leq \epsilon
\end{align}
\textit{Proof of Proposition 4.4:} Throughout the proof let $\mathfrak{m}_\mathfrak{p} = \sqrt{\mathfrak{p} -1}$, $\mathfrak{p} \in [2,\infty)$, let $\mathbb{F}^d_t \subseteq \mathcal{F}$, $d\in \N$, $t\in [0,T]$ satisfy for all $d \in \N$, $t\in [0,T]$ that:
\begin{align}
\mathbb{F}^d_t = \begin{cases}
\bigcap_{s\in[t,T]} \sigma \left(\sigma \left(W^{d,0}_r: r \in [0,s]\right) \cup \{A\in \mathcal{F}: \mathbb{P}(A)=0\}\right) & :t<T \\
\sigma \left(\sigma \left(W^{d,0}_s: s\in [0,T]\right) \cup \{ A \in \mathcal{F}: \mathbb{P}(A)=0\}\right) & :t=T
\end{cases}
\end{align}
Observe that (58) guarantees that $\mathbb{F}^d_t \subseteq \mathcal{F}$, $d\in \N$, $t\in [0,T]$ satisfies that:
\begin{enumerate}[label = (\Roman*)]
\item it holds for all $d\in \N$ that $\{ A \in \mathcal{F}: \mathbb{P}(A) = 0 \} \subseteq \mathbb{F}^d_0$
\item it holds for all $d \in \N$, $t\in [0,T]$, that $\mathbb{F}^d_t = \bigcap_{s \in (t,T]}\mathbb{F}^d_s$.
\end{enumerate}
Combining item (I), item (II), (58) and Hutzenthaler [31, Lemma 2.17] assures us that for all $d \in \N$ it holds that $W^{d,0}:[0,T] \times \Omega \rightarrow \R^d$ is a standard $\left(\Omega, \mathcal{F}, \mathbb{P}, \left(\mathbb{F}^d_t\right)_{t\in [0,T]}\right)$-Brownian Brownian motion. In addition $(58)$ ensures that it is the case that for all $d\in N$, $x\in \R^d$ it holds that $[0,T] \times \Omega \ni (t,\omega) \mapsto x + W^{d,0}_t(\omega) \in \R^d$ is an $\left(\mathbb{F}^d_t\right)_{t\in [0,T]}/\mathcal{B}\left(\R^d\right)$-adapted stochastic process with continuous sample paths.
\medskip
This and the face that for all $d\in \N$, $t\in [0,T]$, $x\in \R^d$ it holds that $a_d(t,x) = 0$, and the fact that for all $d\in \N$, $t \in [0,T]$, $x$,$v\in \R^d$ it holds that $b_d(t,x)v = v$ yield that for all $d \in \N$, $x\in \R^d$ it holds that $[0,T] \times \Omega \ni (t,\omega) \mapsto x+W^{d,0}_t(\omega) \in \R^d$ satisfies for all $t\in [0,T]$ it holds $\mathbb{P}$-a.s. that:
\begin{align}
x+W^{d,0}_t = x + \int^t_0 0 ds + \int^t_0 dW^{d,0}_s = x + \int^t_0 a_d(s,x+W^{d,0}_s) ds + \int^t_0 b_d(s,x+W^{d,0}_s) dW^{d,0}_s
\end{align}
\medskip
This and Hutzenthaler et. al. [31, Lemma 2.6] (applied for every $d \in \N$, $x \in \R^d$ with $d \curvearrowleft d$, $m \curvearrowleft d$, $T \curvearrowleft T$, $C_1 \curvearrowleft d$, $ C_2 \curvearrowleft 0$, $\mathbb{F} \curvearrowleft \mathbb{F}^d$, $ \xi \curvearrowleft x, \mu \curvearrowleft a_d, \sigma \curvearrowleft b_d, W \curvearrowleft W^{d,0}, X \curvearrowleft \left(\left[0,T\right] \times \Omega \ni (t, \omega) \mapsto x+W^{d,0}_t(\omega) \in \R^d\right)$ in the notation of [31, Lemma 2.6] ensures that for all $r\in [0,\infty)$, $d\in \N$, $x\in \R^d$, $t \in [0,T]$ it holds that
\begin{align}
\E \left[\left| \left| x+W^{d,0}_t\right| \right|^r\right] \leq \max\{T,1\} \left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{r}{2}} + (r+1)d^{\frac{r}{2}}\right) \exp \left(\frac{r(r+3)T}{2}\right) < \infty
\end{align}
\medskip
This, the triangle inequality, and the fact that for all $v$,$w\in [0,\infty)$, $r\in (0,1]$, it holds that $(v+w)^r \leq v^r + w^r$ assure that for all $\p \in [2,\infty)$, $d\in \N$, $x \in \R^d$ it holds that:
\begin{align}
\sup_{s\in [0,T]} \left(\E \left[\left(1+ \left| \left| x+W^{d,0}_s \right| \right|^q\right)^\p\right]\right)^\frac{1}{\p} \\
\leq 1 + \sup_{s\in [0,T]} \left(\E \left[\left| \left| x+W^{d,0}_{s}\right| \right| ^{q\p}\right]\right)^{\frac{1}{\p}} \\
\leq 1 + \sup_{s \in [0,T]} \left(\max\{T,1\} \left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q\p(q\p+3)T}{2}\right) \right)^\frac{1}{\p} \\
\leq 1 + \max\{T^\frac{1}{\p},1\} \left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q(q\p+3)T}{2}\right) \\
\leq \left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{q(q\p+3)T}{2}+\frac{T}{\p}\right) \\
\leq \left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)
\end{align}
\medskip
Given that for all $d\in \N$, $x \in [-L,L]^d$ it holds that $\left| \left| x \right| \right| \leq Ld^{\frac{1}{2}}$, this demonstrates for all $\p \in [2,\infty)$, $d\in \N$, it holds that:
\begin{align}
L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \sup_{s \in T} \left( \E \left[\left(1+\left| \left| x + W^{d,0}_s \right| \right| ^q\right)^\p\right]\right)^{\frac{1}{\p}}\right) \\
\leq L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \left[\left(\left(1+\left| \left| x\right| \right|^2\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)\right] \right)\\
\leq L\left( \frac{ \mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \\
\end{align}
Combining this with \textbf{Corollary 3.15} tells us that:
\begin{align}
\left( \E \left[ \left| u_d(t,x)-U^{d,0}_{n}(t,x) \right| ^\p \right] \right) ^{\frac{1}{\p}} \\
\leq L\left(\frac{\mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\sup_{x \in [-L,L]^d} \sup_{s \in T} \left( \E \left[\left(1+\left| \left| x + W^{d,0}_s \right| \right| ^q\right)^\p\right]\right)^{\frac{1}{\p}}\right) \\
\leq L\left( \frac{\mathfrak{m}_\p}{m^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right)
\end{align}
\medskip
This and the fact that for all $d \in \N$ and $\epsilon \in (0,\infty)$ and the fact that $\mathfrak{m}_\p \leq 2$ it holds that there exists an $\mathfrak{M} \in \R$ such that $\mathscr{M}_\epsilon \geq \mathfrak{M}$ and $\mathscr{M}_\epsilon$ a square number forces:
\begin{align}
L\left( \frac{\mathfrak{m}_\p}{\mathscr{M}_\epsilon^\frac{1}{2}}\right)\left(\left(1+L^2d\right)^{\frac{q\p}{2}} + (q\p+1)d^{\frac{q\p}{2}}\right) \exp \left(\frac{\left[q(q\p+3)+1\right]T}{2}\right) \leq \epsilon
\end{align}
\medskip
Observe that (53) indicates that for all $d,j \in \N$, and $n\in \N_0$ it holds that:
\begin{align}
\C_d \leq \alpha d^{\mathfrak{d}}m
\end{align}
\medskip
Combining this with Corollary 4.3 from HJKP21, applied for every $d\in \N$ with $\gamma \curvearrowleft 0$, $\beta \curvearrowleft m$, $\alpha_0 \curvearrowleft \alpha d^\mathfrak{d}$, $\alpha_1 \curvearrowleft \alpha d^\mathfrak{d}$, $ \left(x_n\right)_{n\in \N_0} \curvearrowleft \left(\C_{d,n,j}\right)_{n\in \N_0}$
\begin{align}
\mathscr{M}_\epsilon \C_d \leq \alpha d^\mathfrak{d} m \mathscr{M}_\epsilon \leq \alpha d^\mathfrak{d} \mathscr{M}_\epsilon^2 \\
\C_d \leq \alpha d^\mathfrak{d}\mathscr{M}_\epsilon
\end{align}
From
\newpage
\section{The case of generic parabolic partial differential equations}
We can easily extend the work for the heat equation to generic parabolic partial differential equations.
\subsection{The case without $f$}
\textbf{Lemma 4.1:} 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) + \left(\Delta_x u_d\right) \left(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 = s + \int^s_t \sqrt{2} dW^d_r = x + \sqrt{2} 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}
\textbf{Setting R3.2} Let $d,m \in \mathbb{N}$ $T,L, \mathfrak{L},p \in [0,\infty)$, $\mathfrak{p} \in [2,\infty)$ $\mathfrak{m} = \mathfrak{k}_{\mathfrak{p}}\sqrt{\mathfrak{p}-1}$, $\Theta = \bigcup_{n\in \mathbb{N}}\mathbb{Z}^n$, $g \in C(\mathbb{R}^d,\mathbb{R})$, assume $\forall t \in [0,T],x\in \mathbb{R}^d$ that:
\begin{align}
\max\{|g(x)|\} \leq \mathfrak{L}(1+||x||^p)
\end{align}
and let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space. Let $W^{\theta}: [0,T] \times \Omega \rightarrow \mathbb{R}^d$, $\theta \in \Theta$ be independent standard brownian motions, $(W^\theta)_{\theta \in \Theta}$ are independent, let $u \in C([0,T] \times \mathbb{R}^d,\mathbb{R})$ satisfy $\forall t \in [0,T]$,$x\in \mathbb{R}^d$, that $\mathbb{E}[|g(x+W^0_{T-t})] < \infty$ and:
\begin{align}
u(t,x) &= \mathbb{E}[g(x+W^0_{T-t})]
\end{align}
and let let $U^\theta_n:[0,T] \times \mathbb{R}^d \times \Omega \rightarrow \mathbb{R}$, $\theta \in \Theta$ satisfy $\forall n \in \mathbb{N}_0$,$\theta \in \Theta, t \in [0,T],x\in \mathbb{R}^d$, that:
\begin{align}
U^\theta_n(t,x) = \frac{}{m^n}\left[\sum^{m^n}_{k=1}g(x+W^{(\theta,0,-k)})\right]
\end{align}
\end{document}