Brownian Motion and Stochastic Calculus

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/Part 1: Stochastic Processes/Chapter 2: Diffusion Operators

2.2 Diffusion Operators

Definition 2.2.1.label Let $d \in \nat$, then a 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 $f, g \in C_{c}^{\infty}(\real^{d}; \real)$.

Lemma 2.2.2.label 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,

\[\int Lg d\mu = 0\]

Proof. Let $\eta \in C_{c}^{\infty}(\real^{d}; \real)$ such that $\eta|_{\supp{g}}= 1$, then

\[\int g Lf d\mu = \int g L(\eta f)d\mu = \int (\eta f)Lgd\mu = \int f Lgd\mu\]

$\square$

Lemma 2.2.3.label Let $L$ be a diffusion operator, then for any $f, g \in C_{c}^{\infty}(\real^{d}; \real)$,

\[\dpn{Df, ADg}{\real^d}= \frac{1}{2}[L(fg) - fLg - gLf]\]

Proof. Firstly,

\begin{align*}D(fg)&= fDg + gDf \\ D^{2}(fg)&= fD^{2}g + 2Df \otimes Dg + gD^{2}f \\\end{align*}

so

\[L(fg) = fLg + gLf + 2\dpn{Df, ADg}{\real^d}\]

$\square$

Definition 2.2.4.label 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 energy form of $L$.

Proposition 2.2.5 ([Theorem 4.6, Proposition 4.11, Bau14]).label Let

\[Lu = \dpn{A, D^2u}{\real^{d \times d}}+ \dpn{b, Du}{\real^d}\]

be a diffusion operator, then:

  1. (1)

    $L$ admits a self-adjoint extension.

  2. (2)

    If $L$ is elliptic with smooth coefficients and there exists $\seq{h_n}\in C_{c}(\real^{d}; [0, 1])$ such that:

    1. (a)

      $h_{n} \upto 1$ pointwise on $\real^{d}$ as $n \to \infty$.

    2. (b)

      $||\dpn{Dh_n, ADh_n}{\real^{d}}||_{L^\infty(\real^d)}\to 0$ as $n \to \infty$.


    then $L$ is essentially self-adjoint.

Proof. (1): Since $A$ is semipositive, $-L$ is a semipositive operator. Therefore $L$ admits a Friedrichs extension by Theorem 2.1.5.

(2): By Lemma 2.1.7, 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 $f = 0$.$\square$

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Brownian Motion and Stochastic Calculus

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