Proposition 2.7.1 ([Proposition 4.47, Bau14]).label 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
\[\begin{cases}\partial_{t} u = Lu&t > 0 \\ u(0, x) = f(x)&\forall x \in \real^{d}\end{cases}\]
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.$\square$
Lemma 2.7.2.label 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
\[\dpn{f^2g, Lg}{L^2(\mu)}\]
Proof. Since $\mu$ is a symmetrising measure, $\dpn{f^2g, Lg}{L^2(\mu)}= \dpn{L(f^2g), g}{L^2(\mu)}$.
By Lemma 2.2.3,
\[L(f^{2}g) = 2\dpn{D(f^2), ADg}{\real^d}+ f^{2}Lg + gL(f^{2})\]
$\square$
Proposition 2.7.3.label Let $v \in C^{\infty}(\real^{d} \times \real)$ such that:
(1)
For any $(x, t) \in \real^{d} \times \real$, $\partial_{t} v(x, t) \le Lv(x, t)$.
(2)
For every $x \in \real^{d}$, $v(x, 0) = 0$.
(3)
For any $t > 0$, $\norm{v(\cdot, t)}_{L^2(\mu)}< \infty$.
then $v = 0$.
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 $t > 0$,
\[\int_{\real^d}h^{2}v(\cdot, t)Lv(x, t)\mu(dx) = \int_{\real^d}L(h^{2}v)(x, t)v(\cdot, t)\mu(dx)\]
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^{2}v)(x, t)v(x, t)\mu(dx)\end{align*}
By the product rule,
\[L(h^{2}v \cdot v) = h^{2}vLv + L(h^{2}v)v + 2\dpn{AD(h^2v), v}{\real^d}\]
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^{2}v \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,
\[D(h^{2}v) = 2hv \cdot Dh + h^{2}Dv\]
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.$\square$