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.$\square$