Proof that $\Delta$ generates analytic semigroup First off, I apologize for asking a question which I'm sure has been studied to death, but I can't seem to find an answer with google. 
I want to see a proof that the Laplace operator $\Delta$ with domain $W^{2,p}(\Omega) \cap W^{1,p}_0(\Omega)$ generates an analytic semigroup on $L^p(\Omega)$ where $\Omega$ is a reasonably nice domain and $1 < p < \infty$. I'm assuming this boils down to the applying the following theorem:
If $A$ is closed and densely defined then $A$ generates an analytic semigroup iff there exists $\omega \in R$ such that the half plane Re $\lambda > \omega$ is contained in the resolvent set of $A$ and there is a constant $C$ such that the resolvent $R_{\lambda}(A)$ satisfies 
$$\|R_\lambda(A)\| \le C/|\lambda - \omega|$$
for all Re $\lambda > \omega$.
That $\Delta$ is closed and densely defined is clear to me, but how do I prove the remaining 2 hypotheses?
Thanks in advance.
 A: A way to show this for $\mathbb{R}^n$ (I'm not sure how to generalize it to a nice subset $\Omega \subset \mathbb{R}^n$) which is maybe a little bit easier is to use Corollary 4.10 of 


*

*Klaus-Jochen Engel, Rainer Nagel: "A Short Course on Operator Semigroups",


which states that if A is the generator of a strongly continuous group, then $A^2$ generates an analytic semigroup of angle $\frac{\pi}{2}$.
The latter part means that the semigroup is defined for parameter values
$$
| \arg z | \lt \frac{\pi}{2}
$$
Since in one dimension
$$
f \mapsto f'
$$
is the generator of the translations, it follows immediatly from this corollary that
$$
f \mapsto f''
$$
generates an analytic semigroup. In several dimensions you'll have to note that the semigroups of translations $T_1, ..., T_n$ along the coordinate axes of a cartesian coordinate system commute, as do the resolvents of the generators $A_1, ..., A_n$. 
The definition of the translation groups is:
$$
T_i(t) (f(x)) := f(x_1, ..., x_i + t , ..., x_n)
$$
According to the corollary, every $A^2_i$ generates an analytic semigroup $U_i (z)$, and the $U_i$ also commute. Therefore we can define
$$
U(z) := U_1(z) \cdot \cdot \cdot U_n(z)
$$
and get that it is an analytic semigroup. It's generator contains at least all functions that are twice continuously differentiable and is on this domain given by
$$
A^2 f = (A_1^2 + ... + A^2_n) f = \triangle f
$$ 
