Lemma 2.4.3 | Brownian Motion and Stochastic Calculus

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)
For any $f \in E_{\lambda}$, $|f| \in E_{\lambda}$ with $\norm{|f|}_{\lambda} \le \norm{f}_{\lambda}$.
(2)
For any $f \in E_{\lambda}$, $f \wedge 1 \in E_{\lambda}$ with $\norm{f \wedge 1}_{\lambda} \le \norm{f}_{\lambda}$.
(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*}

$\square$