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Merged old notes, part 1.

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refs.bib
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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}
}

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src/index.tex
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\part{Diffusion Processes}
\label{part:diffusion}
\input{./operator/index.tex}
\input{./diffusion/index.tex}
\input{./calculus/index.tex}
\input{./sde/index.tex}
\input{./mal/index.tex}

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\section{$h$-Brownian Motions}
\label{section:h-brownian}
\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}
\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}
\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}

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\chapter{Unknown Chapter}
\label{chap:mal}
\input{./brownian.tex}

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\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}

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\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}

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\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}

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\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}

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\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}

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\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}

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\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}

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\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}