# 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?

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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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I like your argument, but it only works for $\Omega=\mathbb R^n$. – Hendrik Vogt Jul 7 '11 at 11:22
@Hendrik Vogt: Ugh, right, I'll add this as a disclaimer. – Tim van Beek Jul 7 '11 at 15:46
@Tim, thanks thats a good (and quite beautiful) find. I think this argument might work also if you simply extend functions in $L^p$ by 0. However, I'm interested in perturbing the semigroup generated by $\Delta$ so in view of Kato's result regarding this it would be nice to explicitly see the spectral properties of $\Delta$ needed. – user12014 Jul 7 '11 at 19:29
@PZZ: No, I don't think that simply extending by $0$ will help you here. What kind of perturbations do you have in mind, and what kind of properties do you want to obtain? – Hendrik Vogt Jul 8 '11 at 12:59
@Hendrik I will be perturbing with a multiplication operator. The main point is that in order to get bounds on the fractional powers of the generator we need bounds on its resolvent of the form mentioned in the OP. In the case of the perturbation these can be explicitly obtained once they are known for the original operator. – user12014 Jul 11 '11 at 5:32