Let
- $(E,\mathcal E)$ be a measurable space
- $\mathcal M_b(E,\mathcal E):=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$
- $(P_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$ and $$P_tf:=\int P_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ for $f\in\mathcal M_b(E,\mathcal E)$ and $t\ge0$
- $\mu$ be a probability measure on $(E,\mathcal E)$ subinvariant with respect to $(P_t)_{t\ge0}$
It's easy to see that $(P_t)_{t\ge0}$ is a contraction semigroup on $\left(\mathcal M_b(E,\mathcal E),\left\|\;\cdot\;\right\|_{L^2(\mu)}\right)$ and hence has a unique extension to a contraction semigroup on $L^2(\mu)$. Let $(\mathcal D(L),L)$ denote the generator of that semigroup, $\mathcal A\subseteq\mathcal D(L)$ be closed under multiplication and $$\Gamma(f,g):=\frac12(L(fg)-fLg-gLf)\;\;\;\text{for }f,g\in\mathcal A.$$ Assume $\mu$ is reversible with respect to $(P_t)_{t\ge0}$ and hence $$\mathcal E(f,g):=\int\Gamma(f,g)\:{\rm d}\mu=-\langle f,Lg\rangle_{L^2(\mu)}=-\langle Lf,g\rangle_{L^2(\mu)}\;\;\;\text{for all }f,g\in\mathcal A\tag2.$$
In the book Analysis and Geometry of Markov Diffusion Operators the authors write the following:
How can we verify the first equality (stressing the fact that $f\in L^2(\mu)$ and not $f\in\mathcal D(A)$)?
Clearly, $$\varphi:L^2(\mu)\to L^1(\mu)\;,\;\;\;f\mapsto f^2$$ is Fréchet differentiable with derivative $${\rm D}\varphi:L^2(\mu)\to\mathfrak L(L^2(\mu),L^1(\mu))\;,\;\;\;f\mapsto(g\mapsto 2fg)$$ and hence it should be just differentiation under the integral sign. However, don't we need the orbit $$\operatorname{orb}f:[0,\infty)\to L^2(\mu)\;,\;\;\;f\mapsto P_tf$$ to be differentiable for that? That should require $f\in\mathcal D(L)$ ...
If the semigroup $(P_t)$ is symmetric in $L^2(\mu)$, then the generator $A$ is self-adjoint. In this case, $P_t$ maps into $D(A)$ for all $t>0$. In particular, $t\mapsto P_t f$ is differentiable on $(0,\infty)$ for all $f\in L^2(\mu)$. This can be verified using the spectral theorem:
Denote by $\nu_f$ the spectral measure of $f\in L^2$. For $t>0$ one has $$ \int_{(-\infty,0]}\lambda^2e^{2t\lambda}\,d\nu_f(\lambda)\leq \frac{1}{e^2t^2}\nu_f((-\infty,0])=\frac{\|f\|_2^2}{e^2t^2}. $$ Thus $P_t f=e^{tA}f\in D(A)$ and $\|AP_t f\|_2^2\leq e^{-2}t^{-2}\|f\|_2^2$.
Similarly, the equality $\partial_t \|P_t f\|_2^2=-2\mathcal{E}(P_t f)$ can be verified (just write both sides as integrals with respect to the spectral measure and use the dominated convergence theorem to check that you can interchange integration and differentiation).