Housekeeping.

.gitignore

@@ -0,0 +1 @@
+spec.db

preamble.sty

@@ -110,7 +110,6 @@
 \newcommand{\proj}{\text{proj}}
 \newcommand{\tr}{\text{tr}}
 \newcommand{\spec}[1]{\text{Spec}\paren{#1}}
-\newcommand{\ba}{\mathbf{A}}
 
 % Dual Pairings & Inner Products
 
@@ -151,7 +150,6 @@
 \newcommand{\cm}{\mathcal{M}}
 \newcommand{\cf}{\mathcal{F}}
 \newcommand{\ce}{\mathcal{E}}
-\newcommand{\ci}{\mathcal{I}}
 \newcommand{\pow}[1]{\mathcal{P}\paren{#1}}
 \newcommand{\cb}{\mathcal{B}}
 \newcommand{\cn}{\mathcal{N}}
@@ -167,7 +165,6 @@
 \newcommand{\cd}{\mathcal{D}}
 \newcommand{\calr}{\mathcal{R}}
 \newcommand{\scp}{\mathscr{P}}
-\newcommand{\wien}{\mathcal{W}}
 
 % Jokes
 \newcommand{\lol}{\boxed{\text{LOL.}}}
@@ -224,3 +221,31 @@
 
 % Real or Complex Numbers
 \newcommand{\RC}{\bracs{\real, \complex}}
+
+
+% AWS
+\newcommand{\ci}{\mathcal{I}}
+\newcommand{\ba}{\mathbf{A}}
+
+
+% Words
+\newcommand{\hldr}{Hölder\ }
+
+% Wiener Space
+\newcommand{\wien}{\mathcal{W}}
+\newcommand{\sps}{{\Theta(\real^d)}} % Sample Path Space
+\newcommand{\ccm}{{\mathbf{H}^1(\real^d)}} % Classical Cameron-Martin Space
+
+% Dual Pairings & Inner Products
+
+
+\newcommand{\floor}[1]{\lfloor #1 \rfloor}
+\newcommand{\ceil}[1]{\lceil #1 \rceil}
+\newcommand{\seqfn}[1]{\bracsn{#1}_1^{n}}
+\newcommand{\seqfm}[1]{\bracs{#1}_1^{m}}
+\newcommand{\seqfmz}[1]{\bracs{#1}_0^{m}}
+\newcommand{\rad}[1]{\text{rad}\!\left(#1\right)}
+\newcommand{\fp}{\mathfrak{p}}
+\newcommand{\fq}{\mathfrak{q}}
+
+

refs.bib

@@ -37,3 +37,37 @@
   year={1989},
   publisher={Taylor \& Francis}
 }
+@book{conway,
+  title={A Course in Functional Analysis},
+  author={Conway, J.B.},
+  isbn={9783540960423},
+  lccn={84010568},
+  series={Graduate Texts in Mathematics},
+  url={https://books.google.ca/books?id=Z9JFAQAAIAAJ},
+  year={1985},
+  publisher={Springer New York}
+}
+
+@book{stroock, place={Cambridge}, edition={3}, title={Probability Theory, An Analytic View}, publisher={Cambridge University Press}, author={Stroock, Daniel W.}, year={2024}} 
+
+@book{folland,
+  title={Real Analysis: Modern Techniques and Their Applications},
+  author={Folland, G.B.},
+  isbn={9781118626399},
+  series={Pure and Applied Mathematics: A Wiley Series of Texts, Monographs and Tracts},
+  url={https://books.google.ca/books?id=wI4fAwAAQBAJ},
+  year={2013},
+  publisher={Wiley}
+}
+
+@book{measure,
+  title={Measure Theory},
+  author={Bogachev, V.I.},
+  number={v. 1},
+  isbn={9783540345145},
+  lccn={2006933997},
+  series={Measure Theory},
+  url={https://books.google.com/books?id=CoSIe7h5mTsC},
+  year={2007},
+  publisher={Springer Berlin Heidelberg}
+}

spec.toml

@@ -1,6 +1,6 @@
 database = "spec.db"
 document = "document.tex"
-siteTitle = "Unnamed Website"
+siteTitle = "Brownian Motion and Stochastic Calculus"
 
 [compiler]
 compileAll = false

src/aws/aws.tex

@@ -0,0 +1,88 @@
+\section{Abstract Wiener Spaces}
+\label{section:aws}
+
+\begin{lemma}[{{\cite[Lemma 8.2.2]{stroock}}}]
+    \label{lem:banach-hilbert}
+
+    Let $E$ be a separable Banach space over $\real$ and $H$ be a Hilbert space over $\real$ that is continuously embedded into $E$, then
+    \begin{enumerate}
+        \item The adjoint of the inclusion $\iota^*: E^* \to H$ is injective and continuous with respect to the weak* topology on $E$ and the weak topology on $H$. In addition, if $L \subset E^*$ is weak*-dense, then $\iota^*(L)$ is dense in $H$.
+        \item For each $x \in E$, $x \in H$ if and only if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E < \infty$. In which case, $\norm{x}_H = \sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E$.
+        \item For each weak*-dense subspace $L \subset E^*$, there exists $\bracs{x^*_n}_1^\infty \subset L$ such that $\bracsn{\iota^*x^*_n}_1^\infty \subset H$ is an orthonormal basis. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(1)$: Since $\iota: H \to E$ is continuous, $\iota^*: E^* \to H$ is continuous with respect to the strong and weak* topologies. As $H \subset E$ is dense, $\iota^*: E^* \to H^*$ is injective.
+
+    Let $h \in H$ such that $h \perp \iota^*(L)$, then by the weak*-density of $L \subset E^*$, $\angles{h, x^*}_E = 0$ for all $x^* \in E$. Therefore $h = 0$, and $\iota^*(L)$ is dense in $H$.
+
+    $(2)$: By $(1)$, the image $\iota^*(E^*) \subset H$ is dense. Hence if $x \in H$, then
+    \[
+    \norm{x}_H = \sup_{\substack{h \in H \\ \norm{h}_H = 1}}\angles{x, h}_H = \sup_{\substack{h \in \iota^*(E^*) \\ \norm{h}_H = 1}}\angles{x, h}_H = \sup_{\substack{x^* \in E^* \\ \norm{\iota^*x^*}_H = 1}}\angles{x, x^*}_E
+    \]
+    On the other hand, if $\sup_{x^* \in E^*, \norm{\iota^*x^*}_H = 1}\angles{x, x^*}_E < \infty$ then the map $\iota^*(E^*) \to \real$ with $\iota^*x^* \mapsto \angles{x, x^*}_E$ is well-defined as $\iota^*$ is injective, and continuous by assumption. By density of $\iota^*(E^*) \subset H$, there exists $h \in H$ such that $\anglesn{h, \iota^*x^*}_H = \angles{h, x^*}_{E} = \angles{x, x^*}_E$ for all $x \in E^*$. In which case, $x = h \in H$.
+
+    $(3)$: Since $E^*$ is separable with respect to the weak* topology, there exists a linearly independent set $\seq{x_n^*} \subset L$ whose span is dense in $E$. By $(1)$, $\seq{\iota^*x_n^*} \subset H$ has dense span in $H$. Using the Gram-Schmidt process, there exists an orthonormal basis $\seq{h_n} \subset \text{span}\bracs{\iota^*x_n^*: n \in \nat}$ for $H$. Since $\iota^*$ is injective and $L$ is a subspace, $\bracsn{(\iota^*)^{-1}(h_n)}_1^\infty \subset L$ is the desired sequence.
+\end{proof}
+
+\begin{definition}[Abstract Wiener Space]
+    \label{define:aws}
+
+    Let $E$ be a separable Banach space over $\real$, $H \subset E$ be a Hilbert space over $\real$ that is continuously embedded into $E$, and $\wien: \cb(E) \to [0, 1]$ be a Gaussian measure, then the triple $(H, E, \wien)$ is an \textbf{abstract Wiener space} if
+    \[
+    \wh \wien(x^*) = e^{-\norm{\iota^* x^*}_H^2/2} \quad \forall x^* \in E^*
+    \]
+\end{definition}
+
+\begin{theorem}[{{\cite[Theorem 8.3.2]{stroock}}}]
+    \label{thm:gaussian-aws}
+
+    Let $E$ be a separable Banach space over $\real$, and $\wien: \cb(E) \to [0, 1]$ be a Borel probability measure, then the following are equivalent:
+    \begin{enumerate}
+        \item $\wien$ is a non-degenerate centred Gaussian measure.
+        \item There exists a Hilbert space $(H, \inp_H)$ contained in $E$ such that $(H, E, \wien)$ is an abstract Wiener space.
+    \end{enumerate}
+    If $(G, \inp_G)$ is another Hilbert space in $E$ such that $(G, E, \wien)$ is an abstract Wiener space, then $G = H$ and $\inp_G = \inp_H$.
+\end{theorem}
+\begin{proof}[Proof of equivalence.]
+    $(1) \Rightarrow (2)$: The requirement on the characteristic function of $\wien$ defines the desired inner product for us, thus the work lies in finding an appropriate subspace and completing it as a Hilbert space. To this end, observe that since each $x^* \in E^*$ viewed as a random variable is Gaussian, the inner product can be taken in $L^2(\wien; \real)$, where we can take the completion, and then recover an element of $E$ by integrating against the identity. 
+    
+    We now begin the construction. By Fernique's theorem, $\int_E \norm{x}_E^2 d\wien(x) < \infty$. Thus there exists a bounded linear map $R: E^* \to L^2(\wien; \real)$ defined by identifying each $x^* \in E^*$ as a $L^2$-random variable on $E$. In addition, since $\text{Id}_E$ itself is an $E$-valued random variable, there exists a bounded linear map
+    \[
+    I: L^2(\wien; \real) \to E \quad Y \mapsto \int_E Y(x)xd\wien(x)
+    \]
+    then for any $x^* \in E^*$ and $Y \in L^2(\wien; \real)$, 
+    \[
+    \angles{IY, x^*}_E = \int_E Y(x)\angles{x, x^*}d\wien(x) = \angles{Y, Rx^*}_{L^2(\wien; \real)}
+    \]
+    In particular, $IY = 0$ if and only if $Y \perp R(E^*)$, and $I|_{\overline{R(E^*)}}$ is injective. 
+    
+    Let $H = I(\overline{R(E^*)})$ and define $\angles{x, y}_H = \anglesn{I^{-1}x, I^{-1}y}_{L^2(\wien; \real)}$. Since $\wien$ is non-degenerate, for any $x^* \in E^*$, $\angles{IRx^*, x^*}_E = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)} > 0$. Thus $H$ is dense in $E$ with $IR = \iota^*: E^* \to H$ corresponding to the adjoint of the inclusion.
+
+    Lastly, since $\wien$ is a centred Gaussian measure, for any $x^* \in E^*$, $\wh \wien (x^*) = e^{-\norm{Rx^*}_{L^2(\wien; \real)}^2/2} = e^{-\normn{\iota^*x^*}_{H}^2/2}$.
+
+    $(2) \Rightarrow (1)$: Since $\wh \wien(x^*) = e^{-\normn{\iota^*x^*}_H^2/2}$ and $\iota^*$ is injective, $x^*$ is a non-degenerate centred Gaussian. Thus $\wien$ is non-degenerate.
+\end{proof}
+\begin{proof}[Proof of uniqueness.]
+    Let $R: E^* \to L^2(\wien; \real)$ and $(H, \inp_H)$ be as in the proof of $(1) \Rightarrow (2)$, and $\iota^*_G: E^* \to G$ be the adjoint of the inclusion $\iota_G: G \to E$. Since $\wh \wien(x^*) = e^{-\norm{\iota_G^*x^*}_G^2/2}$ for all $x^* \in E^*$,
+    \[
+    \angles{\iota_G^*x^*, \iota_G^*x^*}_G = \angles{Rx^*, Rx^*}_{L^2(\wien; \real)}
+    \]
+    and by polarisation, for any $y^* \in E^*$,
+    \[
+    \angles{\iota_G^*x^*, y^*}_E = \angles{\iota_G^*x^*, \iota_G^*y^*}_G = \angles{Rx^*, Ry^*}_{L^2(\wien; \real)} = \angles{IRx^*, y^*}_E
+    \]
+    Therefore $\iota_G^*x^* = IRx^*$ for all $x^* \in E^*$, and the following diagram commutes
+    \[
+    \begin{CD}
+    E^* @>R>> \overline{R(E^*)} \\
+    @A{(\iota_G^*)^{-1}}AA @VV{I}V \\
+    \iota_G^*(E^*) @>{\iota}>> H
+    \end{CD}
+    \]    
+    By density of $\iota^*(E^*)$, the composition $IR(\iota_G^*)^{-1}: \iota_G^*(E^*) \to H$ admits a unique extension to $G$ as an isometry. Therefore $G = H$ with the same inner product.
+\end{proof}
+
+
+
+

src/aws/classical.tex

@@ -0,0 +1,307 @@
+\section{Classical Wiener Space}
+\label{sec:classical}
+
+Let $C([0, \infty); \real^d)$ be the space of $\real^d$-valued continuous functions on $[0, \infty)$, equipped with the topology of uniform convergence on compact sets. The space $C_0((0, \infty); \real^d)$ of $\real^d$-valued continuous functions on $(0, \infty)$ that vanish at $0$ and $\infty$ is equipped with the uniform norm.
+
+
+\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
+    \label{lem:path-space}
+
+    For each $\theta \in C([0, \infty); \real^d)$, let
+    \[
+    \norm{\theta}_{\sps} = \sup_{t \ge 0}\frac{\abs{\theta(t)}}{1 + t}
+    \]
+    and
+    \[
+    \sps = \bracs{\theta \in C([0, \infty); \real^d): \theta(0) = 0, \lim_{t \to  \infty}\frac{\theta(t)}{t} = 0}
+    \]
+    then
+    \begin{enumerate}
+        \item $\norm{\cdot}_{\sps}: C([0, \infty): \real^d) \to [0, \infty]$ is lower semicontinuous. 
+        \item $\iota: \sps \to C([0, \infty); \real^d)$ is continuous.
+        \item The mapping $\Lambda: \sps \to C_0((0, \infty); \real^d)$ defined by $(\Lambda \theta)(t) = \theta(t)/(1 + t)$ is an isometric isomorphism, and $\sps$ is a separable Banach space.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(1)$: Let $R \ge 0$, then
+    \[
+    \bracs{\theta \in C([0, \infty); \real^d): \norm{\theta}_{\sps} \le R} = \bigcap_{t \ge 0}\bracs{\theta \in C([0, \infty); \real^d): \frac{\abs{\theta(t)}}{1 + t} \le R}
+    \]
+    where each $\bracs{\theta \in C([0, \infty); \real^d): \abs{\theta(t)}/(1+t) \le R}$ is a closed subset of $C([0, \infty); \real^d)$, and their intersection is also closed. Thus $\norm{\cdot}_{\sps}$ is lower semicontinuous. 
+
+    $(2)$: Let $K \subset [0, \infty)$ be compact, then $M = \sup_{t \in K}(1 + t) < \infty$. Thus $\norm{\theta|_K}_u \le M\norm{\theta}_{\sps}$.
+
+    $(3)$: Since $\theta(0) = 0$ and $\limv{t}\theta(t)/t = 0$ for all $\theta \in \sps$,
+    \[
+    \lim_{t \to 0}(\Lambda \theta)(t) = \limv{t}(\Lambda \theta)(t) = 0
+    \]
+    and $\norm{\Lambda \theta}_{u, (0, \infty)} = \norm{\theta}_{\sps}$. On the other hand, for any $f \in C_0((0, \infty); \real^d)$, let $\Lambda^{-1}f$ be defined by $(\Lambda^{-1}f)(0) = 0$ and $(\Lambda^{-1}f)(t) = f(t)(1 + t)$ for $t > 0$. Since $\lim_{t \to 0}f(t) = \limv{t}f(t) = 0$,
+    \[
+    \lim_{t \to 0^+}(\Lambda^{-1}f)(t) = \limv{t}\frac{(\Lambda^{-1}f)(t)}{t} = 0
+    \]
+    so $\Lambda^{-1}f \in \sps$ with $\normn{\Lambda^{-1}f}_{\sps} = \norm{f}_{u, (0, \infty)}$.
+
+    As $C_0((0, \infty); \real^d)$ is a separable Banach space, so is $\sps$.
+\end{proof}
+
+\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
+    \label{lem:thetadual}
+
+    For each $\phi \in \sps^*$, there exists a unique $\real^d$-valued Borel measure $\lambda$ on $[0, \infty)$ such that
+    \begin{enumerate}
+        \item $\lambda(0) = 0$.
+        \item $\angles{\theta, \phi}_{\sps} = \int_{[0, \infty)} \theta d\lambda$ for all $\theta \in \sps$.
+        \item $\norm{\phi}_{\sps^*} = \int_{[0, \infty
+    )}(1 + t) d\abs{\lambda}(t) < \infty$.
+    \end{enumerate}
+    If a finite $\real^d$-valued Borel measure $\lambda$ on $[0, \infty)$ satisfies the above, then $\lambda$ defines a linear functional on $\sps^*$.
+\end{lemma}
+\begin{proof}
+    Let $\Lambda: \sps \to C_0((0, \infty); \real^d)$ be the isomorphism defined above, then $\phi \circ \Lambda^{-1} \in C_0((0, \infty); \real^d)^*$. By the Riesz representation theorem, there exists a vector measure $\lambda_0 \in M((0, \infty); \real^d)$ such that
+    \[
+    \anglesn{\Lambda^{-1}f, \phi}_{\sps} = \anglesn{f, \phi \circ \Lambda^{-1}}_{C_0((0, \infty); \real^d)} = \int_{(0, \infty)}f d\lambda_0
+    \]
+    and $\norm{\lambda_0}_{\text{var}} = \normn{\phi \circ \lambda^{-1}}_{\sps}$. Thus for any $\theta \in \sps$, 
+    \[
+    \angles{\theta, \phi}_{\sps} = \anglesn{\Lambda \theta, \phi \circ \Lambda^{-1}}_{C_0((0, \infty); \real^d))} = \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\lambda_0(t)
+    \]
+    and 
+    \[
+    \lambda: \cb([0, \infty)) \to \real^d \quad A \mapsto \int_{A \setminus \bracs{0}}\frac{1}{1 + t}d\lambda_0(t)
+    \]
+    is the desired measure.
+\end{proof}
+
+\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
+    \label{lem:sigma-algebra}
+
+    The space $\sps$ is a Borel set in $C([0, \infty); \real^d)$, and $\cb(\sps)$ is the restriction of the Borel $\sigma$-algebra on $C([0, \infty); \real^d)$ to $\sps$.
+\end{lemma}
+\begin{proof}
+    Firstly, $\lim_{t \to \infty}\theta(t)/t = 0$ if and only if for every $n \in \nat$, there exists $k \in \nat$ such that $\abs{\theta(s)}/s \le 1/n$ for all $s \ge k$. Thus
+    \begin{align*}
+    \Theta_\infty &= \bracs{\theta \in C([0, \infty); \real^d): \limv{t}\frac{\theta(t)}{t} =0} \\
+    &= \bigcap_{n \in \nat}\bigcup_{k \in \nat}\underbrace{\bigcap_{s \ge k}\bracs{\theta \in C([0, \infty); \real^d): \frac{\abs{\theta(s)}}{s} \le \frac{1}{n}}}_{\text{closed}}
+    \end{align*}
+    and
+    \[
+    \Theta_0 = \bracs{\theta \in C([0, \infty); \real^d): \theta(0) = 0}
+    \]
+    are Borel. Thus $\sps = \Theta_0 \cap \Theta_\infty \cap \Theta_b$ is Borel in $C([0, \infty); \real^d)$.
+
+    Now, by lower semicontinuity of $\norm{\cdot}_{\sps}$, every closed set in $\sps$ is the intersection of $\sps$ and a Borel set in $C([0, \infty), \real^d)$. Therefore $\cb(\sps)$ contains the induced $\sigma$-algebra from $\cb(C([0, \infty); \real^d))$. On the other hand, the inclusion $\iota: \sps \to C([0, \infty); \real^d)$ is continuous. Thus $\cb(\sps)$ contains the induced $\sigma$-algebra.
+\end{proof}
+
+\begin{lemma}[{{\cite[Lemma 8.1.1]{stroock}}}]
+    \label{lem:path-justify}
+
+    Let $(\Omega, \cf, \bp)$ be a probability space and $(B_t, \bracs{\cf_t: t \ge 0})$ be a standard Brownian motion, then
+    \begin{enumerate}
+        \item Almost every sample path of $B$ is in $\sps$.
+        \item $\ev[\norm{B}_{\sps}^2] \le 32d$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(1)$: By the Brownian Strong Law of Large Numbers, $B(\omega) \in \sps$ for almost every $\omega$.
+
+    $(2)$: By Doob's maximal inequality, for any $r \ge 0$,
+    \begin{align*}
+    \ev\braks{\norm{B}_{u, [0, 2]}^2}^{1/2} &\le 2\ev\braks{\norm{B_r}_{\real^d}^2}^{1/2} \\
+    \ev\braks{\norm{B}_{u, [0, 2]}^2} &\le 4\ev\braks{\norm{B_r}_{\real^d}^2}
+    \end{align*}
+    and for any $\theta \in \sps$,
+    \begin{align*}
+    \norm{\theta}_{\sps}^2 &\le \sum_{n = 0}^\infty \sup_{t \in [0, 2^n]}\frac{\abs{\theta(t)}^2}{(1 + t)^2} \le \sum_{n = 0}^\infty \frac{1}{2^{2(n-1)}}\norm{\theta}_{u, [0, 2^n]}^2\\
+    &\le \sum_{n = 0}^\infty 4^{1 - n}\norm{\theta}_{u, [0, 2^n]}^2
+    \end{align*}
+    Thus
+    \begin{align*}
+        \ev[\norm{B}_{\sps}^2] &\le \sum_{n = 0}^\infty 4^{1 - n}\ev\braks{\norm{\theta}_{u, [0, 2^n]}^2}\le \sum_{n = 0}^\infty 4^{1 - n}\ev\braks{\norm{B_{2^n}}_{\real^d}^2} \le 32d
+    \end{align*}
+\end{proof}
+
+\begin{definition}[Characteristic Function {{\cite[Lemma 8.1.2]{stroock}}}]
+    \label{def:charfunc}
+
+    Let $(E, \norm{\cdot})$ be a separable Banach space over $\real$, and $\mu$ be a Borel probability measure on $E$. Define
+    \[
+    \wh \mu: E^* \to \complex \quad \wh x^* \mapsto \int_E e^{i\angles{x, x^*}_E}d\mu(x)
+    \]
+    as the \textbf{characteristic function} of $\mu$, then
+    \begin{enumerate}
+        \item $\wh \mu$ is \textit{sequentially}\footnote{Most sources I found only claimed and proved the sequential case. Since Stroock used Dominated Convergence Theorem, I assume it should be sequential too.} continuous with respect to the weak* topology on $E^*$.
+        \item If $\nu$ is a Borel probability measure on $E$ such that $\wh \mu = \wh \nu$, then $\nu = \mu$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    $(1)$: Dominated Convergence Theorem.
+
+    $(2)$: Let $\bracsn{x_j^*}_1^n \subset E^*$ and $\mu_{(x_1^*, \cdots, x_n^*)}$ be the distribution of $(x_1^*, \cdots, x_n^*): E \to \real^n$ under $\mu$, then the characteristic function $\mu_{(x_1^*, \cdots, x_n^*)}$ is
+    \begin{align*}
+        \wh \mu_{(x_1^*, \cdots, x_n^*)}(\xi) &= \int_{\real^n}e^{i\angles{\xi, x}_{\real^n}}d\mu_{(x_1^*, \cdots, x_n^*)}(x) \\
+        &= \int_{E}e^{i\angles{x, \sum_{j = 1}^n \xi_j x_j^*}}d\mu(x) = \wh \mu\paren{\sum_{j = 1}^n \xi_jx_j^*}
+    \end{align*}
+    Thus $\wh \mu$ uniquely determines the joint characteristic function of $(x_1^*, \cdots, x_n^*)$. So for any $\bracsn{x_j^*}_1^n \subset E^*$ and Borel sets $\seqf{B_j} \subset \cb(\real)$, 
+    \[
+    \mu\paren{\bigcap_{j = 1}^n (x_j^*)^{-1}(B_j)} = \nu\paren{\bigcap_{j = 1}^n (x_j^*)^{-1}(B_j)}
+    \]
+    As these cylinder sets form a $\pi$ system that generates $\cb(E)$, $\mu = \nu$ by Dynkin's Uniqueness Theorem.
+\end{proof}
+
+\begin{lemma}[{{\cite[Equation 8.1]{stroock}}}]
+    \label{lem:classical-ft}
+
+    Let $\wien: \cb(\sps) \to [0, 1]$ be the classical Wiener measure on $(\sps, \cb(\real^d))$, then for every $\lambda \in \sps^*$,
+    \begin{enumerate}
+        \item $\lambda$ viewed as a random variable on $\sps$ is a centred Gaussian.
+        \item $\ev^\wien[\abs{\lambda}^2] = \iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.
+        \item $\wh \wien(\lambda) = \exp\braks{-\frac{1}{2}\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)}$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    $(1)$, Finite: First suppose that there exists $0 = t_0 \le t_1 < \cdots < t_n < \infty$ and $\seqf{\xi_j} \subset \real^d$ such that
+    \[
+    \angles{\theta, \lambda}_{\sps} = \sum_{j = 1}^n \angles{\theta(t_j), \xi_j}_{\real^d}
+    \]
+    for all $\theta \in \sps$. This allows decomposing
+    \begin{align*}
+        \angles{\theta, \lambda}_{\sps} &= \sum_{j = 1}^n \sum_{k = 1}^{j}\angles{\theta(t_k) - \theta(t_{k - 1}), \xi_j}_{\real^d} \\
+        &= \sum_{k = 1}^n \angles{\theta(t_k) - \theta(t_{k - 1}), \sum_{j = k}^n\xi_j}_{\real^d}
+    \end{align*}
+    as a sum of independent centred Gaussian random variables. Thus $\theta \mapsto \angles{\theta, \lambda}_{\real^d}$ is a centred Gaussian random variable.
+
+    $(1)$, with Riemann-Stieltjes: Let $\mu: \cb((0, \infty)) \to \real$ be a finite signed measure, and $F(t) = \mu((0, t])$ be the distribution function of $\mu$, then $\norm{F}_{\text{var}} = \norm{\mu}_{\text{var}} < \infty$. 
+    
+    For any indicator function of the form $\one_{(a, b]}$ with $0 \le a < b < \infty$, 
+    \[
+    \int_0^\infty \one_{(a, b]}(t)dF(t) = F(b) - F(a) = \mu((0, t]) = \int_{(0, \infty)}\one_{(a, b]}(t)d\mu(t)
+    \]
+    where the integral on the left is a Riemann-Stieltjes integral. As simple functions that vanish at $0$ are dense in $C_0((0, \infty); \real)$, and
+    \[
+    \abs{\int_0^\infty \phi(t)dF(t)} \le \norm{\phi}_u\norm{F}_{\text{var}} \quad \abs{\int_{(0, \infty)}\phi(t)d\mu(t)} \le \norm{\phi}_u\norm{\mu}_{\text{var}}
+    \]
+    for any simple function $\phi: (0, \infty) \to \real$,
+    \[
+    \int_0^\infty \phi(t)dF(t) = \int_{(0, \infty)}\phi(t)d\mu(t)
+    \]
+    for all $\phi \in C_0((0, \infty); \real)$. 
+
+    As seen in the proof of \autoref{lem:thetadual}, there exists a $\real^d$-valued finite Borel measure $\mu: \cb((0, \infty)) \to \real^d$ such that for any $\theta \in \sps$,
+    \[
+    \angles{\theta, \lambda}_{\sps} = \int_{(0, \infty)} \Lambda \theta d\mu = \int_{(0, \infty)} \frac{\theta(t)}{1 + t}d\mu(t)
+    \]
+    Let $F(t) = \mu((0, t])$, then since $\Lambda \theta \in C_0((0, \infty); \real)$, 
+    \[
+    \angles{\theta, \lambda}_{\sps} = \int_0^\infty \frac{\theta(t)}{1 + t}dF(t)
+    \]
+    can be expressed as a Riemann-Stieltjes integral. Thus
+    \[
+    \angles{\theta, \lambda}_{\sps} = \lim_{t \to \infty}\limv{n}\sum_{j = 1}^{2^n}\theta(2^{-n}jt)[F(2^{-n}jt) - F((2^{-n} - 1)jt)]
+    \]
+    As the above holds for all $\theta \in \sps$, $\lambda$ viewed as a random variable is a\footnote{I'm pretty sure I can just use one single limit here, but just to be safe I nested it.} sequential limit of Gaussian random variables, so it is also Gaussian.
+
+    $(1)$, with Approximations: As seen in \autoref{lem:thetadual}, there exists a $\real^d$-valued Borel measure $\mu: \cb((0, \infty)) \to \real^d$ such that
+    \[
+    \angles{\theta, \lambda}_{\sps} = \int_{(0, \infty)}\frac{\theta(t)}{1 + t}d\mu(t) \quad \forall \theta \in \sps
+    \]
+    Since there exists $\real^d$-valued Borel measures $\seq{\mu_k}$ with finite support such that $\mu_k \Rightarrow \mu$, each $\theta \mapsto \int \frac{\theta(t)}{1 + t}d\mu_k(t)$ is Gaussian random variable and $\theta \mapsto \angles{\theta, \lambda}_{\sps}$ is a pointwise limit of Gaussian random variables, thus it is also Gaussian.
+
+    $(2)$: Firstly, for any $0 \le s, t < \infty$ and $\xi, \eta \in \real^d$,
+    \[
+    \ev^\wien(\angles{\theta(s), \xi} \cdot \angles{\theta(t), \eta}) = (s \wedge t)\angles{\xi, \eta}
+    \]
+    Let $F: [0, \infty) \to \real^d$ be measurable such that $d\lambda = Fd\abs{\lambda}$, then
+    \begin{align*}
+    \angles{\theta, \lambda}_{\sps}^2 &= \braks{\int_{[0, \infty)} \theta(t) d\lambda(t)}^2 = \int_{[0, \infty)}\theta(t)\int_{[0, \infty)} \theta(s)d\lambda(s)d\lambda(t) \\
+    &= \int_{[0, \infty)^2} \angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t)
+    \end{align*}
+    By Fubini's theorem, the expectation can be pushed inside the integral:
+    \begin{align*}
+        \ev^\wien[\abs{\lambda}^2] &= \iint_{[0, \infty)^2} \ev^\wien\braks{\angles{\theta(s), F(s)}_{\real^d}\angles{\theta(t), F(t)}_{\real^d}}d\abs{\lambda}(s)d\abs{\lambda}(t) \\
+        &= \iint_{[0, \infty)^2} (s \wedge t)\angles{F(s), F(t)}_{\real^d}d\abs{\lambda}(s)d\abs{\lambda}(t) \\
+        &= \iint_{[0, \infty)^2} s \wedge t d\lambda(s)d\lambda(t)
+    \end{align*}
+
+    $(3)$: $\lambda$ is a Gaussian random variable with variance $\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.
+\end{proof}
+
+\begin{definition}[Classical Cameron-Martin Space]
+    \label{define:cameron-martin-classical}
+
+    Let $\ccm$ be the set of absolutely continuous functions $\theta: [0, \infty) \to \real^d$ such that $\theta(0) = 0$ and $D\theta \in L^2([0, \infty); \real^d)$, equipped with the inner product
+    \[
+    \angles{\theta, \eta}_{\ccm} = \angles{D\theta, D\eta}_{L^2([0, \infty); \real^d)} \quad \forall \theta, \eta \in \ccm
+    \]
+    then
+    \begin{enumerate}
+        \item For each $\theta \in \ccm$, $\norm{\theta}_{\sps} \le \frac{1}{2}\norm{\theta}_{\ccm}$. 
+        \item $\ccm$ is a dense subspace of $\sps$.
+    \end{enumerate}
+    Thus $\ccm$ is continuously embedded in $\sps$ as a dense subspace, known as the \textbf{classical Cameron-Martin space} for the classical Wiener measure. 
+\end{definition}
+\begin{proof}
+    $(1)$: Let $\theta \in \ccm$, then for each $t \ge 0$,
+    \[
+        \abs{\theta(t)} = \abs{\int_{[0, \infty)} \one_{[0, t)} D\theta ds} \le \normn{\one_{[0, t)}}_{L^2([0, \infty))}\normn{D\theta}_{L^2([0, \infty); \real^d)} = \sqrt{t} \norm{\theta}_\ccm
+    \]
+    Thus
+    \[
+    \frac{\abs{\theta(t)}}{1 + t} \le \norm{\theta}_\ccm \frac{\sqrt{t}}{1 + t} \le \frac{\norm{\theta}_\ccm}{2}
+    \]
+
+    $(2)$: Follows from the density of $C_c^\infty(\real^d)$ in $C_0((0, \infty); \real^d)$ and thus in $\sps$.
+\end{proof}
+
+\begin{lemma}
+    \label{lem:classical-cameron-martin}
+
+    Let $\lambda \in \sps^*$, then there exists $h_\lambda \in \ccm$ such that $\angles{\theta, \lambda}_\sps = \angles{\theta, h_\lambda}_\ccm$ for all $\theta \in \ccm$, which is given by
+    \[
+    h_\lambda(t) = \int_{(0,t]}\lambda((s, \infty))ds
+    \]
+    with
+    \[
+    \norm{h_\lambda}_\ccm^2 = \iint_{[0, \infty)}(s \wedge t)d\lambda(s)d\lambda(t) = \ev^\wien(\abs{\lambda}^2)
+    \]
+    thus
+    \[
+    \wh \wien(\lambda) = \exp\braks{-\frac{\norm{h_\lambda}_\ccm}{2}^2}
+    \]
+    Moreover, for each $\theta \in \sps$,
+    \[
+    \angles{\theta, \lambda}_\sps = \int_0^\infty Dh_\lambda d\theta
+    \]
+    is an improper Riemann-Stieltjes integral.
+\end{lemma}
+\begin{proof}
+    Firstly, $\lambda$ is bounded, so $h_\lambda$ is absolutely continuous and $Dh_\lambda(s) = \lambda((s, \infty))$ by the Lebesgue differentiation theorem. From here,
+    \begin{align*}
+        \norm{h_\lambda}_\ccm^2 &= \int_{[0, \infty)}\abs{\lambda((t, \infty))}^2dt = \int_{[0, \infty)}\iint_{[t, \infty)^2}d\lambda(r)d\lambda(s)dt \\
+        &= \iiint_{[0, \infty)^3}\one_{\bracs{r, s \ge t}}(r, s, t)d\lambda(r)d\lambda(s)dt \\
+        &= \iint_{[0, \infty)^2} \int_{[0, \infty)}\one_{\bracs{t \le r \wedge s}}(r, s, t)dtd\lambda(r)d\lambda(s) \\
+        &= \iint_{[0, \infty)^2}s \wedge t d\lambda(r)d\lambda(s) = \ev^\wien[\abs{\lambda}^2]
+    \end{align*}
+    so $h_\lambda \in \ccm$ and
+    \begin{align*}
+        \angles{h, \lambda}_\sps &= \int_{[0, \infty)} h d\lambda = \int_{[0, \infty)} \int_{[0, t]}Dh(s)dsd\lambda(t) \\
+        &= \int_{[0, \infty)} \int_{[0, \infty)}\one_{\bracs{s \le t}}(s, t)Dh(s)d\lambda(t) ds \\
+        &= \int_{[0, \infty)}Dh(s)\lambda((s, \infty))ds = \angles{\theta, h_\lambda}_\ccm
+    \end{align*}
+    Now let $F(t) = \lambda((0, t])$, then $Dh_\lambda(t) = \lambda((0, \infty)) - F(t)$ and
+    \[
+    \int_0^\infty \theta(t) d\lambda(t) = \int_0^\infty \theta(t)dF(t) = -\int_0^\infty \theta(t) dDh_\lambda(t)
+    \]
+    can be expressed as improper Riemann-Stieltjes integrals. Using the by parts formula for Riemann-Stieltjes integrals, 
+    \[
+    \int_0^T \theta(t) dDh_\lambda(t) = \theta(T)Dh_\lambda(T) - \int_0^TDh_\lambda(t)d\theta(t)
+    \]
+    where since $\theta(T)/T \to 0$ as $T \to \infty$ and $(1 + t) \in L^1(\abs{\lambda})$,
+    \[
+    \limv{T}\abs{\theta(T)}\abs{\lambda}((T, \infty)) \le \limv{T}\frac{\abs{\theta(T)}}{1 + T}\int_{(0, \infty)}(1 + t)d\abs{\lambda}(t) = 0
+    \]
+    Therefore
+    \[
+    \angles{h, \lambda}_\sps = -\int_0^\infty \theta(t)dDh_\lambda(t) = \int_0^\infty Dh_\lambda(t)d\theta(t)
+    \]
+\end{proof}
+

src/aws/cm.tex

@@ -0,0 +1,93 @@
+\section{Cameron-Martin Space}
+\label{section:cm-space}
+
+\begin{proposition}[{{\cite[Theorem 8.2.6]{stroock}}}]
+    \label{prop:aws-compact}
+    Let $(H, E, \wien)$ be an abstract Wiener space, then the operators
+    \[
+    \begin{CD}
+        E^* @>{\iota^*}>> H @>{\iota}>> E
+    \end{CD}
+    \]
+    are compact. In particular, $H \in \cb(E)$.
+\end{proposition}
+\begin{proof}
+    It is sufficient to show that both operators map weakly convergent sequences to strongly convergent sequences. 
+
+    Let $\seq{x_n^*} \subset E^*$ such that $x_n^* \to x \in E^*$ weakly, then since $\wh \wien: E^* \to \complex$ is continuous with respect to sequential weak* convergence, 
+    \[
+    e^{-\norm{\iota^*x_n^* - \iota^*x^*}_H^2/2} = \wh \wien (x^*_n - x^*) \to 1
+    \]
+    as $n \to \infty$. Thus $\iota^* x_n^* \to \iota^*x^*$ strongly in $H$.
+
+    To see the compactness of $\iota$, let $\seq{h_n} \subset H$ such that $h_n \to 0$ weakly. By the Uniform Boundedness Principle, there exists $R \ge 0$ such that $\sup_{n \in \nat}\norm{h_n}_H \le R$. Let $\seq{x_n^*} \subset E^*$ be a weak*-dense subset of $E^*$ with $\norm{x_n^*}_{E^*} = 1$ for all $n \in \nat$. Let $\eps > 0$, then by compactness of $\seq{\iota^*x^*_n} \subset H$, there exists $N \in \nat$ such that $\seq{\iota^*x^*_n} \subset \bigcup_{n = 1}^N B_H(\iota^*x^*_n, \eps)$. In which case, 
+    \[
+        \norm{h_n}_E =  \sup_{n \in \nat}\angles{h_n, x_n^*}_{E} \le 2\max_{1 \le k \le N}\angles{h_n, x_k^*}_E + R\eps 
+    \]
+    Given that $h_n \to 0$ weakly, $\max_{1 \le k \le N}\angles{h_n, x_k^*}_E \to 0$ as $n \to \infty$. As $\eps > 0$ is arbitrary, $\norm{h_n}_E \to 0$ as $n \to \infty$.
+\end{proof}
+
+\begin{proposition}
+    Let $(H, E, \wien)$ be an abstract Wiener space. If $E$ is infinite-dimensional, then $\wien(H) = 0$.
+\end{proposition}
+\begin{proof}
+    Let $\seq{x_n^*} \subset E^*$ such that $\seq{\iota^* x_n^*}$ is an orthonormal basis for $H$. For each $n \in \nat$, let $X_n(x) = \dpb{x, x_n^*}{E}$, then $\seq{X_n}$ is a family of independent Gaussian random variables with variance $1$. Thus $\sum_{n \in \nat} X_n^2 = \infty$ almost surely. By (2) of \autoref{lem:banach-hilbert}, $\wien(H) = 0$.
+\end{proof}
+
+
+\begin{theorem}
+    \label{thm:paley-wiener}
+    Let $(H, E, \wien)$ be an abstract Wiener space, then 
+    \begin{enumerate}
+        \item There exists a unique linear isometry $\ci: H \to L^2(\wien; \real)$ such that $\ci(\iota^*x^*) = \dpb{\cdot, x^*}{E}$ for all $x^* \in E^*$.
+        \item $\bracs{\ci(h): h \in H}$ is a Gaussian family.
+    \end{enumerate}
+    The map in (1) is the \textbf{Paley-Wiener map}.
+\end{theorem}
+\begin{proof}
+    Let $x^* \in E^*$, then $\dpb{\cdot, x^*}{E}$ is a centred Gaussian random variable with variance $\norm{\iota^*x^*}_H^2$. By (3) of \autoref{lem:banach-hilbert}, $\iota^*(E^*) \subset H$ is dense. Thus $\ci$ extends uniquely into a continuous linear isometry.
+\end{proof}
+
+\begin{theorem}[Cameron-Martin, {{\cite[Theorem 8.2.7]{stroock}}}]
+    \label{thm:cameron-martin}
+    Let $(H, E, \wien)$ be an abstract Wiener space and $g \in E$.  Denote $\tau_g: E \to E$ with $x \mapsto x + g$ as the translation map, then:
+    \begin{enumerate}
+        \item If $g \in H$, then $(\tau_g)_* \ll \wien$ and
+        \[
+        \frac{d(\tau_g)_*\wien}{d\wien}(x) = e^{\ci (g) - \norm{g}_H^2/2}
+        \]
+        \item If $g \not\in H$, then $(\tau_g)_* \wien \perp \wien$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $\mu = (\tau_g)_* \wien$, then for any $x^* \in E^*$,
+    \[
+    \wh \mu(x^*) = \ev^\wien\braks{e^{i\angles{x + g, x^*}_E}} = e^{i\dpb{g, x^*}{E} - \norm{\iota^*x^*}_H^2/2}
+    \]
+    On the other hand, for any $x, y \in H$, $\ci(x)$ and $\ci(y)$ are jointly Gaussian, so for any $\xi, \eta \in \complex$,
+    \[
+    \ev^\wien\braks{e^{\xi \ci(x) + \eta \ci(y)}} = \exp\braks{\frac{1}{2}\paren{\xi^2\norm{x}_H^2 + 2\xi \eta \dpb{x, y}{H} + \eta^2\norm{y}_H^2}}
+    \]
+    In particular, if $\xi = 1$, $\eta = i$, $x = g$, and $y = x^*$, then
+    \[
+    \ev^\wien\braks{e^{\ci (g) + ix^*}} = e^{i\dpb{g, x^*}{E} - \norm{\iota^*x^*}_H^2/2} = \wh \mu(x^*)
+    \]
+
+    (2): Let $x^* \in E^*$ with $\norm{\iota^*x^*}_H = 1$, and $\mathcal F = \sigma(x^*)$, then
+    \[
+    \frac{d\mu|_{\cf}}{d\wien_\cf}(x) = e^{\dpb{g, x^*}{E}\dpb{x, x^*}{E} - \dpb{g, x^*}{E}^2/2}
+    \]
+    If $\mu = \lambda + Fd\wien$ is the Lebesgue decomposition of $\mu$ with respect to $\wien$, then
+    \[
+    \frac{d\mu|_{\cf}}{d\wien_\cf} \ge \ev^\wien[F|\cf]
+    \]
+    Given that $\norm{\iota^*x^*}_H = 1$,
+    \[
+    e^{\dpb{g, x^*}{E}^2}= \ev^\wien\braks{\paren{\frac{d\mu|_{\cf}}{d\wien_\cf}}^2} \ge \ev^\wien(F^2)
+    \]
+    where $\ev^\wien(F^2)$ does not depend on $x^*$. If $F \ne 0$, then the above bound implies that 
+    \[
+    \sup_{\substack{x^* \in E^* \\ ||\iota^* x^*||_H = 1}}|\dpb{g, x^*}{E}^2| < \infty
+    \]
+    Thus $g \in H$.
+\end{proof}

src/aws/donsker.tex

@@ -0,0 +1,135 @@
+\section{Donsker's Theorem}
+\label{section:donsker}
+
+In this section, equip $\Omega = C(\real_{\ge 0}; \real)$ with the topology of uniform convergence on compact sets. 
+
+For every $\delta > 0$ and $N \in \nat$, denote
+\[
+V_{\delta, n}(\omega) = \sup\bracs{\abs{\pi_s(\omega) - \pi_t(\omega)}| s, t \in [0, N], \abs{s - t} \le \delta}
+\]
+then
+
+\begin{theorem}[Arzela-Ascoli]
+    \label{thm:arzela-ascoli}
+    Let $K \subset \Omega$, then $K$ is compact if and only if the following holds:
+    \begin{enumerate}
+        \item $\pi_0(K)$ is precompact.
+        \item For each $N \in \nat$,
+        \[
+        \lim_{\delta \to 0}\sup_{\omega \in K}V_{\delta, N}(\omega) = 0
+        \]
+    \end{enumerate}
+\end{theorem}
+
+\begin{theorem}[Prokhorov]
+    \label{thm:prokhorov}
+    Let $X$ be a Polish space and $\mathcal{P} \subset \mathbf{M}_1(X, \cb(X))$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $\alpha \in (0, 1)$, there exists $K \subset X$ compact such that
+        \[
+        \inf_{\mu \in \mathcal{P}}\mu(K) \ge \alpha
+        \]
+        \item $\mathcal{P}$ is weakly precompact. 
+    \end{enumerate}
+\end{theorem}
+
+\begin{proposition}
+    \label{prop:weakly-precompact-theta}
+
+    Let $\seq{\mu_n} \subset \mathbf{M}_1(\Omega)$, then $\seq{\mu_n}$ is weakly precompact if and only if the following holds:
+    \begin{enumerate}
+        \item $\seq{(\pi_0)_*\mu_n}$ is tight. 
+        \item For any $N \in \nat$ and $\eta > 0$,
+        \[
+        \lim_{\delta \to 0}\sup_{n \in \nat}\mu_n(\bracs{V_{\delta, N} > \eta}) = 0
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Using Prokhorov's theorem, it is sufficiently to show that $\seq{\mu_n}$ is tight. 
+    
+    To this end, let $\eps > 0$, then there exists $K \subset \real$ compact such that
+    \[
+    \sup_{n \in \nat}\mu_n(\bracs{\pi_0 \not\in K}) \le \eps
+    \]
+    Fix $N \in \nat$, then for each $k \in \nat$, there exists $\delta(N, k) > 0$ such that
+    \[
+    \sup_{n \in \nat}\mu_n\paren{\bracs{V_{\delta(N, k), N} > \frac{1}{k}}} \le 2^{-N-k} \cdot \eps
+    \]
+    In which case, 
+    \[
+    \sup_{n \in \nat}\mu_n\paren{\bracs{\omega \in \Omega \bigg | \forall k \in \nat, V_{\delta(N, k), N}(\omega) \le \frac{1}{k}}^c} \ge 1 - 2^{-N}\eps
+    \]
+    Let $K'$ be the intersection of $\bracs{\pi_0 \in K}$ and the above compact set, then $K'$ is precompact by the Arzela-Ascoli theorem, and $\sup_{n \in \nat}\mu_n(\Omega \setminus K') \le 2\eps$.
+\end{proof}
+
+\begin{definition}
+    \label{define:remainder}
+    Let $t \ge 0$ and $n \in \nat$, then
+    \[
+    \floor{t}_n = \max\bracs{\frac{k}{n} \bigg |k \in \nat_0, \frac{k}{n} \le t}
+    \]
+    is the \textbf{$n$-adic floor} of $t$, respectively, and
+    \[
+    r_n(t) = t - \fl{t}_n
+    \]
+    is the \textbf{$n$-adic remainder} of $t$.
+\end{definition}
+
+\begin{definition}
+    \label{define:linearisation}
+    Let $f: \bracs{k/n|k \in \nat_0} \to \real$, then the function
+    \[
+    L_nf: \real_{\ge 0} \to \real \quad t \mapsto (1 - nr_n(t))f(\fl{t}_n) + nr_n(t)f(\fl{t}_n + n^{-1})
+    \]
+    is the \textbf{$n$-adic linearisation} of $f$. 
+\end{definition}
+
+
+\begin{theorem}[Donsker]
+    \label{thm:donsker}
+    Let $\seq{S_n}$ be the symmetric random walk on $\integer$. For each $n \in \nat$, let $S^n_t = L_nP_n$ be the $n$-adic linearisation of
+    \[
+    P_n: \bracs{k/n|k \in \nat_0} \to \real \quad \frac{k}{n} \mapsto \frac{S_k}{\sqrt{n}}
+    \]
+    Let $B$ be the standard Brownian motion, then $S^n \to B$ in distribution as $n \to \infty$.
+\end{theorem}
+\begin{proof}
+     For any $0 \le s < t \le <\infty$, 
+    \[
+    S^n_t - S^n_s = \frac{S_{n\fl{t}_n} - S_{n\fl{s}_n}}{\sqrt{n}} + \frac{R_{n, s, t}}{\sqrt{n}}
+    \]
+    where $\sup_{n \in \nat}\norm{R_{n, s, t}}_{L^\infty} < \infty$\footnote{I believe that 3 is an easy bound, but I cannot be bothered to check.}, so $R_{n, s, t}/\sqrt{n} \to 0$ in $L^\infty$ as $n \to \infty$. 
+    
+    Suppose that $S^n$ \textit{converges} in distribution. By the above calculation, it is sufficient to consider the convergence in distribution of $(S_{n\floor{t}_n} - S_{n\floor{s}_n})/\sqrt{n}$. 
+
+    Since $\sqrt{n}(\fl{t}_n - \fl{s}_n)/\sqrt{n} \to (t - s)$ as $n \to \infty$, for any $0 \le t_1 < \cdots < t_k < \infty$, 
+    \[
+    \frac{1}{\sqrt{n}}\paren{S_{n\fl{t_2}_n} - S_{n\fl{t_1}_n}, \cdots, S_{n\fl{t_k}_n} - S_{n\fl{t_{k-1}}_n}} \to (X_1, \cdots, X_k)
+    \]
+    in distribution by the central limit theorem, where $\bracs{X_j}_1^k$ are independent random variables with $X_j \sim \gamma_{0, t_{j+1} - t_{j}}$. Therefore $S^n \to B$ in distribution. 
+
+    To see that $S^n$ does converge in distribution, it is sufficient to show that it is tight. Given that $S^n_0 = 0$ for all $n \in \nat$, it is sufficient to check (2) of \autoref{prop:weakly-precompact-theta}. 
+
+    Let $\lambda > 0$. To achieve the appropriate order of bound, apply Doob's maximal inequality to $S_n^4$.
+    \begin{align*}
+        \bp\bracs{\max_{1 \le k \le n}|S_k| > \lambda \sqrt{n}} &= \bp\bracs{\max_{1 \le k \le n}|S_k^4| > \lambda^4n^2} \le \frac{1}{\lambda^4n^2}\ev\braks{|S_n^4|} \\
+        &=  \frac{1}{\lambda^4 n^2}\braks{n + {n \choose 2}{4 \choose 2}} \\
+        &= \frac{3n^2 - 2n}{\lambda^4n^2} \le \frac{3}{\lambda^4}
+    \end{align*}
+    For any $\delta > 0$ and $\eps > 0$, if $3/\sqrt{N} < \eps$, then for any $0 \le s < t$ with $t - s < \delta$, and $n \ge N$,
+    \begin{align*}
+    \bp\bracs{|S_t^n - S^n_s| \ge 2\eps} &\le \bp\bracs{|S_{n\floor{t}_n} - S_{n\floor{s}_n}| \ge \frac{\eps}{\sqrt{\delta}}\sqrt{n \delta}}\\
+    &\le \frac{3 \delta^2}{\eps^4}
+    \end{align*}
+    In particular, for any $\eta > 0$, a choice of $\delta = \eta \eps^4/3$ yields
+    \[
+    \bp\bracs{|S_t^n - S^n_s| \ge 2\eps} \le \frac{3 \delta^2}{\eps^4} = \frac{\eta^2\eps^4}{3} = \eta \delta
+    \]
+    Partition the interval $[0, N]$ into $\lceil N/\delta \rceil$ pieces of of intervals of length $\delta$, then
+    \[
+    \bp\bracs{\sup_{\substack{0 \le s < t \\ t - s < \delta}}|S_t^n - S_s^n| \ge 2\eps} \le (1 + N/\delta)\eta \delta = \eta \delta + N\eta
+    \]
+\end{proof}
+
+

src/aws/gaussian.tex

@@ -0,0 +1,97 @@
+\section{Gaussian Measures}
+\label{sec:gaussian}
+
+\begin{definition}
+    \label{define:gaussian}
+
+    Let $E$ be a separable Banach space over $\real$ and $\wien$ be a Borel measure on $E$, then $\wien$ is a \textbf{Gaussian measure} if $x \mapsto \angles{x, x^*}_E$ is a centred Gaussian random variable for each $x^* \in E^*$. If each $x \mapsto \angles{x, x^*}_E$ is non-degenerate, then $\wien$ is \textbf{non-degenerate}.
+\end{definition}
+
+\begin{lemma}
+    \label{lem:shift}
+
+    Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X, X': \Omega \to E$ be i.i.d Gaussian random variables, then $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ have the same distribution as $(X, X')$.
+\end{lemma}
+\begin{proof}
+    Let $x^*, y^* \in E^*$, $Y(\omega) = \angles{X(\omega), x^*}_E$ and $Y'(\omega) = \anglesn{X'(\omega), x^*}_E$, then $Y$ and $Y'$ are independent Gaussian random variables, so
+    \begin{align*}
+        \ev^\bp\braks{e^{i\angles{X, x^*}_E + i\anglesn{X', y^*}_E}} &= \ev^\bp\braks{e^{i\angles{X, x^*}_E}}\ev^\bp\braks{e^{i\anglesn{X', y^*}_E}} \\
+        &= \exp\braks{-\frac{1}{2}\paren{\ev^\bp\braks{\angles{X, x^*}_E^2} + \ev^\bp\braks{\angles{X', y^*}_E^2}}}
+    \end{align*}
+    On the other hand, 
+    \begin{align*}
+        &\ev^\bp\braks{e^{i\anglesn{(X + X')/\sqrt{2}, x^*}_E + i\anglesn{(X - X')/\sqrt{2}, y^*}_E}} \\
+        &= \ev^\bp\braks{e^{i\anglesn{X/\sqrt{2}, x^* + y^*}_E}}\ev^\bp\braks{e^{i\anglesn{X'/\sqrt{2}, x^* - y^*}_E}} \\
+        &= \exp\braks{-\frac{1}{4}\paren{\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2}}}
+    \end{align*}
+    where
+    \begin{align*}
+        \angles{X, x^* + y^*}_E^2 &= \angles{X, x^*}_E^2 + \angles{X, y^*}_E^2 + 2\angles{X, x^*}_E\angles{X, y^*}_E \\
+        \angles{X', x^* - y^*}_E^2 &= \angles{X', x^*}_E^2 + \angles{X', y^*}_E^2 - 2\angles{X', x^*}_E\angles{X', y^*}_E
+    \end{align*}
+    Since $X$ and $X'$ are identically distributed, 
+    \begin{align*}
+        &\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2} \\
+        &= \ev^\bp\braks{\angles{X, x^*}_E^2 + \angles{X', x^*}_E^2} + \ev^\bp\braks{\angles{X, y^*}_E^2 + \angles{X', y^*}_E^2} \\
+        &+2\ev^\bp\braks{\angles{X, x^*}_E\angles{X, y^*}_E} - \ev^\bp\braks{2\angles{X', x^*}_E\angles{X', y^*}_E} \\
+        &= 2\ev^\bp\braks{\angles{X, x^*}_E^2} + 2\ev^\bp\braks{\angles{X', y^*}_E^2}
+    \end{align*}
+    Thus the characteristic functions of $(X, X')$ and $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ are equal. By \autoref{def:charfunc}, their distributions are equal.
+\end{proof}
+
+\begin{theorem}[{{Fernique, \cite[Theorem 8.2.1]{stroock}}}]
+    \label{thm:fernique}
+
+    Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X: \Omega \to E$ be a centred Gaussian random variable. Let $R > 0$ such that $\bp\bracs{\norm{X}_E \le R} \ge 9/10$, then
+    \[
+    \ev^\bp\braks{e^{\frac{\norm{X}_E^2}{18R^2}}} \le e^{1/2} + \sum_{n = 0}^\infty \paren{\frac{e}{3}}^{2^n}
+    \]
+\end{theorem}
+\begin{proof}
+    By enlarging the sample space, let $X': \Omega \to E$ be independent from $X$ and with the same distribution as $X$. For any $0 \le s \le t < \infty$,
+    \begin{align*}
+        &\bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t} \\
+        &= \bp\bracs{\norm{X}_E \le s}\bp\bracsn{\normn{X'}_E \ge t} = \bp\bracs{\norm{X}_E \le s, \normn{X'}_E \ge t} \\
+        &= \bp\bracs{\norm{X - X'}_E \le \sqrt{2}s, \normn{X + X'}_E \ge \sqrt{2}t} \\
+        &\le \bp\bracs{\abs{\norm{X}_E - \normn{X'}_E} \le \sqrt{2}s, \norm{X}_E + \normn{X'}_E \ge \sqrt{2}t}
+    \end{align*}
+    Here, if $\abs{\norm{X}_E - \normn{X'}_E} \le \sqrt{2}s$, then
+    \[
+    \sqrt{2}t \le \norm{X}_E + \norm{X'}_E \le 2(\norm{X}_E \wedge \norm{X'}_E) + \sqrt{2}s
+    \]
+    Thus $\norm{X}_E + \norm{X'}_E \ge \sqrt{2}(t - s)$, so
+    \begin{align*}
+        \bp\bracs{\norm{X}_E \le s}\bp\bracs{\norm{X}_E \ge t} &\le \bp\bracs{\norm{X}_E \wedge \norm{X'}_E \ge (t - s)/\sqrt{2}} \\
+        &= \bp\bracs{\norm{X}_E \ge (t - s)/\sqrt{2}}^2
+    \end{align*}
+    Let $t_0 = R$, and $t_n = R + \sqrt{2}t_{n - 1}$ for all $n \in \nat$. From here, for each $n \in \nat$,
+    \begin{align*}
+        \bp\bracs{\norm{X}_E \le R}\bp\bracs{\norm{X}_E \ge t_n} &\le \bp\bracs{\norm{X}_E \ge (t_n - R)/\sqrt{2}}^2 \\
+        &= \bp\bracs{\norm{X}_E \ge t_{n-1}}^2 \\
+        \frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}} &\le \braks{\frac{\bp\bracs{\norm{X}_E \ge t_{n - 1}}}{\bp\bracs{\norm{X}_E \le R}}}^2
+    \end{align*}
+    Thus
+    \begin{align*}
+    \frac{\bp\bracs{\norm{X}_E \ge t_n}}{\bp\bracs{\norm{X}_E \le R}} &\le \braks{\frac{\bp\bracs{\norm{X}_E \ge R}}{\bp\bracs{\norm{X}_E \le R}}}^{2^n} \le 9^{-2^n} \\
+    \bp\bracs{\norm{X}_E \ge t_n} &\le 9^{-2^{n}} \le 3^{-2^{n+1}}
+    \end{align*}
+    so for any $K > 0$,
+    \begin{align*}
+        \ev^\bp\braks{e^{\norm{X}_E^2/K}} &\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K}\bp\bracs{\norm{X}_E \ge t_n} \\
+        &\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K} \cdot 9^{-2^n}\le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{t_{n+1}^2/K} \cdot 3^{-2^{n+1}}
+    \end{align*}
+    Now, since
+    \begin{align*}
+    t_N &= R + \sqrt{2}t_{N - 1} = R + \sqrt{2}(R + \sqrt{2}t_{N - 2}) \\
+    &= R + \sqrt{2}R + \sqrt{2}^2(R + \sqrt{2}t_{N - 3}) = R\sum_{n = 0}^{N}\sqrt{2}^n = \frac{R(1 - \sqrt{2}^{N+1})}{1 - \sqrt{2}} \\
+    &\le 3R\sqrt{2}^{N+1} = 3R \cdot 2^{(N+1)/2}
+    \end{align*}
+    so if $K = 18R^2$,
+    \[
+    \frac{t^2_{n+1}}{K} \le \frac{9R^2 \cdot 2^{(N+2)}}{18R^2} = 2^{N+1}
+    \]
+    Thus
+    \[
+        \ev^\bp\braks{e^{\norm{X}_E^2/K}} \le e^{t_0^2/K} + \sum_{n = 0}^\infty e^{2^{n+1}} \cdot 3^{-2^{n+1}} \le e^{1/2} + \sum_{n = 0}^\infty \paren{\frac{e}{3}}^{2^n}
+    \]
+\end{proof}

src/aws/index.tex

@@ -0,0 +1,10 @@
+\chapter{Abstract Wiener Spaces}
+\label{chap:aws}
+
+\input{./classical.tex}
+\input{./gaussian.tex}
+\input{./aws.tex}
+\input{./cm.tex}
+\input{./wiener.tex}
+\input{./donsker.tex}
+

src/aws/wiener.tex

@@ -0,0 +1,57 @@
+\section{Wiener's Construction}
+\label{section:wiener-construction}
+
+
+\begin{theorem}[{{\cite[Theorem 8.3.1]{stroock}}}]
+    \label{thm:wiener-consurtcion}
+
+    Let $H$ be an infinite-dimensional, separable Hilbert space over $\real$, $E$ be a Banach space with $H$ continuously embedded as a dense subspace.
+    
+    Let $\seq{h_n} \subset H$ be an orthonormal basis and $\seq{X_n}$ be mutually independent $\gamma_{0, 1}$-distributed random variables on a probability space $(\Omega, \cf, \bp)$. If
+    \begin{enumerate}
+        \item $\sum_{n = 1}^\infty X_n h_n$ converges in $E$ almost surely.
+        \item $S: \Omega \to E$ is given by
+        \[
+        S(\omega) = \begin{cases}
+            \sum_{n = 1}^\infty X_n(\omega) h_n &\text{The series converges in } $E$ \\
+            0 &\text{Otherwise}
+        \end{cases}
+        \]
+        \item $\wien = S_* \bp$.
+    \end{enumerate}
+    then $(H, E, \wien)$ is an abstract Wiener space.
+\end{theorem}
+\begin{proof}
+    Let $S_n = \sum_{k = 1}^n X_kh_k$, then
+    \[
+    \wh \wien(x^*) = \limv{n}\ev^\bp\braks{e^{\dpb{S_n, x^*}{E}}} = \limv_{n}\prod_{k = 1}^n e^{-\dpb{\iota^*x^*, h_k}{H}^2/2} = e^{-\norm{\iota^*x^*}_H^2/2}
+    \]
+\end{proof}
+
+\begin{theorem}
+    \label{thm:wiener-structure}
+
+    Let $(H, E, \wien)$ be an abstract Wiener space and $\seq{h_n}$ be an orthonormal sequence, then
+    \begin{enumerate}
+        \item For any $p \in [1, \infty)$,
+        \[
+        \ev^\wien\braks{\sup_{n \in \nat}\norm{\sum_{k = 1}^n \ci(h_k)h_k}_E^p} < \infty
+        \]
+        \item For $\wien$-almost every $x \in E$,
+        \[
+        \sum_{n = 1}^\infty \ci(h_k)(x)h_k = \ev^\wien\braks{x|\sigma(\bracs{\ci(h_n): n \in \nat})}
+        \]
+        \item $\sum_{n = 1}^\infty \ci(h_n)h_n$ is $\wien$-independent of $x - \sum_{n = 1}^\infty \ci(h_n)h_m$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Let $n \in \nat$, $\cf_n = \sigma(\bracs{\ci(h_k): 1 \le k \le n})$, and $S_n = \sum_{k = 1}^n \ci(h_k)h_k$, then for any $x^* \in E^*$, $\dpb{x - S_n(x), x^*}{E} \perp h_k$ in $L^2(\wien; E)$ for all $k \le n$. Since $\bracs{\ci(h_n): n \in \nat}$ is a Gaussian family, $x - S_n(x)$ is independent of $\seqf{h_k}$ and $\cf_n$. 
+
+    (1): By \autoref{thm:fernique}, $x \in L^p(\wien; E)$ for all $p \in [1, \infty)$, so (1) holds.
+    
+    (2): $S_n = \ev^\wien[x|\cf_n]$, and $S_n \to \ev^\wien[x|\sigma(\bracs{\ci(h_n): n \in \nat})]$ $\wien$-almost surely.
+
+    (3): Since $x - S_n(x)$ is independent of $\seq{h_k}$ for each $n \in \nat$, $x - S$ is independent of $S$.
+\end{proof}
+
+

src/calculus/index.tex

@@ -1,4 +1,4 @@
-\chapter{Integral Manifolds/Shenanigans}
+\chapter{Integral Manifolds}
 \label{chap:integral}
 
 \input{./integrable.tex}

src/index.tex

@@ -1,6 +1,14 @@
-\part{Diffusion Processes}
+\part{Stochastic Processes}
 \label{part:diffusion}
 
+\input{./aws/index.tex}
+\input{./operator/index.tex}
+
+
+\part{Stochastic Calculus}
+\label{part:calculus}
+
 \input{./diffusion/index.tex}
-\input{./calculus/index.tex}
+\input{./sde/index.tex}
-\input{./sde/index.tex}
+\input{./mal/index.tex}
+\input{./calculus/index.tex}

src/mal/bridge.tex

@@ -0,0 +1,40 @@
+\section{The Brownian Bridge}
+\label{section:brownian-bridge}
+
+
+\begin{definition}[Brownian Bridge]
+\label{definition:brownian-bridge}
+    Let $(\Omega, \cf, \bp)$ be a probability space, $a \in \real^d$, and $X: \Omega \to C([0, 1]; \real^d)$ be a Gaussian process, then $X$ is a \textbf{Brownian bridge} from $0$ to $a$ such that:
+    \begin{enumerate}
+        \item For each $t \in [0, 1]$, $\ev(X_t) = at$.
+        \item For each $s, t \in [0, 1]$, $\text{Cov}(X_s, X_t) = (s \wedge t - st)I$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{theorem}[{{\cite[Theorem 40.3]{Rogers}}}]
+\label{theorem:brownian-bridge}
+    Let $a \in \real^d$, then the following are equivalent definitions of the distribution of the Brownian bridge from $0$ to $a$. 
+    \begin{enumerate}
+        \item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$, 
+        \[
+        X_t = B_t + t(a - B_1)
+        \]
+        \item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$, 
+        \[
+        X_t = at + (1 - t)B_{t/(1-t)}
+        \]
+        \item $X$ is a continuous process such that
+        \[
+        Y_t = X_t - at + \int_0^1 \frac{X_s - as}{1 - s}ds
+        \]
+
+        is a Brownian motion. 
+        \item The distribution of $X$ is the $h$-transform of the classical Wiener measure on $C([0, 1]; \real^d)$, where
+        \[
+        h(t, x) = \frac{1}{[2\pi(1 - t)]^{d/2}}\exp\braks{-\frac{\norm{x - a}_{\real^d}^2}{2(1 - t)}}
+        \]
+    \end{enumerate}
+    
+\end{theorem}
+
+

src/mal/brownian.tex

@@ -0,0 +1,47 @@
+\section{Girsanov Transforms}
+\label{section:girsanov}
+
+\begin{theorem}[Cameron-Martin-Girsanov, {{\cite[Theorem 38.5]{Rogers}}}]
+\label{theorem:cameron-martin}
+    Let $\wien$ be the classical Wiener measure on $C([0, \infty); \real^d)$ and $\mathcal{V}$ be equivalent to $\wien$. For each $t \ge 0$, let $\bracs{X_t \ge 0}$ be the canonical process, $\mathcal{F}_t^\circ = \sigma(\bracs{X_s|0 \le s \le t})$, $\bracs{\mathcal{F}_t}$ be its $\wien/\mathcal{V}$-augmentation, then there exists a previsible $\real^n$-valued process $\bracs{c_t|t \ge 0}$ such that for each $t \ge 0$, 
+    \[
+    \frac{d \mathcal{V}}{d\wien} \bigg |_{\cf_t} = \exp\braks{\int_0^t c_s dX_s - \frac{1}{2}\int_0^t |c_s|^2ds}
+    \]
+
+    and
+    \[
+    \tilde X_t = X_t - \int_0^t c_sds
+    \]
+
+    is a Brownian motion under $\mathcal{V}$. 
+
+    Conversely, for any previsible $\real^n$-valued process $\bracs{c_t|t \ge 0}$, 
+    \[
+    \zeta_t = \exp\braks{\int_0^t c_sdX_s - \frac{1}{2}\int_0^t |c_s|^2ds}
+    \]
+
+    is a uniformly integrable martingale with respect to 
+\end{theorem}
+
+\begin{theorem}[{{\cite[Theorem 38.9]{Rogers}}}]
+\label{theorem:density-brownian}
+    Let $\wien$ be the classical Wiener measure on $C([0, \infty); \real^d)$ and $\bracs{X_t|t \ge 0}$ be the canonical process. For each $t \ge 0$, let $\mathcal{F}_t = \sigma(\bracs{X_s|0 \le s \le t})$, then for any $\bracs{\mathcal{F}_t}$-previsible process $\bracs{c_t|t \ge 0}$ such that 
+    \[
+    \zeta_t = \exp\braks{\int_0^t c_sdX_s - \frac{1}{2}\int_0^t |c_s|^2ds}
+    \]
+
+    is a martingale, then
+    \begin{enumerate}
+        \item There exists a unique measure $\mathcal{V}$ on $(\Omega, \mathcal{F}_\infty)$ such that for each $t \ge 0$, 
+        \[
+        \frac{d \mathcal{V}}{d \mathcal{W}} \bigg | \cf_{t^+} = \zeta_t
+        \]
+        \item The process
+        \[
+        \tilde X_t = X_t - \int_0^t \gamma_s ds
+        \]
+
+        is a $\bracs{\mathcal{F}_{t^+}}$-Brownian motion. 
+    \end{enumerate}
+\end{theorem}
+

src/mal/h.tex

@@ -0,0 +1,93 @@
+\section{$h$-Transforms}
+\label{section:h-transform}
+
+
+\begin{definition}[Mean Value Property]
+\label{definition:mean-value-property}
+    Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$. For each $0 \le s \le t \le T$ and $x \in \real^d$, let $P_{s, t}(x, \cdot)$ be its transition distribution, and $h: (0, \infty) \times \real^d \to [0, \infty)$ be a measurable function, then $h$ satisfies the \textbf{mean-value property} with respect to $\bp$ if for any $0 \le s \le t \le T$ and $x \in \real^d$, 
+    \[
+    h(s, x) = \int_{\real^d}h(t, y)P_{s, t}(x, dy)
+    \]
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:h-density-martingale}
+    Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, and $A_0 = \bracs{h(0, \cdot) > 0}$, then the process
+    \[
+    \one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
+    \]
+
+    is a $\bracs{\mathcal{F}_t}$-martingale with respect to $\bp$. 
+\end{lemma}
+\begin{proof}
+    Let $t \in [0, T]$ and $\mu$ be the initial distribution of $\bp$, then
+    \[
+        \ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}} = 
+        \int_{A_0} \frac{1}{h(0, x)}\int_{\real^d} h(t, y)P_{0, t}(x, dy) \mu(dx) = \mu(A_0)
+    \]
+
+    so the given process is integrable. Now, since $h$ satisfies the mean-value property, 
+    \begin{align*}
+        \ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(T, X_T)}{h(0, X_0)} \bigg | \cf_t} &= \frac{\one_{\bracs{X_0 \in A_0}}}{h(0, X_0)} \cdot \ev_\bp\braks{h(t, X_t)| \cf_t} \\
+        &= \one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
+    \end{align*}
+    
+    
+\end{proof}
+
+\begin{proposition}
+\label{proposition:space-time-regular-measure}
+    Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\mu$ be its initial distribution, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, then there exists a probaiblity measure $\bp^h$ on $D([0, \infty); \real^d)$ such that:
+     \begin{enumerate}
+         \item For each $t \ge 0$, 
+         \[
+         \frac{d \bp^h}{d \bp} \bigg |_{\cf_{t^+}} = \frac{\one_{\bracs{X_0 \in A_0}}}{\mu(A_0)}\frac{h(t, X_t)}{h(0, X_0)}
+         \]
+
+         where $A_0 = \bracs{h(0, \cdot) > 0}$. 
+
+         \item Under $\bp^h$, $X_t$ is a Markov process with transition function
+         \[
+         P_{s, t}^h(x, dy) = \begin{cases}
+            \frac{h(t, y)}{h(s, x)}P_{s, t}(x, dy) &h(s, x) > 0 \\
+            0 & h(s, x) = 0
+         \end{cases}
+         \]
+         \item If $\bp$ is the classical Wiener measure, then the process
+         \[
+         \tilde X_t = X_t - \int_0^t \frac{\partial_x h(s, X_s)}{h(s, X_s)}ds
+         \]
+
+         is a $\bp^h$-Brownian motion. 
+     \end{enumerate}
+
+     The distribution $\bp^h$ is the \textbf{$h$-transform} of $\mathbf{P}$.
+     
+\end{proposition}
+
+
+\begin{theorem}[Reuter, {{\cite[Theorem 39.66]{Rogers}}}]
+\label{theorem:reuter}
+    Let $(\Omega, \cf, \bp)$ be a filtered probability space, $X: \Omega \to C([0, \infty); \real^d)$ be a standard $\bracs{\mathcal{F}_t}$-Brownian motion with drift $\mu$ starting at $0$. If $\tau = \inf\bracs{t \ge 0: |X_t| = 1}$, then $X_\tau$ and $\tau$ are independent. 
+\end{theorem}
+\begin{proof}
+    If $\mu = 0$, then $X_\tau$ and $\tau$ are independent by rotational invariance. 
+    
+    Otherwise, let $\wien$ be the classical Wiener measure on $C([0, \infty); \real^d)$, $\mathcal{V}$ be the distribution of $X$, and $Y$ be the canonical process on $C([0, \infty); \real^d)$. For each $t \ge 0$, let $\mathcal{G}_t = \sigma(\bracs{Y_s|0 \le s \le t})$, then by the Cameron-Martin formula, 
+    \[
+    h(t, Y_t) = \frac{d \mathcal{V}}{d \wien} \bigg |_{\mathcal{G}_t} = \exp\braks{\angles{\mu, Y_t}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 t}
+    \]
+
+    Since $h(t \wedge \tau, X_{t \wedge \tau})$ is a uniformly integrable martingale, 
+    \[
+    \frac{d \mathcal{V}}{d\wien} \bigg |_{\mathcal{G}_{T^+}} = h(\tau, Y_\tau)
+    \]
+
+    In which case, for any measurable functions $f: \mathbb{S}^{d} \to [0, 1]$ and $g: [0, \infty) \to [0, 1]$, 
+    \begin{align*}
+        \ev^{\mathcal{V}}[f(Y_\tau)g(\tau)] &= \ev^{\wien}\braks{f(Y_\tau)g(\tau)\exp(\angles{\mu, Y_\tau}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau)} \\
+        &= \ev^\wien[f(Y_\tau)e^{\angles{\mu, Y_\tau}_{\real^d}}]\ev^\wien[g(\tau)e^{- \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau}] \\
+        &= \ev^{\mathcal{V}}(f(Y_\tau))\ev^{\mathcal{V}}(g(\tau))
+    \end{align*}
+\end{proof}

src/mal/index.tex

@@ -0,0 +1,6 @@
+\chapter{$h$-Transforms}
+\label{chap:mal}
+
+\input{./brownian.tex}
+\input{./h.tex}
+\input{./bridge.tex}

src/operator/dif.tex

@@ -0,0 +1,125 @@
+\section{Diffusion Operators}
+\label{section:diffusion-op}
+
+
+\begin{definition}
+    \label{definition:diffusion-op}
+    Let $d \in \nat$, then a \textbf{diffusion operator} on $\real^d$ is a differential operator of the form
+    \[
+    Lu = \dpn{A, D^2u}{\real^{d \times d}} + \dpn{b, Du}{\real^d}
+    \]
+    where $A \in C(\real^d; \real^{d \times d})$ is symmetric and non-negative, and $b \in C(\real^d; \real^d)$.
+
+    It is assumed that there exists a Borel measure $\mu: \cb(\real) \to [0, \infty]$ equivalent to the Lebesgue measure such that
+    \[
+    \int g Lf d\mu = \int f Lg d\mu
+    \]
+    for all $f, g \in C_c^\infty(\real^d; \real)$.
+\end{definition}
+
+\begin{lemma}
+    \label{lem:diffusion-symmetric}
+    Let $L$ be a diffusion operator, then for any $f \in C^\infty(\real^d; \real)$ and $g \in C_c^\infty(\real^d; \real)$, 
+    \[
+    \int g Lf d\mu = \int f Lg d\mu
+    \]
+    In particular, 
+    \[
+    \int Lg d\mu = 0
+    \]
+\end{lemma}
+\begin{proof}
+    Let $\eta \in C_c^\infty(\real^d; \real)$ such that $\eta|_{\supp{g}} = 1$, then
+    \[
+    \int g Lf d\mu = \int g L(\eta f)d\mu = \int (\eta f)Lgd\mu = \int f Lgd\mu
+    \]
+\end{proof}
+
+\begin{lemma}
+    \label{lem:energy-form}
+
+    Let $L$ be a diffusion operator, then for any $f, g \in C_c^\infty(\real^d; \real)$, 
+    \[
+    \dpn{Df, ADg}{\real^d} = \frac{1}{2}[L(fg) - fLg - gLf]
+    \]
+\end{lemma}
+\begin{proof}
+    Firstly, 
+    \begin{align*}
+    D(fg) &= fDg + gDf \\ 
+    D^2(fg) &= fD^2g + 2Df \otimes Dg + gD^2f \\
+    \end{align*}
+    so
+    \[
+        L(fg) = fLg + gLf + 2\dpn{Df, ADg}{\real^d}
+    \]
+\end{proof}
+
+\begin{definition}
+    \label{def:energy-form}
+    Let $L$ be a diffusion operator. For each $f, g \in C_c^\infty(\real^d; \real)$, 
+    \begin{align*}
+    \dpn{f, g}{\ce} &= \dpn{Df, ADg}{L^2(\real^d, \mu; \real^d)} \\
+    &= -\dpn{f, Lg}{L^2(\real^d, \mu)} = -\dpn{g, Lf}{L^2(\real^d, \mu)}
+    \end{align*}
+    is the \textbf{energy form} of $L$. 
+\end{definition}
+
+\begin{proposition}[{{\cite[Theorem 4.6, Proposition 4.11]{Baudoin}}}]
+    \label{prop:diffusion-sa}
+
+    Let  
+    \[
+    Lu = \dpn{A, D^2u}{\real^{d \times d}} + \dpn{b, Du}{\real^d}
+    \]
+    
+    be a diffusion operator, then:
+    \begin{enumerate}
+        \item $L$ admits a self-adjoint extension.
+        \item If $L$ is elliptic with smooth coefficients and there exists $\seq{h_n} \in C_c(\real^d; [0, 1])$ such that:
+        \begin{enumerate}
+            \item[(a)] $h_n \upto 1$ pointwise on $\real^d$ as $n \to \infty$.
+            \item[(b)] $||\dpn{Dh_n, ADh_n}{\real^{d}}||_{L^\infty(\real^d)} \to 0$ as $n \to \infty$.
+        \end{enumerate}
+        then $L$ is essentially self-adjoint.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Since $A$ is semipositive, $-L$ is a semipositive operator. Therefore $L$ admits a Friedrichs extension by \autoref{thm:friedrichs}.
+
+    (2): By \autoref{lem:injective-eigen-ess}, it is sufficient to show that there exists $\lambda > 0$ such that $(\lambda - L^*)$ is injective. 
+    
+    In fact, $(\lambda - L^*)$ is injective for all $\lambda > 0$. Fix $\lambda > 0$ and $u \in L^2(\real^d)$ such that $L^*u = \lambda u$. In this case, since $\mu$ is equivalent to the Lebesgue measure, $Lu = \lambda u$ as distributions. By the Elliptic Regularity Theorem, $u \in C^\infty(\real^d) \cap L^2(\real^d)$.
+
+    For any $h \in C_c^\infty(\real^d; \real)$, 
+    \begin{align*}
+        \dpn{Df, AD(h^2f)}{L^2(\real^d; \real^d )}&= -\dpn{f, L(h^2f)}{L^2(\real^d)} \\
+        &=- \dpn{Lf, h^2f}{L^2(\real^d)} \\
+        &= -\dpn{L^*f, h^2f}{L^2(\real^d)} \\
+        &= -\lambda \dpn{f^2, h^2}{L^2(\real^d)} \le 0
+    \end{align*}
+    Since
+    \begin{align*}
+        \dpn{Df, AD(h^2f)}{L^2(\real^d; \real^d)} &= \dpn{Df, h^2ADf}{L^2(\real^d; \real^d)} \\
+        &+ \dpn{Df, 2fhADh}{L^2(\real^d; \real^d)} 
+    \end{align*}
+    By the Schwarz inequality, 
+    \[
+    \dpn{Df, ADh}{\real^d} \le \sqrt{\dpn{Df, ADf}{\real^d}} \cdot \sqrt{\dpn{Dh, ADh}{\real^d}}
+    \]
+    so
+    \begin{align*}
+        \dpn{Df, h^2ADf}{L^2(\real^d; \real^d)} &\le 2\norm{\dpn{Dh, ADh}{\real^d}}_{L^\infty(\real^d)}^{1/2} \\
+        &\times \normn{fh{\dpn{Df, ADf}{\real^d}^{1/2}}}_{L^2(\real^d)}
+    \end{align*}
+    where
+    \[
+    \normn{fh{\dpn{Df, ADf}{\real^d}^{1/2}}}_{L^2(\real^d)} \le \norm{f}_{L^2}\dpn{Df, h^2ADf}{L^2(\real^d;\real^d)}^{1/2}
+    \]
+    Therefore
+    \[
+    \dpn{Df, h^2ADf}{L^2(\real^d; \real^d)} \le 4\norm{\dpn{Dh, ADh}{\real^d}}_{L^\infty(\real^d)} \cdot \norm{f}_{L^2}^2
+    \]
+
+    Substituting $\seq{h_n}$ for $h$ and sending $n \to \infty$ yields that $\norm{Df}_{L^2(\real^d; \real^d)} = 0$. Therefore $f = 0$.
+\end{proof}

src/operator/hsg.tex

@@ -0,0 +1,114 @@
+\section{The Heat Semigroup}
+\label{sec:heat-semigroup}
+
+\begin{definition}[Normal Operator]
+    \label{def:normal-op}
+    Let $H$ be a Hilbert space and $T$ be a closed, densely defined operator, then $T$ is \textbf{normal} if $T^*T = TT^*$.
+\end{definition}
+
+\begin{theorem}[Spectral Theorem I, {{\cite[The Spectral Theorem]{conway}}}]
+    \label{thm:spectral}
+    Let $H$ be a Hilbert space and $T$ be a normal operator, then there exists a unique spectral measure $P: \cb_\complex \to L(H; H)$ such that:
+    \begin{enumerate}
+        \item $T = \int z P(dz)$.
+        \item The mapping $f \mapsto \int f(z) P(dz)$ is a $*$-homomorphism.
+    \end{enumerate}
+\end{theorem}
+
+\begin{definition}
+    \label{define:heat-semigroup}
+    Let $L$ be an essentially self-adjoint diffusion operator and $P$ be the spectral measure associated with $-L$. For each $t \ge 0$, let
+    \[
+    \bp_t = \int_\complex e^{-tz}P(dz)
+    \]
+    then:
+    \begin{enumerate}
+        \item For any $f \in L^2(\real^d)$, $\norm{\bp_tf}_{L^2(\real^d, \mu)} \le \norm{f}_{L^2(\real^d, \mu)}$. 
+        \item For each $t \ge 0$, $\bp_t$ is self-adjoint. 
+        \item $\bracs{\bp_t|t \ge 0}$ is a $C_0$-semigroup with $L$ as its generator.
+    \end{enumerate}
+    The family $\bracs{\bp_t|t \ge 0}$ is the \textbf{heat semigroup associated with} $L$.
+\end{definition}
+
+\begin{proposition}
+    \label{prop:heat-estimate}
+    Let $L$ be an essentially self-adjoint elliptic diffusion operator with smooth coefficients, and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup, then:
+    \begin{enumerate}
+        \item For any precompact open set $U \subset \subset \real^d$ and $k \in \nat$,
+        \[
+        \norm{P_tf}_{H^k(K)} \le C_{K, k, \mu}\paren{1 + \frac{1}{t^k}}\norm{f}_{L^2(\real^d)}
+        \]
+        In particular, by the Sobolev Embedding theorem, 
+        \[
+        \sup_{x \in K}|\bp_tf(x)| \le C_{K}\paren{1 + \frac{1}{t^{\lfloor d/2 \rfloor + 1}}}\norm{f}_{L^2(\real^d)}
+        \]
+        \item For any $f \in L^2(\real^d; \real)$, the mapping $(t, x) \mapsto \bp_tf(x)$ is smooth on $(0, \infty) \times \real^d$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For any $k \in \nat$, 
+    \[
+    L^k\bp_t = \int z^k e^{-tz}P(dz)
+    \]
+    Hence
+    \[
+    \normn{L^k \bp_t}_{L(L^2(\mu), L^2(\mu))} \le \sup_{z \ge 0}z^ke^{-z t} \le \frac{k^k}{t^k}e^{-k} \le \frac{C_k}{t^k}
+    \]
+    By the Elliptic Regularity Theorem, 
+    \begin{align*}
+        \norm{P_tf}_{H^k(K)} &\le C_K\sum_{\ell = 0}^k \normn{L^k\bp_tf}_{L^2(\real^d)} \\
+        &\le C_{K, k, \mu}\paren{1 + \frac{1}{t^k}}\norm{f}_{L^2(\real^d)}
+    \end{align*}
+
+    (2): By (1), $\bp_t f \in H^k_{\text{loc}}$ for all $k \in \nat$, so $\bp_t f \in C^\infty$. For any $s, t > 0$ and $k \in \nat$,
+    \[
+    \norm{L^k\bp_t - L^k\bp_s}_{L(L^2(\mu), L^2(\mu))} \le \sup_{z \ge 0}|{z^ke^{-z}(e^{-t} - e^{-s})}|
+    \]
+    so $(t, x) \mapsto \bp_t f(x)$ is jointly continuous. By a very wasteful chain of Sobolev shenanigans, $(t, x) \mapsto \bp_t f(x)$ is smooth.  
+\end{proof}
+
+\begin{theorem}
+    \label{thm:heat-markov}
+    Let $L$ be an essentially self-adjoint elliptic diffusion operator and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup, then there exists $p \in C^\infty(\real \times \real^d \times \real^d; \real)$ such that 
+    \begin{enumerate}
+        \item For any $f \in L^2(\real^d; \real)$ and $x \in \real^d$,
+        \[
+        \bp_t f(x) = \int p(t, x, y)f(y) \mu(dy)
+        \]
+        \item For any $x, y \in \real^d$ and $t > 0$, $p(t, x, y) = p(t, y, x)$.
+        \item \textbf{Chapman-Kolmogorov Relation}: For any $x, y \in \real^d$ and $s, t > 0$,
+        \[
+        p(s + t, x, y) = \int p(t, x, z)p(s, z, y)\mu(dz)
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For each $t > 0$ and $x \in \real^d$, the mapping $f \mapsto \bp_tf(x)$ is a continuous linear functional on $L^2(\mu)$. Hence there exists a continuous mapping
+    \[
+    p: \real \times \real^d \to L^2(\mu)
+    \]
+    such that
+    \[
+    \bp_t f(x) = \int p(t, x, y)f(y)\mu(dy)
+    \]
+    
+    (2): Since $\bp_t$ is self-adjoint on $L^2(\mu)$, $p$ is symmetric. 
+
+    (3): Since $\bracs{\bp_t|t \ge 0}$ is a semigroup, $p$ satisfies the Chapman-Kolmogorov relation. 
+
+    \textbf{Smoothness:} For any $f \in L^2(\mu)$, 
+    \[
+    (t, x) \mapsto \int p(t, x, y) f(y)dy
+    \]
+    is smooth. Thus, viewed as a map $\real \times \real^d \to L^2(\mu)$, $(t, x) \mapsto p(t, x, \cdot)$ is differentiable with respect to the weak topology on $L^2(\mu)$.
+
+    Since the inner product $L^2(\mu) \times L^2(\mu) \to \real$ is smooth, by the Chapman-Kolmogorov relation, for any $t > 0$,
+    \[
+    p(t, x, y) = \int p(t/2, x, z)p(t/2, z, y)\mu(dz)
+    \]
+    is continuous. 
+\end{proof}
+
+
+
+

src/operator/hy.tex

@@ -0,0 +1,197 @@
+\section{The Hille-Yosida Theorem}
+\label{sec:hy}
+
+
+\begin{definition}
+    \label{define:generator}
+    Let $E$ be a Banach space and $\bracs{T_t|t \ge 0}$ be a strongly continuous contraction semigroup on $E$. Let
+    \[
+    D = \bracs{x \in E \bigg | \lim_{t \to 0}\frac{T_tx - x}{t} \text{ exists}}
+    \]
+    then:
+    \begin{enumerate}
+        \item For each $t \ge 0$, let
+        \[
+        A_t: E \to E \quad x \mapsto \int_0^t T_sx ds
+        \]
+        then for any $x \in E$, $A_tx \in D$ with
+        \[
+        A(A_tx) = T_tx - x
+        \]
+        \item $D$ is dense in $E$.
+        \item The operator
+        \[
+        A: D(A) \to E \quad x \mapsto \lim_{t \to 0}\frac{T_tx - x}{t}
+        \]
+        is closed.
+    \end{enumerate}
+    and operator $A$ in $(2)$ is the \textbf{generator} of $\bracs{T_t|t \ge 0}$.
+\end{definition}
+\begin{proof}
+    (1): For any $r \in (0, t)$,
+    \begin{align*}
+        \frac{T_rA_tx - A_tx}{r} &= \frac{1}{r} \braks{T_r\int_0^tT_sx ds - \int_0^tT_sxds} \\
+        &= \frac{1}{r} \braks{\int_r^{t+r}T_sx ds - \int_0^tT_sxds} \\
+        &= \frac{1}{r}\braks{\int_t^{t+r}T_sx ds - \int_0^rT_sxds}
+    \end{align*}
+    By the Fundamental Theorem of Calculus, 
+    \[
+    \lim_{r \downto 0}\frac{T_rA_tx - A_tx}{r} = T_tx - x
+    \]
+    so for any $x \in E$ and $t > 0$, $A_tx \in D$.
+
+    (2): By continuity of $s \mapsto T_sx$ and the Fundamental Theorem of Calculus, $A_tx/t \to x$ strongly as $t \downto 0$. Therefore $D$ is dense in $E$.
+
+    (3): Let $\seq{x_n} \subset D$, $x \in E$, and $y \in E$ such that $x_n \to x$ and $Ax_n \to y$ as $n \to \infty$. By the Fundamental Theorem of Calculus, 
+    \[
+    T_tx_n - x_n = \int_0^t T_sAx_n ds
+    \]
+    By the Dominated Convergence Theorem, 
+    \[
+    T_tx - x = \int_0^tT_syds
+    \]
+    Using the Fundamental Theorem of Calculus again, 
+    \[
+    \lim_{t \downto 0}\frac{T_tx - x}{t} = y
+    \]
+    Hence $y \in D$ with $Ax = y$.
+\end{proof}
+
+
+\begin{theorem}[Hille-Yosida]
+    \label{thm:hy}
+    Let $E$ be a Banach space and $A: D(A) \to E$ be a densely defined closed operator, then $A$ generates a strongly continuous contraction semigroup if and only if:
+    \begin{enumerate}
+        \item $\sigma(A) \subset (-\infty, 0]$.
+        \item $\normn{(\lambda - A)^{-1}}_{L(E; E)} \le \lambda^{-1}$ for all $\lambda > 0$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$. 
+
+    If the following integral does exist, then
+    \[
+    \int_0^\infty e^{-t(\lambda - A)}dt = (\lambda - A)^{-1}
+    \]
+    which suggests that
+    \[
+    R_\lambda = \int_0^\infty e^{-\lambda t}T_tdt
+    \]
+    is the inverse of $(\lambda - A)$. Since $t \mapsto T_t$ is continuous and $\norm{T_t}_{L(E; E)} \le 1$ for all $t \ge 0$, the integral converges absolutely as a Riemann integral with $\norm{R_\lambda}_{L(E; E)} \le \lambda^{-1}$.
+
+    For any $x \in E$ and $h > 0$, 
+    \begin{align*}
+        \frac{T_hR_\lambda x - R_\lambda x}{h} &= \int_0^\infty e^{-\lambda t}\frac{T_hT_tx - T_tx}{h}dt \\
+        &= \frac{e^{\lambda h}}{h}\int_h^\infty e^{-\lambda s}(T_s - T_{s - h})xds \\
+        &= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds} \\
+        &- \frac{e^{\lambda h}}{h}\int_h^\infty e^{-\lambda s}T_{s - h}xds \\
+        &= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds} \\
+        &- \frac{1}{h}\int_0^\infty e^{-\lambda s}T_{s}xds \\
+        &= \frac{e^{\lambda h} - 1}{h}R_\lambda x - \frac{e^{\lambda h}}{h}\int_0^he^{-\lambda s}T_sxds
+    \end{align*}
+    By the Fundamental Theorem of Calculus, 
+    \[
+    \lim_{h \downto 0}\frac{e^{\lambda h}}{h}\int_0^he^{-\lambda s} = x
+    \]
+    Therefore
+    \begin{align*} 
+    AR_\lambda x &= \lambda R_\lambda x - x \\
+    (\lambda - A)R_\lambda x &= x \\
+    (\lambda - A)R_\lambda &= \text{Id}
+    \end{align*}
+    On the other hand, since $A$ is closed, 
+    \[
+    AR_\lambda x = \int_0^\infty e^{-\lambda t}T_tAxdt = R_\lambda Ax
+    \]
+    Hence $R_\lambda(\lambda - A) = \text{Id}$ as well, so $R_\lambda = (\lambda - A)^{-1}$.
+
+    Now suppose that $A$ is a densely defined closed operator satisfying assumptions (1) and (2). For each $\lambda > 0$, let
+    \[
+    A_\lambda = -\lambda + \lambda^2(\lambda - A)^{-1}
+    \]
+    then for any $x \in D(A)$,
+    \begin{align*}
+    A_\lambda x &= -\lambda(\lambda-A)(\lambda-A)^{-1}x + \lambda^2(\lambda-A)^{-1}x \\
+    &= \lambda A(\lambda-A)^{-1}x = \lambda(\lambda-A)^{-1}Ax
+    \end{align*}
+    By assumption (2), for any $y \in E$,
+    \begin{align*}
+        \lambda(\lambda - A)^{-1}y - y &= \lambda(\lambda - A)^{-1}(y - \lambda^{-1}(\lambda-A)y) \\
+        &= (\lambda - A)^{-1}(\lambda y - (\lambda-A)y) \\
+        &= (\lambda-A)^{-1}Ay \\
+        \lim_{\lambda \to \infty}\lambda(\lambda - A)^{-1}y - y &= 0
+    \end{align*}
+    Therefore for any $x \in D(A)$,
+    \[
+    \limv{\lambda}A_nx = x
+    \]
+    For each $\lambda > 0$, define
+    \[
+    T^\lambda_t = \sum_{k = 0}^\infty \frac{t^kA_\lambda^k}{k!}
+    \]
+    then $\bracsn{T^\lambda_t|t \ge 0}$ is a (very) strongly continuous semigroup with
+    \begin{align*}
+        T_t^\lambda &=  e^{-\lambda t}\sum_{k = 0}^\infty \frac{\lambda^{2k}t^k(\lambda - A)^{-k}}{k!} \\
+        \normn{T_t^\lambda}_{L(E; E)} &\le e^{-\lambda t}\sum_{k = 0}^\infty \frac{\lambda^k t^k}{k!} \le 1
+    \end{align*}
+    so $\bracsn{T^\lambda_t|t \ge 0}$ is a contraction semigroup. 
+
+    Now, for any $x \in D(A)$, $t \ge 0$, and $\lambda, \mu > 0$, by the Fundamental Theorem of Calculus, 
+    \begin{align*}
+        {(T_t^\lambda - T_t^\mu) x} &= \int_0^t \frac{d}{dx}(T_s^\lambda T_{t - s}^\mu x)ds \\
+        &= \int_0^t T_s^\lambda T_{t - s}^\mu (A_\lambda - A_\mu)xds \\
+        \normn{(T_t^\lambda - T_t^\mu) x}_E &\le t\norm{(A_\lambda - A_\mu)x}_E
+    \end{align*}
+    This allows defining
+    \[
+    T_tx = \limv{n}T_t^nx
+    \]
+    for all $x \in D(A)$, and extend to $E$ by density. By continuity, $\bracs{T_t|t \ge 0}$ is a strongly continuous contraction semigroup. 
+
+    Finally, let $B$ be the generator of $\bracs{T_t|t \ge 0}$, then for any $t \ge 0$, $x \in D(A)$, and $\lambda > 0$, by the Dominated Convergence Theorem,
+    \begin{align*}
+    T_t^\lambda x &= x + \int_0^t T_s^\lambda Axds \\
+    T_tx&= x + \int_0^tT_s Axds
+    \end{align*}
+    Hence $B$ is an extension of $A$. On the other hand, $(\lambda - A)$ and $(\lambda - B)$ are both bijective. Therefore $A$ is an extension of $B$, and $A=  B$.
+\end{proof}
+
+\begin{definition}
+    \label{def:dissipative}
+    Let $E$ be a Banach space, $A: E \to E$ be a densely defined, closed operator, then $A$ is \textbf{dissipative} if for every $x \in E$, there exists $\phi \in E^*$ such that:
+    \begin{enumerate}
+        \item $\norm{\phi}_{E^*} = \norm{x}_E$.
+        \item $\dpb{x, \phi}{E} = \norm{x}_E^2$.
+        \item $\phi(Ax) \le 0$.
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+    \label{prop:hy}
+    Let $E$ be a Banach space and $A: E \to E$ be a densely defined, closed operator, then $A$ generates a strongly continuous contraction semigroup if and only if:
+    \begin{enumerate}
+        \item $A$ is dissipative.
+        \item For each $\lambda > 0$, $(\lambda - A)$ is surjective.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$. By the Hahn-Banach theorem, there exists $\phi \in E^*$ with $\norm{\phi}_{E^*} = \norm{x}$, $\dpb{x, \phi}{E} = \norm{x}_E^2$. 
+
+    
+    Since $\bracs{T_t|t \ge 0}$ is a contraction semigroup, 
+    \[
+    \dpb{T_tx, \phi}{E} \le \norm{\phi}_{E^*}\norm{x}_E \le \norm{x}^2 = \dpb{x, \phi}{E}
+    \]
+    At $t = 0$,
+    \[
+    0 \ge \frac{d}{dt}\dpb{T_tx, \phi}{E} = \dpb{Ax, \phi}{E}
+    \]
+
+    Now suppose that $A$ is a dissipative operator with $(\lambda - A)$ being surjective for all $\lambda > 0$. For any $x \in D(A)$, let $\phi \in E^*$ with $\norm{\phi}_{E^*} = \norm{x}_E$, $\dpb{x, \phi}{E} = \norm{x}_E^2$, and $\dpb{Ax, \phi}{E} \le 0$, then 
+    \begin{align*}
+        \lambda \norm{x}_E^2 &\le \lambda \dpb{x, \phi}{E} - \dpb{Ax, \phi}{E} \\
+        &= \dpb{(\lambda - A)x, \phi}{E} \le \norm{(\lambda - A)x}_E\norm{\phi}_{E^*}
+    \end{align*}
+    so $(\lambda - A)$ is coercive with $\normn{(\lambda - A)^{-1}}_{L(E; E)} \le \lambda^{-1}$.
+\end{proof}
+

src/operator/index.tex

@@ -0,0 +1,10 @@
+\chapter{Diffusion Operators}
+\label{chap:diffusion-operators}
+
+\input{./sa.tex}
+\input{./dif.tex}
+\input{./hsg.tex}
+\input{./sm.tex}
+\input{./rt.tex}
+\input{./hy.tex}
+\input{./solve.tex}

src/operator/rt.tex

@@ -0,0 +1,38 @@
+\section{Interpolations}
+\label{sec:interpolation}
+
+\begin{theorem}[Riesz-Thorin]
+    \label{thm:rt}
+    Let $1 \le p_0, p_q, q_0, q_1 \le \infty$ and $\theta \in (0, 1)$. Let $1 \le p, q \le \infty$ with
+    \[
+    \frac{1}{p} = \frac{1 - \theta}{p_0} + \frac{\theta}{p_1} \quad \frac{1}{q} = \frac{1 - \theta}{q_0} + \frac{\theta}{q_1}
+    \]
+    Let $T: (L^{p_0} + L^{q_0}) \to(L^{p_1} + L^{q_1})$ with
+    \[
+    M_0 = \norm{T}_{L(L^{p_0}; L^{q_0})} \quad 
+    M_1 = \norm{T}_{L(L^{p_1}; L^{q_1})}
+    \]
+    then $T$ extends uniquely as a bounded map $L^p \to L^q$ with
+    \[
+    \norm{T}_{L(L^p; L^q)} \le M_0^{1 - \theta}M_1^{\theta}
+    \]
+\end{theorem}
+
+\begin{theorem}[{{\cite[Theorem 4.31]{Baudoin}}}]
+    \label{thm:heat-rt}
+    Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$, and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup. For each $p \in [1, \infty]$ and $t \ge 0$, $\bp_t$ extends uniquely to a mapping $L^p(\mu) \to L^p(\mu)$ with
+    \[
+    \norm{\bp_t f}_{L^p(\mu)} \le \norm{f}_{L^p(\mu)}
+    \]
+\end{theorem}
+\begin{proof}
+    Let $f, g \in L^1(\mu) \cap L^\infty(\mu)$, then
+    \begin{align*}
+        \dpn{\bp_tf, g}{L^2(\mu)} &= \dpn{f, \bp_tg}{L^2(\mu)} \le \norm{f}_{L^1(\mu)}\norm{\bp_tg}_{L^\infty(\mu)} \\
+        &\le \norm{f}_{L^1(\mu)}\norm{g}_{L^\infty(\mu)}
+    \end{align*}
+    so $\norm{\bp_t f}_{L^1(\mu)} \le \norm{f}_{L^1(\mu)}$.
+\end{proof}
+
+
+

src/operator/sa.tex

@@ -0,0 +1,125 @@
+\section{Self-Adjoint Operators}
+\label{sec:sa-op}
+
+\begin{definition}
+    \label{definition:adjoint}
+    Let $H$ be a Hilbert space, $\varphi: H^* \to H$ be the canonical map, and $T: D(T) \to H$ be a densely defined linear operator. Let
+    \[
+    D(T^*) = \bracsn{\psi \in H| \dpb{T\cdot, \psi}{H} \in H^*}
+    \]
+    then the \textbf{adjoint} of $T$ is the mapping
+    \[
+    D(T^*) \to H \quad \phi \mapsto \varphi(\dpb{T\cdot, \psi}{H})
+    \]
+\end{definition}
+
+\begin{definition}[{{\cite[X.2.2]{conway}}}]
+    \label{definition:sa}
+    Let $H$ be a Hilbert space and $T: D(T) \to H$ be densely defined linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item $T^*$ is an extension of $T$. 
+        \item $\angles{Tx, y} = \angles{x, Ty}$ for all $x, y \in D(T)$. 
+        \item $\angles{Tx, x} \in \real$ for all $x \in D(T)$.
+    \end{enumerate}
+    If the above holds, then $T$ is a \textbf{symmetric operator}. 
+    
+    If $T = T^*$, then $T$ is \textbf{self-adjoint}.
+\end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Let $x, y \in D(T)$, then
+    \[
+    \angles{Tx, y}_H = \angles{x, T^*y}_H = \angles{x, Ty}_H
+    \]
+    $(2) \Rightarrow (3)$: For any $x, y \in D(T)$,
+    \[
+    \angles{Tx, x}_H = \overline{\angles{Tx, x}_H} \quad \angles{Tx, x}_H \in \real
+    \]
+    $(3) \Rightarrow (1)$: By the polarisation identity, for any $x, y \in D(T)$,
+    \begin{align*}
+        4\angles{Tx, y}_H &= \sum_{k = 0}^3i^k\angles{T(x + i^ky), x + i^ky}_H \\
+        &= \sum_{k = 0}^3i^k\angles{x + i^ky, T(x + i^ky)}_H \\
+        &= 4\angles{x, Ty}_H
+    \end{align*}
+    Hence $T^*$ is an extension of $T$. 
+\end{proof}
+
+\begin{definition}
+    \label{definition:positive}
+    Let $H$ be a Hilbert space and $T: D(T) \to H$ be a symmetric linear map, then $T$ is \textbf{semipositive} if $\dpb{x, Tx}H \ge 0$ for all $x \in H$, \textbf{positive} if $\dpb{x, Tx}H > 0$ for all $x \in H$, and \textbf{coercive} if there exists $\eps > 0$ such that $\dpb{x, Tx}H \ge \eps \norm{x}_H$ for all $x \in H$.
+\end{definition}
+
+
+\begin{lemma}
+    \label{lem:inverse-self-adjoint}
+
+    Let $H$ be a Hilbert space and $T: D(T) \to T(H)$ be an injective, self-adjoint operator with dense range, then $T^{-1}: T(H) \to D(T)$ is also self-adjoint. 
+\end{lemma}
+\begin{proof}
+    Let $y \in D((T^{-1})^*)$, then there exists $\phi \in H$ such that $\dpb{T^{-1}x, y}H = \dpb{x, \phi}H$ for all $x \in T(H)$. Under the substitution $x = Tz$, this implies that $\dpb{z, y}H = \dpb{Tz, \phi}H$ for all $x \in T(H)$. Hence $y = T^*(\phi) = T(\phi) \in T(H)$.
+\end{proof}
+
+
+\begin{theorem}[Friedrichs Extension, {{\cite[Theorem 4.6]{Baudoin}}}]
+    \label{thm:friedrichs}
+
+    Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive operator, then $T$ admits a self-adjoint extension. 
+\end{theorem}
+\begin{proof}
+    For any $f, g \in D(T)$, let
+    \[
+    \dpb{f, g}{E} = \dpb{f, Tg}{H} + \dpb{f, g}{H}
+    \]
+    then $\dpb{\cdot, \cdot}{E}$ is an inner product on $D(T)$. Let $E$ be the completion of $D(T)$ with respect to the norm induced by $\dpb{\cdot, \cdot}{E}$. Since $\norm{\cdot}_E \ge \norm{\cdot}_H$, the inclusion $\iota: D(T) \to H$ admits a continuous extension to $E$, which maps it into $H$ as a dense subspace.
+
+    To this end, let $\seq{x_n} \subset D(T)$ be a Cauchy sequence in $E$ such that $\norm{x_n}_H \to 0$ as $n \to \infty$, then
+    \begin{align*}
+        \limv{n}\dpb{x_n, x_n}E &= \limv{n}\limv{m}\dpb{x_m, x_n}E \\
+        &= \limv{n}\limv{m}\dpb{x_m, Tx_n}H \\
+        &+ \limv{n}\limv{m}\dpb{x_m, x_n}H = 0
+    \end{align*}
+    Hence $x_n \to 0$ in $E$, and the extension $\iota: E \to H$ is injective.  
+
+    Since $\iota: E \to H$ is bounded and injective, it admits a bounded adjoint $\iota^*: H \to E$. Let $B = \iota \circ \iota^*: H \to H$, then $B$ is a bounded, injective, and self-adjoint.
+
+    Now, let $A = B^{-1}: B(H) \to H$, then $A$ is self-adjoint by \autoref{lem:inverse-self-adjoint}. Moreover, for any $x, y \in B(H)$,
+    \begin{align*}
+        \dpb{x, Ay}{H} &= \dpb{x, (\iota \circ \iota^*)^{-1}y}{H} \\
+        &= \dpb{x, ({\iota^*})^{-1}\iota^{-1}y}H \\
+        &= \dpb{\iota^{-1}x, \iota^{-1}y}E \\
+        &= \dpb{x, (T + 1)y}H
+    \end{align*}
+    So $(A - 1): B(H) \to H$ is the self-adjoint extension of $T$.
+\end{proof}
+
+\begin{definition}[Essentially Self-Adjoint]
+    \label{definition:ess-sa}
+    Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive operator, then $T$ is \textbf{essentially self-adjoint} if the Friedrichs extension is its unique self-adjoint extension. 
+\end{definition}
+
+\begin{lemma}
+    \label{lem:injective-eigen-ess}
+
+    Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive. If there exists $\lambda > 0$ such that $(T^* + \lambda)$ is injective, then $T$ is essentially self-adjoint. 
+\end{lemma}
+\begin{proof}
+    Let
+    \[
+    \dpb{x, y}E = \dpb{x, (T^* + \lambda)y}H
+    \]
+    be the inner product on $D(T^*)$, then for any $y \in D(T^*)$ with $y \perp D(T)$,
+    \[
+    \dpb{x, (T^* + \lambda)y}H = 0 \quad \forall x \in D(T)
+    \]
+    By density of $D(T)$ and injectivity of $T^* + \lambda$, $y = 0$. Hence $D(T)$ is dense in $D(T^*)$ with respect to the inner product $\dpb{\cdot, \cdot}E$.
+
+    Let $S$ be a semipositive self-adjoint extension of $T$, then $D(S) \subset D(T^*)$, so $D(T)$ is dense in $D(S)$ with respect to the inner product $\dpb{\cdot, \cdot}E$. 
+
+    Let $E$ be the completion of $D(T^*)$ with respect to the inner product $\dpb{\cdot, \cdot}E$ and $\iota: E \to H$ be the extension of the inclusion map, then for any $x, y \in D(T) \subset E$, 
+    \[
+        \dpb{x, (S + \lambda)y}H = \dpn{\iota^{-1}x, \iota^{-1}y}E
+    \]
+    Hence $S$ and the Friedrichs extension coincide on $D(T)$. By density of $D(T)$ in $D(S)$, they also coincide on $D(S)$.
+\end{proof}
+
+
+

src/operator/sm.tex

@@ -0,0 +1,126 @@
+\section{The Submarkov Property}
+
+\begin{definition}[Submarkov Property]
+    \label{define:submarkov}
+    Let $\bracs{\bp_t|t \ge 0}$ be a semigroup on $L^2$, then $\bracs{\bp_t|t \ge 0}$ has the \textbf{submarkov} property if for any $f \in L^2$ with $0 \le f \le 1$ (a.e.), $0 \le \bp_t f \le 1$ for all $t > 0$. 
+\end{definition}
+
+\begin{lemma}[Kato Inequality, {{\cite[Lemma 4.24]{Baudoin}}}]
+    \label{lem:kato}
+    Let
+    \[
+    Lu = \dpn{A, D^2u}{\real^{d \times d}} + \dpn{b, Du}{\real^d}
+    \]
+    be a diffusion operator on $\real^d$ with symmetrising measure $\mu$, then for any $u \in C_c^\infty(\real^d; \real)$,
+    \[
+    L(|\mu|) \ge \sgn(u) \cdot Lu
+    \]
+    in the sense of distributions. 
+\end{lemma}
+\begin{proof}
+    First suppose that $u \in C_c^\infty(\real^d; \real)$. For any $\phi \in C^2(\real; \real)$,
+    \begin{align*}
+        D(\phi \circ u) &= (\phi' \circ u) \cdot Du \\
+        D^2(\phi \circ u) &= (\phi' \circ u) \cdot Du \otimes Du + (\phi' \circ u) \cdot D^2u
+    \end{align*}
+    so
+    \[
+    L(\phi \circ u) = (\phi' \circ u) Lu + (\phi'' \circ u)\dpn{Du, ADu}{\real^d}
+    \]
+    In particular, if $\phi$ is also convex, then
+    \[
+    L(\phi \circ u) \ge (\phi' \circ u)Lu
+    \]
+    Now, for each $\eps > 0$, let $\phi_\eps(x) = \sqrt{x^2 + \eps^2}$, then
+    \[
+    L(\phi_\eps \circ u)(x) \ge \frac{u(x)}{\sqrt{x^2 + \eps^2}}Lu(x)
+    \]
+    by the Dominated Convergence Theorem, 
+    \[
+    L|u| \ge \sgn(u) Lu
+    \]
+\end{proof}
+
+\begin{lemma}
+    \label{lem:resolvant}
+    Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$. For any $\lambda > 0$ and $f, g \in C_c^\infty(\real^d; \real)$, let
+    \[
+    \dpn{f, g}{\lambda} = \dpn{f, g}{L^2(\mu)} + \lambda \dpn{Df, ADg}{L^2(\mu; \real^d)}
+    \]
+    and $E_\lambda$ be the completion of $C_c^\infty(\real^d; \real)$ with respect to the above inner product, then
+    \begin{enumerate}
+        \item For any $f \in E_\lambda$, $|f| \in E_\lambda$ with $\norm{|f|}_\lambda \le \norm{f}_\lambda$.
+        \item For any $f \in E_\lambda$, $f \wedge 1 \in E_\lambda$ with $\norm{f \wedge 1}_\lambda \le \norm{f}_\lambda$.
+        \item Let
+        \[
+        R_\lambda = (1 - \lambda L)^{-1}: L^2(\mu) \to D(L)
+        \]
+        then for any $f \in E_\lambda$ and $g \in L^2(\mu)$,
+        \[
+        \dpn{f, R_\lambda g}{\lambda} = \dpn{f, g}{L^2(\mu)}
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $u \in C_c^\infty(\real^d; \real)$, by the Kato inequality,
+    \begin{align*}
+        \dpn{D|u|, AD|u|}{\real^d} &= \frac{1}{2}L(|u| \cdot |u|) - |u|L|u| \\
+        &\le \frac{1}{2}L(u^2) - uLu \\
+        &= \dpn{Du, ADu}{L^2(\mu; \real^d)}
+    \end{align*}
+    Therefore for any $u \in E_\lambda$, $|u| \in E_\lambda$ with $\norm{|u|}_\lambda \le \norm{u}_{\lambda}$.
+
+    (2): For any $f \in C_c^\infty(\real^d; \real)$, let $\eta \in C_c^\infty(\real^d; [0, 1])$ with $\eta|_{\supp{f}} = 1$, then $\eta \in D(L) \subset E_\lambda$ with
+    \[
+    f \wedge 1 = f \wedge \eta = \frac{1}{2}(f + \eta + |f - \eta|)
+    \]
+    Despite the deceptive appearance of the above expression, $\norm{f \wedge 1}_\lambda \le \norm{f}_\lambda$. Hence the inclusion extends by continuity to all $f \in E_\lambda$.
+
+    (3): For any $f \in E_\lambda$ and $g \in L^2(\mu)$,
+    \begin{align*}
+        \dpn{f, R_\lambda g}{\lambda} &= \dpn{f, R_\lambda g}{L^2(\mu)} - \lambda \dpn{f, LR_\lambda g}{L^2(\mu)} \\
+        &= \dpn{f, (1 - \lambda L)R_\lambda g}{L^2(\mu)} = \dpn{f, g}{L^2(\mu)}
+    \end{align*}
+\end{proof}
+
+
+\begin{theorem}[{{\cite[Theorem 4.25]{Baudoin}}}]
+    \label{thm:heat-positive}
+    Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$ and heat semigroup $\bracs{\bp_t|t \ge 0}$, then for any $f \in L^2(\mu)$ with $f \ge 0$ (a.e.), $\bp_tf \ge 0$.
+\end{theorem}
+\begin{proof}
+    By the Dominated Convergence Theorem and the Spectral Theorem, for any $f \in L^2(\mu)$,
+    \[
+    \bp_tf = \limv{n}\paren{1 - \frac{t}{n}L}^{-n}f
+    \]
+    Thus it is sufficient to show that $(1 - \lambda L)^{-1}$ satisfies the positivity criterion for all $\lambda > 0$. 
+
+    Let $g \in L^2(\mu)$ with $g \ge 0$ (a.e.) and $f = R_\lambda g$, then
+    \[
+        \norm{f}_\lambda^2 = \dpn{f, g}{L^2(\mu)} \le \dpn{|f|, g}{L^2(\mu)} = \dpn{|f|, f}{\lambda} \le \norm{f}_\lambda^2
+    \]
+    so $f = |f|$, and $R_\lambda g \ge 0$.
+\end{proof}
+
+\begin{theorem}[{{\cite[Theorem 4.27]{Baudoin}}}]
+    \label{thm:heat-bound}
+    Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$, then for any $f \in L^2(\mu) \cap L^\infty(\mu)$,
+    \[
+    \norm{\bp_t f}_\infty \le \norm{f}_\infty
+    \]
+\end{theorem}
+\begin{proof}
+    Similar to the previous theorem, it is sufficient to show that $(1 - \lambda L)^{-1}$ satisfies the bound for all $\lambda > 0$. 
+
+    To this end, let $\lambda > 0$, $g \in L^2(\mu) \cap L^\infty(\mu)$ with $0 \le g \le 1$ (a.e.), $f = R_\lambda g$, and $h = 1 \wedge f$, then
+    \begin{align*}
+    0 &\le \norm{f - h}_\lambda^2  = \norm{f}_\lambda^2 - 2\dpn{f, h}{\lambda} + \norm{h}_\lambda^2 \\
+    &\le \dpn{f, g}{L^2(\mu)} - 2\dpn{f, h}{\lambda} + \norm{h}_{L^2(\mu)}^2 - \dpn{h, \lambda Lh}{L^2(\mu)} \\
+    &= \dpn{f, g}{L^2(\mu)} + \norm{g - h}_{L^2(\mu)}^2 - \norm{g}_{L^2(\mu)}^2- \dpn{h, \lambda Lh}{L^2(\mu)} \\
+    &\le \dpn{f, g}{L^2(\mu)} + \norm{g - f}_{L^2(\mu)}^2 - \norm{g}_{L^2(\mu)}^2- \dpn{f, \lambda Lf}{L^2(\mu)} \\
+    &\le \norm{f}_{L^2(\mu)}^2 - \dpn{f, g}{L^2(\mu)}- \dpn{f, \lambda Lf}{L^2(\mu)} \\
+    &= \norm{f}_{L^2(\mu)}^2 - \dpn{f, f}{\lambda}- \dpn{f, \lambda Lf}{L^2(\mu)} \\
+    &= \norm{f}_{L^2(\mu)} - \norm{f}_{L^2(\mu)} = 0
+    \end{align*}
+    so $f = h$ a.e., and $0 \le f \le 1$ (a.e.).
+\end{proof}

src/operator/solve.tex

@@ -0,0 +1,122 @@
+\section{Solution of the Cauchy Problem}
+\label{sec:solve}
+
+\begin{proposition}[{{\cite[Proposition 4.47]{Baudoin}}}]
+    \label{prop:cauchy-solve}
+    Let $L$ be an essentially self-adjoint diffusion operator on $\real^d$ with smooth coefficients, symmetrising measure $\mu$, and heat semigroup $\bracs{\bp_t|t \ge 0}$, then for any $f \in L^2(\mu; \real)$, the function
+    \[
+    u: [0, \infty) \times \real^d \quad (t, x) \mapsto (\bp_t f)(x)
+    \]
+    is smooth on $(0, \infty) \times \real^d$ and solves the Cauchy problem
+    \[
+    \begin{cases}
+        \partial_t u = Lu &t > 0 \\
+        u(0, x) = f(x) &\forall x \in \real^d
+    \end{cases}
+    \]
+\end{proposition}
+\begin{proof}
+    For any $\phi \in \cd((0, \infty) \times \real^d)$, using integration by parts and the self-adjointness of $\bp_t$,
+    \begin{align*}
+        &\iint_{\real^d \times [0, \infty)} \phi [(\partial_t - L)u]\mu(dx)dt \\
+        &= \iint_{\real^d \times [0, \infty)}(-\partial_t - L)\phi u\mu(dx)dt \\
+        &= \iint_{\real^d \times [0, \infty)}(-\partial_t - L)\phi(x, t) (\bp_t f)(x)\mu(dx)dt \\
+        &= \iint_{\real^d \times [0, \infty)}\bp_t[(-\partial_t - L)\phi(x, t)] f(x)\mu(dx)dt
+    \end{align*}
+    Since $L$ is the generator of $\bracs{\bp_t|t \ge 0}$, by the product rule,
+    \[
+    \partial_t(\bp_t \phi) = L\phi + \bp_t \partial_t \phi
+    \]
+    so
+    \[
+    -\bp_t\partial_t \phi = -\partial_t(\bp_t\phi) + L\phi
+    \]
+    and
+    \[
+    \iint_{\real^d \times [0, \infty)} \phi [(\partial_t - L)u]\mu(dx)dt = \partial_t (\bp_t \phi) f(x)\mu(dx)dt = 0
+    \]
+    so $\partial_t u = Lu$ in the sense of distributions. As $u \in C^\infty((0, \infty) \times \real^d)$, $\partial_t u = Lu$ pointwise.
+\end{proof}
+
+\begin{lemma}
+    \label{lem:truncate-smooth}
+
+    Let
+    \[
+    Lu = \dpn{A, D^2u}{\real^{d \times d}} + \dpn{b, Du}{\real^d}
+    \]
+
+    be an essentially self-adjoint diffusion operator on $\real^d$ with smooth coefficients and symmetrising measure $\mu$.
+
+    Let $f \in \cd(\real^d)$ and $g \in C^\infty(\mu) \cap L^2(\mu)$, then
+    \[
+    \dpn{f^2g, Lg}{L^2(\mu)}
+    \]
+\end{lemma}
+\begin{proof}
+    Since $\mu$ is a symmetrising measure, $\dpn{f^2g, Lg}{L^2(\mu)} = \dpn{L(f^2g), g}{L^2(\mu)}$.
+
+    By \autoref{lem:energy-form},
+    \[
+    L(f^2g) = 2\dpn{D(f^2), ADg}{\real^d} + f^2Lg + gL(f^2)
+    \]
+\end{proof}
+
+
+
+\begin{proposition}
+    \label{prop:heat-uniqueness}
+    Let $v \in C^\infty(\real^d \times \real)$ such that:
+    \begin{enumerate}
+        \item For any $(x, t) \in \real^d \times \real$, $\partial_t v(x, t) \le Lv(x, t)$.
+        \item For every $x \in \real^d$, $v(x, 0) = 0$.
+        \item For any $t > 0$, $\norm{v(\cdot, t)}_{L^2(\mu)} < \infty$.
+    \end{enumerate}
+    then $v = 0$.
+\end{proposition}
+\begin{proof}
+    For any $T > 0$ and $\eta \in \cd(\real^d)$, 
+    \begin{align*}
+    &\int_0^T\int_{\real^d}h^2(x)v(x, t)Lv(x, t)\mu(dx)dt \\
+    &\ge \int_0^T\int_{\real^d}h^2(x)v(x, t)\partial_t v(x, t)\mu(dx)dt = \frac{1}{2}\norm{hv}_{L^2(\mu)}
+    \end{align*}
+    On the other hand, for any $t > 0$, 
+    \[
+    \int_{\real^d}h^2v(\cdot, t)Lv(x, t)\mu(dx) = \int_{\real^d}L(h^2v)(x, t)v(\cdot, t)\mu(dx)
+    \]
+    so
+    \begin{align*}
+        &\int_{\real^d}h^2(x)v(x, t)Lv(x, t)\mu(dx) \\
+        &= \frac{1}{2}\int_{\real^d}h^2(x)v(x, t)Lv(x, t)\mu(dx) \\
+        &+ \frac{1}{2}\int_{\real^d}L(h^2v)(x, t)v(x, t)\mu(dx)
+    \end{align*}
+    By the product rule, 
+    \[
+    L(h^2v \cdot v) = h^2vLv + L(h^2v)v + 2\dpn{AD(h^2v), v}{\real^d}
+    \]
+    Thus
+    \begin{align*}
+        &\int_{\real^d}h^2(x)v(x, t)Lv(x, t)\mu(dx) \\
+        &= \frac{1}{2}\int_{\real^d}L(h^2v \cdot v) - 2\dpn{AD(h^2v), Dv}{\real^d}d\mu \\
+        &= - 2\int_{\real^d}\dpn{AD(h^2v), Dv}{\real^d}d\mu
+    \end{align*}
+    Using the product rule again,
+    \[
+    D(h^2v) = 2hv \cdot Dh + h^2Dv
+    \]
+    where given that
+    \[
+    \dpn{Ah^2Dv, Dv}{L^2} \le C\norm{v}_{L^2}^2 \norm{\dpn{ADh, Dh}{\real^d}}_{L^\infty}
+    \]
+    \begin{align*}
+        &\int 2hv \cdot \dpn{ADh, Dv}{\real^d} d\mu\\
+        &\le \int 2hv \cdot \sqrt{\dpn{ADh, Dh}{\real^d}\dpn{ADv, Dv}{\real^d}} d\mu \\
+        &= 2\int v \cdot \sqrt{\dpn{ADh, Dh}{\real^d}\dpn{Ah^2Dv, Dv}{\real^d}} d\mu \\
+        &\le 2\norm{v}_{L^2}\norm{\dpn{ADh, Dh}{\real^d}}_{L^\infty}^{1/2}\normn{\dpn{Ah^2Dv, Dv}{\real^d}^{1/2}}_{L^2} \\
+        &\le C\norm{v}_{L^2}^2 \norm{\dpn{ADh, Dh}{\real^d}}_{L^\infty}^{3/2}
+    \end{align*}
+    Choosing a family $\seq{h_n} \subset \cd(\real^d; [0, 1])$ with $h_n \upto 1$ and $\norm{\dpn{ADh_n, Dh_n}{\real^d}}_{L^\infty} \to 0$ as $n \to \infty$ yields the desired estimate.
+\end{proof}
+
+
+