heat kernel $p_t(x_0,y) \in D(\Delta) \cap L^\infty$ for a manifold with Ricci curvature bounded below? X is an n-dim Riemannian manifold with the Dirichlet form
 $$
\varepsilon (u,v) =-\int_X \langle \nabla u,\nabla v \rangle
 $$
for $u,v \in W^{1,2}(X)$.
Let $P_t$ and $p_t(x,y)$ be the associate semigroup and heat kernel respectively. 
If there exists $f \in L^2(x)$ such that $$
\varepsilon (u,v) =-\int_X \langle \nabla u,\nabla v \rangle= \int_X  f\phi dvol
$$
Then we say $\Delta u=f$. Denote $$D(\Delta)=\{u:u \in W^{1,2}(X),\Delta u \in L^2(X)\}$$(Obviously, the definition of $\Delta$ and $D(\Delta)$ coincide with the generator and the domain of the generator of the Dirichlet form).
Suppose $Ric(X) \ge -(n-1)$, then for any $h \in L^2(X)$, can we get $P_t h \in D(\Delta) \cap L^\infty$? 
Fix $x=x_0$, then $p_t(x_0,y) \in D(\Delta) \cap L^\infty$ and 
$\Delta p_t(x_0,y) \in D(\Delta) \cap L^\infty$?
 A: Yes.
It follows from the spectral theorem that for any $h \in L^2$, we have $P_t h \in D(\Delta)$.  
I think this is easiest to see from the multiplication operator form of the spectral theorem; i.e., there is a measure space $(\hat{X}, \mu)$, a measurable function $g : \hat{X} \to \mathbb{R}$,  and a unitary operator $U : L^2(X) \to L^2(\hat{X})$ such that $\Delta = U^{-1} M_g U$.  Here $M_g$ is the multiplication operator on $L^2(\hat{X})$ defined by $M_g v = gv$ with domain $D(M_g) = \{ v \in L^2(\hat{X}) : gv \in L^2(\hat{X})\}$.  Since $\Delta$ is a negative definite operator, $g \le 0$ almost everywhere.
Now it is easy to check that $P_t = U^{-1} M_{e^{tg}} U$.  But notice that $g e^{tg}$ is a bounded function.  Thus for any $v \in L^2(\hat{X})$, we have $M_{e^{tg}} v \in D(M_g)$.  Using $U$ to move back to $L^2(X)$, we see that for any $h \in L^2(X)$, $P_t h \in D(\Delta)$.
A similar argument shows that $P_t h \in D(\Delta^\infty)$.
To show $P_t h \in L^\infty$, one could use the heat kernel bounds shown by Li and Yau:

Li, Peter; Yau, Shing-Tung. 
  On the parabolic kernel of the Schrödinger operator.
  Acta Math. 156 (1986), no. 3-4, 153–201.

They show, among other things, that in a complete Riemannian manifold with Ricci curvature bounded from below, we have Gaussian upper estimates for the heat kernel.  In particular, for each $t$, $p_t$ is bounded.  From this it is obvious that $P_t$ maps $L^1$ into $L^\infty$. 
Now $P_t$ also maps $L^\infty$ into $L^\infty$, so it maps $L^1 + L^\infty$ into $L^\infty$; in particular it maps $L^2$ into $L^\infty$.
Since $P_t$ maps $L^\infty$ into $L^\infty$ and is symmetric, it also maps $L^1$ into $L^1$.  So it maps $L^1$ into $L^1 \cap L^\infty$; in particular it maps $L^1$ into $L^2$.  In particular, since $p_t(x_0, \cdot) = P_{t/2} [p_{t/2}(x_0, \cdot)]$, where $p_{t/2}(x_0, \cdot) \in L^1$, we have $p_t(x_0, \cdot) \in L^2$.  Then by using $p_t(x_0, \cdot) = P_{t/2} [p_{t/2}(x_0, \cdot)]$ again, we get $p_t(x_0, \cdot) \in D(\Delta)$.
Finally, we note that for $h \in D(\Delta)$, we have $\Delta P_t h = P_t \Delta h$.  So we have
$$\Delta [p_t(x_0, \cdot)] = \Delta P_{t/2} [p_{t/2}(x_0, \cdot)]   = P_{t/2} \Delta [p_{t/2}(x_0, \cdot)].$$
Since $p_{t/2}(x_0, \cdot) \in D(\Delta)$ as previously argued, we have $\Delta p_{t/2}(x_0, \cdot) \in L^2$.  Since $P_{t/2}$ maps $L^2$ into $L^\infty \cap D(\Delta)$, we have $\Delta [p_t(x_0, \cdot)] \in L^\infty \cap D(\Delta)$ as desired.
