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Thread: derivative of operator valued function

  1. #1
    Member Mauritzvdworm's Avatar
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    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?
    Last edited by Mauritzvdworm; Jun 8th 2011 at 11:52 PM.
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  2. #2
    Member Mauritzvdworm's Avatar
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    for some reason the $\displaystyle \delta(a)$ did not want to show in the above equation, however, here is what I have done up to now

    $\displaystyle \tau_{r+h}(\delta(a))-\tau_r(\delta(a))$

    $\displaystyle =U_{r+h}\delta(a)U^*_{r+h}-U_r\delta(a)U^*_{r}$

    $\displaystyle =U_r\left[U_h\delta(a)U^*_h-\delta(a)\right]U^*_r$

    $\displaystyle =U_r\left[U_h\left(\left.\frac{d}{dr}\tau_r(a) \right|_{r=0}\right) U^*_h-\left.\frac{d}{dr}\tau_r(a)\right|_{r=0}\right]U^*_r$

    $\displaystyle =U_r\left[\left.\frac{d}{dr}\tau_{r+h}(a) \right|_{r=0}-\left.\frac{d}{dr}\tau_r(a)\right|_{r=0}\right]U^*_r$

    $\displaystyle =U_r\left[\left.\frac{d}{dr}\left(\tau_{r+h}(a)-\tau_r(a) \right) \right|_{r=0} \right]U^*_r$

    bringing the $\displaystyle h$ into account I have something similar to what I need, however, the second part does not want to play along..

    Any Ideas?
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