Dear all,

I have some difficulties with the following assertion in the book of Kirillov.

Let $G$ be a connected Lie group, and T a given (!) representation of G on a Banach space V.

Let $V^\omega$ be the space of analytic vectors.
Then T induces a representation of $U(\mathcal{G})$ on $V^\omega$, that I denote $T^.$. It is claimed that from $T^.$, we can reconstruct $T$.

If we were in the finite dimensional representations case, I would have no problem. Unfortunately, I don't see how to adapt the proof to the Banach case.

Actually, I don't see how to construct the action of $G$ on $V^\omega$.

By definition, for all $\xi\in V^\omega$, the map $g\mapsto T(g)\xi$ is analytic, so in a neighborhood $U_\xi$ of the neutral element $e$ of $G$, we can reconstruct this map out of $T^.$. That is, we can reconstruct $T()\xi$ on $U_\xi$. But I don't see how to extend this to $G$.

In the finite dimensional case, one doesn't have to take neighborhood depending on $\xi$, since the given $T$ is analytic as map between Lie groups.

Thanks