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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.

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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 $

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    @PZZ: One way to approach this is as follows: In $L^2$ everything is clear (since the operator is self-adjoint), and then there are known results that the semigroup extends to an analytic semigroup on $L^p$. (If the boundary is nice enough, you'll get the domain you describe in the question, but I think this is not really important.) There is extensive literature on the subject, so I don't quite know where to start. What kind of project is it where you need this?2011-07-14