Let $X$ be a complex Banach space. Let $(\sigma\_t)\_{t \in \mathbb{R}}$ be a 1-parameter group of linear isometries of $X$ which is strongly continuous i.e. $t \mapsto \sigma\_t(x)$ is continuous for each $x \in X$. An element $x \in X$ is said to be **entire** (for the given flow $\sigma$) if $t \mapsto \sigma\_t(x)$ admits an entire extension: that is, extends to a complex differentiable map $\mathbb{C} \to X$. There is also the equivalent "weak" definition which uses the dual $X^\*$ to frame things in terms of complex-valued maps. By a routine smoothing argument, the entire elements form a dense subspace of $X$. 


Now, associated to $\sigma$, is its generator $D$. This is the closed operator $D : X \to X$ whose domain is all the $x$ for which $t \mapsto \sigma\_t(x)$ is differentiable at $t=0$ (which then implies $t \mapsto \sigma\_t(x)$ is a $C^1$ map) given on its domain by
$$D(x) = \frac{d}{dt} \sigma\_t(x) \bigg|\_{t =0 } .$$
Inductively, one gets that the domain of $D^k$ is all the $x$ for which $t \mapsto \sigma\_t(x)$ is $C^k$ and, on this domain,
$$D^k(x) = \frac{d^k}{dt^k} \sigma\_t(x) \bigg|\_{t = 0}.$$
Now if $x$ is entire then, from general theory of power series, the entire extension of $t \mapsto \sigma\_t(x)$ is given by the norm-convergent series
$$ \sum\_{n=0}^\infty \frac{1}{n!}z^n D^n(x).$$
My question is about the converse:



> 
> **Question:** If $x \in X$ is such that the series $\sum\_{n=0}^\infty \frac{1}{n!}z^n D^n(x)$ has infinite radius of convergence (hence defines an entire mapping $\mathbb{C} \to X$), is it then true that $\sigma\_t (x) = \sum\_{n=0}^\infty \frac{1}{n!}z^n D^n(x)$ for all $t \in \mathbb{R}$ so that $x$ is actually entire?
> 
> 
> 


Let me explain why I don't think this is immediate. To my eye, the natural line of reasoning would be to look at the derivatives of $t \mapsto \sigma\_t(x)$ and $t \mapsto \sum\_{n=0}^\infty \frac{1}{n!} t^n D^n(x)$. A bit of calculation shows that both solve the same initial value problem:
\begin{align\*}
\frac{d}{dt} f(t) = D(f(t)) && f(0) = x.
\end{align\*}
The trouble is, however, that $D$ is not a bounded operator. So there is not, to my knowledge, a good uniqueness theorem for this initial value problem.


Ultimately, my goal here is to understand how to recover $\sigma$ from $D$ (if indeed this is possible at this level of generality). I had figured a natural approach would be to try to recover the dense subspace of entire elements since, there, we have an explicit formula for the flow. 



Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?
If I am not mistaken, then this is connected to the notion of analytic vectors and related object.


If I understand your question correctly, it is answered in this [paper](http://www.ams.org/journals/tran/1972-167-00/S0002-9947-1972-0295125-5/) of Chernoff.


Further, for group generators on Banach spaces the analytic vectors are dense and they determine the generator $D$, see Exercise II.3.12(2) in [Engel-Nagel](http://www.fa.uni-tuebingen.de/research/publications/2006/a-short-course-on-operator-semigroups/A_Short_Course_on_Operator_Semigroups.pdf).


