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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
up vote 2 down vote accepted

Arendt-Batty-Hieber-Neubrander: Vector valued Laplace transforms and Cauchy Problems, First Edition, Examle 3.7.6. The Gaussian semigroup. The proof is the same using Fourier multipliers using the excplicite convolution form of the semigroup.

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 $n>1$.

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