I am studying the actions of Banach-Lie groups on Banach manifolds, and I am not able to concretely evaluate the differentiability properties of a specific action.
Let $\mathcal{A}$ be a unital $C^{*}$-algebra, and $\mathcal{A}^{*}$ its topological dual, which is a complex Banach space with norm:
$$ \left|\left|\omega\right|\right|:=\sup\left\{|\omega(\mathbf{A})|\,;\mathbf{A}\in\mathcal{A}\colon||\mathbf{A}||=1\right\} $$ Let us denote with $\mathcal{G}(\mathcal{A})$ the set of all invertible elements in $\mathcal{A}$. This set is a Banach-Lie group with respect to the multiplication operation inherited from $\mathcal{A}$. This group naturally acts on $\mathcal{A}$ as follows:
$$ \mathbf{A}\mapsto c_{\mathbf{G}}(\mathbf{A}):=\mathbf{G}\,\mathbf{A}\,\mathbf{G}^{\dagger}\,. $$ By means of this action, an action on $\mathcal{A}^{*}$ is obtained:
$$ \omega\mapsto \alpha_{\mathbf{G}}(\omega) $$ $$ \left(\alpha_{\mathbf{G}}(\omega)\right)(\mathbf{A}):=\omega\left(c_{\mathbf{G}}(\mathbf{A})\right)=\omega\left(\mathbf{G}\,\mathbf{A}\,\mathbf{G}^{\dagger}\right)\,. $$ It is easy to see that this action is linear and preserves hermiticity.
I am interested in the differentiability properties of this action. By definition, the action is differentiable if the map:
$$ \mathcal{G}(\mathcal{A})\times\mathcal{A}^{*}\rightarrow\mathcal{A}^{*} $$ $$ (\mathbf{G}\,;\omega)\mapsto \alpha_{G}\omega $$ is differentiable. I know the abstract definition of differentiability (Frechet derivative) of a map between Banach manifolds, however, I do not know how to concretely proceed to investigate the subject. Specifically, a map $f:V\rightarrow W$ between Banach space is differentiable at $x_{0}\in V$ if, for all $\epsilon>0$ there exist $\delta_{\epsilon}>0$ and a bounded linear map $Df_{x_{0}}:V\rightarrow W$ such that:
$$ ||f(x-x_{0}) - f(x_{0}) - Df_{x_{0}}(x-x_{0})||<\epsilon $$ whenever $||x-x_{0}||<\delta_{\epsilon}$.
Since $\mathcal{G}(\mathcal{A})$ is only a Banach manifold, I need a coordinates system in order to use the definition of Frechet derivative, and I do not know what kind of coordinates system I should use. Furthermore, even if I knew it, I would need an "educated guess" of the explicit face of the derivative $Df_{x_{0}}$ to actually calculate the limit in the definition of the Frechet derivative.
Are there other, more smart ways to proceed?
Thank You in advance.
EDIT
With a little effort, I was able to prove that the linear map $\alpha_{\mathbf{G}}:\mathcal{A}^{*}\rightarrow\mathcal{A}^{*}$ is bounded, and thus continuous, for all $\mathbf{G}\in\mathcal{G}(\mathcal{A})$. This means that $\alpha_{\mathbf{G}}$ is analytic for all $\mathbf{G}$ (it is a continuous 1-homogeneous polynomial), but I do not know if it is of some relevance.
EDIT 2
According to Andreas Cap's answer, I tried to work out the details without using the $t$ parameter.
Let $U_{\mathbf{G}}\subset\mathcal{G}(\mathcal{A})$ be a neighbourhood of $\mathbf{G}$ such that $\mathbf{G}+\mathbf{H}\in U$ for all $\mathbf{H}\in U$. Then we can write:
$$ \alpha(\mathbf{G}+\mathbf{H}\,;\omega + \tau)(\mathbf{A}) - \alpha(\mathbf{G}\,;\omega)(\mathbf{A})=\omega(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger}) + \omega(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger}) + \tau(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger}) + \tau(\mathbf{G}\mathbf{A}\mathbf{G}^{\dagger}) $$ The linear functional introduced by Andreas Cap is:
$$ D\alpha_{\mathbf{G}\,;\omega}(\mathbf{H}\,;\tau)(\mathbf{A})=\omega(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger}) + \tau(\mathbf{G}\mathbf{A}\mathbf{G}^{\dagger}) $$ and thus:
$$ \alpha(\mathbf{G}+\mathbf{H}\,;\omega + \tau)(\mathbf{A}) - \alpha(\mathbf{G}\,;\omega)(\mathbf{A}) - D\alpha_{\mathbf{G}\,;\omega}(\mathbf{H}\,;\tau)(\mathbf{A})= \tau(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger})+ (\omega+\tau)(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger}) $$ Let us call this functional $\gamma$. The norm of $\gamma$ is:
$$ ||\gamma||=\sup_{\mathbf{A}\neq\mathbf{0}}\frac{|\gamma(\mathbf{A})|}{||\mathbf{A}||} $$ Then:
$$ |\gamma(\mathbf{A})|\leq|\tau(\mathbf{G}\mathbf{A}\mathbf{H}^{\dagger} + \mathbf{H}\mathbf{A}\mathbf{G}^{\dagger})| + |(\omega+\tau)(\mathbf{H}\mathbf{A}\mathbf{H}^{\dagger})|\leq $$ $$ \leq2||\tau||\,||\mathbf{G}||\,||\mathbf{A}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2}\,||\mathbf{A}|| + ||\tau||\,||\mathbf{H}||^{2} \,||\mathbf{A}|| $$ which means:
$$ ||\gamma||=2||\tau||\,||\mathbf{G}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2} + ||\tau||\,||\mathbf{H}||^{2} $$ The action $\alpha$ is differentiable at $(\mathbf{G}\,;\omega)$ if, for all $\epsilon>0$ there exists $\delta_{\epsilon}>0$ such that:
$$ \frac{||\gamma||}{||(\mathbf{H}\,;\tau)||}<\epsilon $$ whenever $||(\mathbf{H}\,;\tau)||<\delta_{\epsilon}$. However, we have:
$$ \frac{||\gamma||}{||(\mathbf{H}\,;\tau)||}=\frac{2||\tau||\,||\mathbf{G}||\,||\mathbf{H}|| + ||\omega||\,||\mathbf{H}||^{2} + ||\tau||\,||\mathbf{H}||^{2}}{||\mathbf{H}|| +||\tau||} $$ and I am not sure that this is less than $\epsilon$.
I think this is easier than it initially looks. First of all, $\mathcal G(\mathcal A)$ is an open subset in the Banach space $\mathcal A$, so there is no need to worry about coordinates. On Banach spaces, differentiation basically works as you are used to in open subsets of $\mathbb R^n$. Moreover, the maps you consider all are induces by linear (or conjugate linear) and bilinear maps.
Since $\mathcal G(\mathcal A)$ is open, any of the curves $G+tB$ stays in $\mathcal G(\mathcal A)$. Now for $t\in\mathbb R$, bilinearity of the multiplication gives $(G+tB)A(G+tB)^\dagger=GAG^\dagger+t(BAG^\dagger+GAB^\dagger)+t^2BAB^\dagger$ and hence $$ (\omega+t\tau)((G+tB)A(G+tB)^\dagger=\omega(GAG^\dagger)+t\big(\omega(BAG^\dagger+GAB^\dagger)+\tau(GAG^\dagger)\big)+O(t^2) $$ which shows that $Df(G,\omega)(B,\tau)$ is the linear functional mapping $A$ to $\omega(BAG^\dagger+GAB^\dagger)+\tau(GAG^\dagger)$.