MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ be a Banach space, and $M=E^*$ (my application has $M$ a von Neumann algebra, but this is unimportant). Let $(\sigma_t)$ be a SOT cts one-parameter group on $E$: so for $t\in\mathbb R$, we have that $\sigma_t$ is an invertible isometry on $E$, and $\sigma_t \sigma_s=\sigma_{t+s}$, and for each $\mu\in E$, the map $\mathbb R\rightarrow E; t\mapsto\sigma_t(\mu)$ is continuous.

Overload notation, and let $(\sigma_t)$ also denote the adjoint group acting on $M$, so that $\langle \sigma_t(x), \mu \rangle = \langle x, \sigma_t(\mu) \rangle$ for $x\in M, \mu\in E$.

We have the notion of the analytic continuation of $(\sigma_t)$. For example, set $S(i) = \{ z=t+iy\in\mathbb C : 0<y<1 \}$ with closure $\overline{S(i)}$. Then $D(\sigma_i)$ consists of those $\mu\in E$ such that there is a bounded continuous map $f:\overline{S(i)}\rightarrow E$ with $f$ analytic on $S(i)$, and with $f(t) = \sigma_t(\mu)$ for $t\in\mathbb R$. Then we set $\sigma_i(\mu) = f(i)$. Similarly we can do this for $\sigma_t$ acting on $M$.

Question: Suppose we have $x,y\in M$ such that $\langle x , \sigma_i(\mu) \rangle = \langle y,\mu \rangle$ for each $\mu\in D(\sigma_i)$. Then is $x\in D(\sigma_i)$ with $\sigma_i(x)=y$?

Now, actually I can prove this. The technicality is that $D(\sigma_i)$ is only dense in $E$, and so you have to be careful with duality arguments (I cannot get a naive argument, using that $\sigma_i$ is a closed operator, to work, for instance). What I have is a long, but not particularly hard, proof. However, surely this should be in the literature, but I cannot find a reference. Does anyone have one? Is there a good general place to look for this sort of thing?

share|cite|improve this question
Have you tried Hille-Phillips? – Theo Buehler Feb 5 '11 at 8:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.