# Extending the domain of the Dirichlet form associated with a symmetric Markov semigroup

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).
• 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 $\left\{x\in H:\int|\psi(\lambda)|^2\:|{\rm d}\langle E_λx,x\rangle_H|<\infty\right\}$? – 0xbadf00d Feb 13 at 13:34
• 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 – 0xbadf00d Feb 13 at 15:39
• 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 ... – 0xbadf00d Feb 13 at 15:39
• 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. – MaoWao Feb 15 at 11:15