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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:

Dirichlet form

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 $f\in\mathcal D(L)$ ...

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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).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Feb 10 at 11:09
  • $\begingroup$ Regarding my first question: In the case where $\psi$ is unbounded, the integral $\int\psi(\lambda)\:{\rm d}\langle E_\lambda x,y\rangle_H$ is defined (for fixed $x,y$) if $$\int|\psi(\lambda)|\:|{\rm d}\langle E_\lambda x,y\rangle_H|<\infty.$$ If I take a look at how they intend to define the operator $\Psi$, it would make more sense to me if we define its domain to be $\left\{x\in H\mid\forall y\in H:\int|\psi(\lambda)|\:|{\rm d}\langle E_\lambda x,y\rangle_H|<\infty\right\}$. Is this domain equal to $\left\{x\in H:\int|\psi(\lambda)|^2\:|{\rm d}\langle E_λx,x\rangle_H|<\infty\right\}$? $\endgroup$ – 0xbadf00d Feb 13 at 13:34
  • $\begingroup$ Just another thought: I get that since $A$ is densely-defined, nonnegative and self-adjoint, it's spectrum $\sigma(A)$ is contained in $[0,\infty)$. (On the other hand, our $A$ is even the generator of a contraction semigroup and hence its resolvent set $\rho(A)$ is contained in $(0,\infty)$; Is this just the same information stated differently (noting that $\sigma(A)=\mathbb R\setminus\rho(A))$ or does it given more information about $\sigma(A)$?). The spectral measure (as I understand it) is now defined on $\mathcal B([0,\infty))$. So, why do we need to extend $\endgroup$ – 0xbadf00d Feb 13 at 15:39
  • $\begingroup$ integrands $\psi:[0,\infty)\to\mathbb R$ by $0$ on $(-\infty,0)$? Moreover, if (as stated in the book), we set $E_\lambda=0$ for all $\lambda<0$, shouldn't we automatically get $\int_{\mathbb R}\psi(\lambda)\:{\rm d}\langle E_\lambda x,y\rangle_H=\int_{[0,\infty)}\psi(\lambda)\:{\rm d}\langle E_\lambda x,y\rangle_H$? Since you're integrating over $(-\infty,0]$, I guess I'm missing something crucial ... $\endgroup$ – 0xbadf00d Feb 13 at 15:39
  • $\begingroup$ Regarding your first comment: No, that's not the same. The problem is that you also want to evaluate expressions of the form $\langle Ax,Ax\rangle$, and your integrability condition is not enough for that. The problem in your second and third comment lies in the typical confusion of signs of the generator. I used the spectral theorem for $A$, the book used it for $-A$ (both operators are sometimes called the generator, depending on the author). You can either define the spectral measure on all of $\mathbb{R}$ or only on a measurable subset containing the spectrum, that does not matter. $\endgroup$ – MaoWao Feb 15 at 11:15

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