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Let $\mathfrak{A}$ be a C${}^*$ algebra and $\mathbb{R}\ni s \mapsto \alpha_s$ a continuous family of its automorphisms. Is it true that

$$ \int d s \, f(s)\, \alpha_s(A) $$

is well defined as a Bochner integral for any $A\in\mathfrak{A}$ and $f \in L^1(\mathbb{R})$.

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    $\begingroup$ What continuity/measurability conditions are you putting on the map $s\mapsto \alpha_s$? You might want to look up the Pettis measurability theorem $\endgroup$
    – Yemon Choi
    Commented Jan 21, 2015 at 20:33
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    $\begingroup$ OK, now if you have continuity in some kind of SOT for the family $(\alpha_s)$ then I think the Pettis measurability theorem should do the job. $\endgroup$
    – Yemon Choi
    Commented Jan 21, 2015 at 21:20
  • $\begingroup$ Alternatively, this feels like the kind of thing that might be discussed in Kadison+Ringrose when they need to introduce some form of modular theory $\endgroup$
    – Yemon Choi
    Commented Jan 21, 2015 at 21:20

1 Answer 1

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Yes.

$\alpha_s(A)$ is a continuous bounded function.

The function $f(s) \alpha_s(A)$ is measurable and because of

$$\int_\mathbb{R} \|f(s) \alpha_s(A)\| ds \le \int_\mathbb{R} |f(s)| ds\, \|A\| < \infty$$

in $L^1$.

Quite elementary, look in the book "Serge Lang, Real and Functional Analysis", Chapter Integration for related things.

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  • $\begingroup$ [deleted comment which was based on a hasty misreading] $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2015 at 0:37

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