Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?

Question

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.

Answer 1

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 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.

Answer 2

It is well known in the theory of operator semigroups that the Cauchy problem for the generator $D$ of a strongly continuous SEMIgroup is always UNIQUELY solvable for initial data in the domain of $D$, that is $[0,\infty)\to X$, $t\mapsto \sigma_t(x)$ is the unique solution of $f'(t)=D(f(t))$ for all $t\ge 0$ and $f(0)=x$. Therefore, if you find by some other means a solution it coincides with $t\mapsto \sigma_t(x)$. This can be found in any book on operator semigroups, i.e. A short course... from K.-J. Engel and R. Nagel. It should be no problem to extend the result to groups of operators (e.g. by considering $\mathbb R_+$ and $\mathbb R_{-}$ separately).

Answer 3

As it turns out, the basic idea is so simple that I had really might as well add a brief summary. As before, $(\sigma_t)$ is a strongly-continuous, isometric flow on a Banach space $X$ and $D$ is its infinitesimal generator. As I mentioned above, on the face of it the domain of $D$ is just the $x$ for which $\frac{d}{dt} \sigma_t(x) |_{t =0}$ exists, but it follows easily from the group law that, if $x \in \operatorname{dom}(D)$, then $t \mapsto \sigma_t(x)$ is $C^1$ and $\frac{d}{dt} \sigma_t(x) = \sigma_t(D(x))$ is satisfied.

Claim: If $f : \mathbb{R} \to \operatorname{dom}(D)$ is a $C^1$ solution to the initial value problem \begin{align*} \frac{d}{dt} f(t) = D(f(t)) && f(0) = x_0 \in \operatorname{dom}(D), \end{align*} then $f(t) = \sigma_t(x_0)$ for all $t \in \mathbb{R}$.

Proof: Since $$ \frac{d}{dt} \sigma_{-t}( f(t)) = - \sigma_t(D(f(t))) + \sigma_t(\frac{d}{dt}f(t))=0,$$ we conclude $\sigma_{-t}(f(t))$ is constant, hence $\sigma_{-t}(f(t)) = \sigma_0(f(0)) = x_0$ for all $t \in \mathbb{R}$. Applying $\sigma_t$ on both sides gives $f(t) = \sigma_t(x_0)$, as desired.

The above claim shows that $\sigma_t(x)$ is determined by $D$ when $x \in \operatorname{dom}(D)$. In particular, this is true when $x$ is entire, which answers my question. Moreover, since the domain of $D$ is norm-dense in $X$, this implies the flow can be recovered from its infinitesimal generator.

Note the proof of the claim is really no different from the standard proof in elementary, single-variable calculus that, if $\frac{dx}{dt} = \delta x(t)$ and $x(0) = x_0$, then $x(t)=x_0e^{ \delta t}$.

Finally, in the Engel and Nagel book which the other answers mention, the IVP in the claim is replaced by the integral equation $$ f(t) = x_0 + D \left( \int_0^t f(s) \ ds \right).$$ Continuous functions $f : \mathbb{R} \to X$ satisfying this equation are called "mild solutions". In the same way, one can check that $t \mapsto \sigma_t(x_0)$ is the unique mild solution. An advantage to this appraoch is that, because of the extra smoothing afforded by integrating before applying $D$, this actually works for every $x_0 \in X$, and not just for $x_0 \in \operatorname{dom}(D)$ as with the IVP.