The $(2N +1)-$dimensional Heisenberg group $\mathbb{H}^N$ is the space $\mathbb{R}^{2N+1} = \left\{ (x,y, \tau ) \right\}\in \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R}$ equipped with the group operation
\begin{equation} (x,y, \tau ) \circ (\tilde{x}, \tilde{y}, \tilde{ \tau } ) = (x+ \tilde{x}, y+\tilde{y}, \tau +\tilde{ \tau } + 2 \left( x \cdot \tilde{y} - \tilde{x} \cdot y\right)) \end{equation}
where $\cdot$ denotes the standard scalar product in $\mathbb{R}^N$. The identity element for $\mathbb{H}^N$ is $0$ and $\eta^{-1} = -\eta$ for any $\eta \in \mathbb{H}^N$. The sub-Laplacian on $\mathbb{H}^N$ is defined as \begin{equation} \Delta_{\mathbb{H}} = \sum_{i=1}^{N} \left( X_i \circ X_i + Y_i \circ Y_i\right), \end{equation} where for $i=1, \cdots, N$, $X_i$, $Y_i$ are vector fields as follows \begin{equation} X_i = \partial_{x_i} - 2y_i \partial_{ \tau } , \quad Y_i = \partial_{y_i} - 2x_i \partial_{ \tau } \end{equation} Many authors have studied the problem $$ u_t - \Delta_{\mathbb{H}} u = u^p , \quad u(0)=u_0.$$ through the operator \begin{equation} Fu(t)= e^{ t\Delta_{\mathbb{H}} }u_0 + \int_{0}^{t} e^{(t-s)\Delta_{\mathbb{H}}} u(s)^p ds. \end{equation}
I'm trying to understand if this solution can be regularized to become a classical solution. To this end, I am reviewing the semigroup theory and Mild solutions regularity criteria. In a Banach space $X$, a mild solution to the problem $u_t - A u = f, \quad u(0)=u_0$, with $A$ be the infinitesimal generator of an analytic semigroup and with $f$ Holder continuous is a classical solution (Theorem 3.1, chapter 4 of Pazy book: Semigroups of Linear Operators and Applications to Partial Differential Equations). So far, I have managed to learn that the Heisenberg Semigroup is analytic in $L^1$, as it has polynomial growth. So to fit this Pazy Theorem I need to know if he is Banach. But a doubt arises: the Koranyi norm $|| (z,t) || = (|z|^4 + 16t^2)^{1/4}$, which is used in general in Heisenberg group, is only homogeneous for the dilations $\delta_r (x,y,t) = (rx,ry,r^2t)$, therefore it is not a norm as we define in metric spaces in general. But I want to know if it is possible to define a Banach space with it, and in this case would the Heisenberg group be a Banach space?
Thanks in advance.