# Element Analytic, C*-dynamical system

good night... I was looking into the Pedersen Book, $C^{*}$-Algebras and their automorphism groups, and found the definition of analytic elements $x\in A$, where $(A,\alpha)$ is a $C^{*}-$dynamical system.

We say that an element $x\in A$ is analytic for $\alpha$ if the function $t\mapsto \alpha_{t}(x)$ has an extension, necessarily unique, to an analytic (entire) vector-valued function $\zeta\rightarrow \alpha_{\zeta}(x)$, $\zeta\in\mathbb{C}$. If $x\in A$ then $$x_n=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}\alpha_t(x)exp(-nt^{2})dt$$ is analytic for $\alpha$.

And it's not clear for me that $x_n$ is well defined, i.e., why $x_n\in A$? and how is the integral defined:

$[1]$ The integration theory for vector-valued functions in a general Banach space, with Riemann sums, or

$[2]$ The Lebesgue integral for Banach spaces is the Bochner integral.

$\alpha_t(x_n)$ extends to $$\alpha_{\zeta}(x_n)=\sqrt{\frac{n}{\pi}}\int_{\mathbb{R}}\alpha_t(x)exp(-n(t-\zeta)^{2})dt$$

Analyticity of $\alpha_{\zeta}(x_n)$ is automatic because $\alpha_t(x)exp(-n(t-\zeta)^{2})$ is continuous and the Fundamental Theorem of Calculus?

I hope you answer me, i really thanks full. Do you recommend some bibliography?

Thanks so much.

In this case, a (improper) Riemann integral will suffice. The reason is that the function you want to integrate is continuous: you need (strong) continuity of $\alpha$ as a basic assumption throuhgout the theory of $C^*$-dynamical systems. Then the Riemann integral is perhaps much easier than the Bochner integral. Unfortunately, this is not completely easy to find in the literature: but things are really as simple as you think they are :) For the holomorphic dependence of $\alpha_\zeta(x_n)$ on $\zeta$ you can argue by the general principle that weakly holomorphc functions are holomorphic. This statement hold for very very general target spaces, much beyond Banach spaces, see e.g. Rudin's book on functional analysis. To check that $\alpha_\zeta(x_n)$ is weakly holomorphic is rather trivial: improper Riemann integrals have the necessary continuity properties which allow to move a continuous linear functional on $A$ inside the integral.