I'm reading a paper that deals with the Neumann Heat Semigroup $e^{t\Delta}$ that until I know is defined like: $e^{t\Delta}=\displaystyle\sum_{k=0}^\infty \frac{(t \Delta)^k}{k!}$ what exactly mean $\Delta$ in this semigroup? is this the Laplace Operator? If $\Delta$ is the Laplacian Operator then the expression $e^{t\Delta}w$ means $e^{t\Delta}(w)$? (here $w\in L^p(X)$) If all before assumptions are correct then what means the $L^p$-norm of a vector $\nabla e^{t\Delta}w$?
Thank you
The definition of the semigroup by the Neumann series is not correct since the Laplace operator is unbounded.
Note that the semigroup $e^{t \Delta}$, for a fixed $t$, is a bounded operator from $L^p(X)$ into $L^p(X)$, so $e^{t \Delta} w=e^{t \Delta} (w) \in L^p(X)$. But for $t>0$, by the analyticity of the heat semigroup, $e^{t \Delta} w \in D(\Delta^m)$ for any $w\in L^p(X)$ and any $m \in \mathbb N$, in particular, $e^{t \Delta} w\in W^{1,p}(X)$ so that you can calculate $\nabla e^{t \Delta} w$ as the weak gradient.