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Math Help - 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 f\in L^2(\mathbb{T}) with \mathbb{T} the unit circle. Define the following mappings

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

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

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

    \tau_r:a\mapsto U_raU^*_r
    where a\in A, a C*-algebra.

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

    Suppose we have a function of the form

    f:\mathbb{R}\rightarrow A

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

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

    I define the following mapping

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

    Now, what I want to show is that \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

    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

    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; June 8th 2011 at 11:52 PM.
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  2. #2
    Member Mauritzvdworm's Avatar
    Joined
    Aug 2009
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    Pretoria
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    for some reason the \delta(a) did not want to show in the above equation, however, here is what I have done up to now

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

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

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

    =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

    =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

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

    bringing the 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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