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Let $E$ be a Banach space, and let $(\sigma_t)$ be a strongly continuous one-parameter group on $E$: so for $t\in\mathbb R$, we have that $\sigma_t$ is a contraction on $E$, $\sigma_t \sigma_s=\sigma_{t+s}$, $\sigma_0$ is the identity, and for each $x\in E$, the map $\mathbb R\rightarrow E; t\mapsto\sigma_t(x)$ is continuous.

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 $x\in E$ such that there is a bounded continuous map $f:\overline{S(i)}\rightarrow E$ with $f$ analytic on $S(i)$, and with $f(t) = \sigma_t(x)$ for $t\in\mathbb R$. Then we set $\sigma_i(x) = f(i)$. Then $\sigma_i$ is a densely defined, closed operator.

Under favourable circumstances, the following is somehow "folklore":

Let $x\in D(\sigma_i)$ with $\sigma_i(x)=x$. Then $\sigma_t(x)=x$ for all $t$.

E.g. if E is a Hilbert space this can be extracted from Stone's Theorem and some spectral theory. What I want to know is if there is a way to prove this using the more abstract framework I have setup (which is more accessible if, e.g. we start considering dual spaces and weakly-continuous groups etc.)

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up vote 5 down vote accepted

Using the semigroup property, you can extend $\sigma_t(x)$ to a bounded entire function. Liouville's theorem will do the rest.

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So I start with an analytic function on the strip-- I extend this by periodicity to the complex plane. The boundary conditions agree, so it's continuous, bounded, analytic on each strip. Something like Morera shows it's analytic everywhere, so Liouville shows it's constant, as required? – Matthew Daws Jul 26 '13 at 14:23
Yes, that is the idea. – Michael Renardy Jul 26 '13 at 14:58

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