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 $1 < p < \infty$.



Generator of Laplace operator as analytic semigroup on $L^1(\mathbb{R}^n)$
[Arendt-Batty-Hieber-Neubrander](http://books.google.hu/books?id=sOnRZFgR374C&lpg=PP1&hl=de&pg=PP1#v=onepage&q&f=false): 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 $n>1$.


