Brownian Motion and Stochastic Calculus

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

2.4 The Submarkov Property

Definition 2.4.1 (Submarkov Property).label Let $\bracs{\bp_t|t \ge 0}$ be a semigroup on $L^{2}$, then $\bracs{\bp_t|t \ge 0}$ has the 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 $t > 0$.

Lemma 2.4.2 (Kato Inequality, [Lemma 4.24, Bau14]).label 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.

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^{2}u\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,

\[L|u| \ge \sgn(u) Lu\]

$\square$

Lemma 2.4.3.label 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

  1. (1)

    For any $f \in E_{\lambda}$, $|f| \in E_{\lambda}$ with $\norm{|f|}_{\lambda} \le \norm{f}_{\lambda}$.

  2. (2)

    For any $f \in E_{\lambda}$, $f \wedge 1 \in E_{\lambda}$ with $\norm{f \wedge 1}_{\lambda} \le \norm{f}_{\lambda}$.

  3. (3)

    Let

    \[R_{\lambda} = (1 - \lambda L)^{-1}: L^{2}(\mu) \to D(L)\]

    then for any $f \in E_{\lambda}$ and $g \in L^{2}(\mu)$,

    \[\dpn{f, R_\lambda g}{\lambda}= \dpn{f, g}{L^2(\mu)}\]

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*}

$\square$

Theorem 2.4.4 ([Theorem 4.25, Bau14]).label 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.), $\bp_{t}f \ge 0$.

Proof. By the Dominated Convergence Theorem and the Spectral Theorem, for any $f \in L^{2}(\mu)$,

\[\bp_{t}f = \limv{n}\paren{1 - \frac{t}{n}L}^{-n}f\]

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 $R_{\lambda} g \ge 0$.$\square$

Theorem 2.4.5 ([Theorem 4.27, Bau14]).label Let $L$ be an essentially self-adjoint diffusion operator with symmetrising measure $\mu$, then for any $f \in L^{2}(\mu) \cap L^{\infty}(\mu)$,

\[\norm{\bp_t f}_{\infty} \le \norm{f}_{\infty}\]

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.).$\square$

Brownian Motion and Stochastic Calculus

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