derivative of operator valued function

Consider the following construction

let $\displaystyle f\in L^2(\mathbb{T})$ with $\displaystyle \mathbb{T}$ the unit circle. Define the following mappings

$\displaystyle \varphi_{r}:x\mapsto x+r$ and

$\displaystyle U_{r}:f\mapsto f\circ\varphi_{r}$

it is easy to show that $\displaystyle U_r$ is unitary and we can then define an anology of time evolution by the mapping

$\displaystyle \tau_r:a\mapsto U_raU^*_r$

where $\displaystyle a\in A$, a C*-algebra.

Let $\displaystyle A^{\infty}:=\left\{a\in A:a \text{ of class }C^{\infty}\right\}$ where $\displaystyle A$ is a C*-algebra. And by class of $\displaystyle C^{\infty}$ I mean the mapping $\displaystyle r\mapsto \tau_r(a)$ with $\displaystyle a\in A$ is differentiable to any order with $\displaystyle r\in\mathbb{R}$.

Suppose we have a function of the form

$\displaystyle f:\mathbb{R}\rightarrow A$

we can define the derivative of the operator valued function as $\displaystyle f'(x)$ satisfying the equation

$\displaystyle 0=\lim_{h\rightarrow\infty}\|\frac{f(x+h)-f(x)}{h}-f'(x) \|$

I define the following mapping

$\displaystyle \delta(a):=\left.\frac{d}{dr}\tau_r(a)\right|_{r=0 }$

Now, what I want to show is that $\displaystyle \delta:A^{\infty}\rightarrow A^{\infty}$.

I am not sure how to get this results, I think that I am going to have to use

$\displaystyle 0=\lim_{h\rightarrow\infty}\|\frac{\tau_{r+h}(a)-\tau_r(a)}{h}-\partial_r\tau_r(a) \|$

but I cannot manipulate the expression in such a way that the above equation pops out. So I want to show that

$\displaystyle 0=\lim_{h\rightarrow\infty}\|\frac{\tau_{r+h}(\del ta(a) )-\tau_r(\delta (a))}{h}-\partial_r\tau_r(\delta(a)) \|$

Any ideas?