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.$\square$