furthermore such that $H_t$ and ${\partial}_t H_t$ are self-adjoint on $L^2$.

Let $u \in \mathcal{D}_{\max}(P)$, i.e. $u \in L^{2}, Pu \in
  L^{2}$. We will show that $u \in \mathcal{D}_{\min}(P)$ as well, and thus $\mathcal{D}_{\min} = \mathcal{D}_{\max}$. Indeed, we will show that $$\label{eq:realdeal}
    H_{t} u \to u \mbox{ in } \mathcal{D}_{\max}, \quad \mbox{i.e.\
      that } H_{t}u \to u
    \mbox{ and } P H_{t} u \to P u \mbox{ in } L^{2}.$$ This suffices to prove that $u \in \mathcal{D}_{\min}$ since $H_{t}u
\in \mathcal{D}_{\min}$ by assumption and $\mathcal{D}_{\min}$ is a closed subspace of $\mathcal{D}_{\max}$ in the graph norm. To prove , we note first that $H_{t}u \to u$ in $L^{2}$ trivially since $H_{t} \to \operatorname{Id}$ in the strong topology on $L^{2}$. Also note that since $u \in \mathcal{D}_{\max}$, $Pu \in
L^{2}$, so $H_{t} Pu \to Pu $ in $L^{2}$ also. Of course, this is not what we want; we want $P H_{t} u \to Pu$, but in fact we claim that $$\label{eq:heat-kernel-commutes}
u \in \mathcal{D}_{\max} \implies P H_{t} u = H_{t} P u,$$ which will establish .

It remains to prove . Note that for $u \in \mathcal{D}_{\max}$ and $v \in L^{2}$, then ${\langle}H_{t} P u , v {\rangle}_{L^2} = {\langle}P u 