# Analytic generator

The Laplace operator is the generator of an analytic semigroup on $L^p(\mathbb R^n)$ for $1 < p < \infty$. Is the same true for $L^1(\mathbb R^n)$? If it is, could someone give a reference? The proof must be different from the case $1 < p < \infty$.

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Your statement is not very precise, since you don't mention the domain of the Laplacian you are considering. Anyway, I think that it is true also for p=1. And I don't see why the proof should necessarily be different than for p>0. How would you prove it in the latter case? –  Hans Feb 18 '12 at 20:12
sorry, I mean p>1 of course. –  Hans Feb 18 '12 at 20:18
I think the OP asks for the whole space. –  András Bátkai Feb 19 '12 at 8:33
@András Bátkai. Thanks for pointing out in your answer in which sense the case $p=1$ is different. What do you mean by "the whole space" here above? I just wanted to say that "the Laplace operator with maximal domain on $L^p(\mathbb R^n)$ is the generator of an analytic semigroup on $L^p(\mathbb R^n)$" would be a precise (and true) statement; while, without specification of the domain, it doesn't make much sense. –  Hans Feb 19 '12 at 18:54
Besides the method used in the reference given in András Bátkai's answer, there is also another way to prove this fact: one could exploit that the Laplacian is "the square of a group generator". A reference for this is for example "One Parameter semigroups for linear evolution equations" by Engel/Nagel. See in particular Corollary 4.9 and Example 4.10. Also in this case there is no substantial difference in the proof between p=1 and p>1, so I was wondering if Martin knows another type of argument which works only for $p>1$ (I would be interested in this). –  Hans Feb 19 '12 at 18:55
Where you find a difference is the question whether the domain of the Laplace is a classical function space. This is only true if $p>1$, then you get a Sobolev space. For $p=1$ the domain is strictly bigger than $W^{2,1}(\mathbb{R}^n)$ for $n>1$.