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Math Help - Linear Operators

  1. #1
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    Linear Operators

    I need help with this:

    Let E a Banach Space and B(E) the Banach space of linear and bounded operators from E to E with the usual norm. Let S $\in$ B(E).

    If the sum $\sum\limits_{n = 1}^\infty  {\frac{{{S^k}}}{{k!}}}$ convergs to the operator $e^S$, show that $Se^S$ = $e^S S$. And if T,S \in B(E) conmute then $e^{S+T}$= $e^S$ $e^T$. In particular prove that $e^S$ is invertible.

    Please somebody give me a hand.
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  2. #2
    Super Member girdav's Avatar
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    For the first question, put S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all n\in\mathbb N we have SS_n=S_nS. Then use the fact that the maps T\mapsto ST and T\mapsto TS are continuous for the usual norm on \mathcal B(E).
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  3. #3
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    Quote Originally Posted by girdav View Post
    For the first question, put S_n :=\sum_{k=0}^n\frac{S^k}{k!}. Show that for all n\in\mathbb N we have SS_n=S_nS. Then use the fact that the maps T\mapsto ST and T\mapsto TS are continuous for the usual norm on \mathcal B(E).
    Sorry to bother again, but I canīt show $S{S_n} = {S_n}S$...
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  4. #4
    Super Member girdav's Avatar
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    Prove it by induction.
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  5. #5
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    Quote Originally Posted by girdav View Post
    Prove it by induction.
    Done!

    Could you explain to me how to use the continuity of T and S?

    Thanks"!
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  6. #6
    Super Member girdav's Avatar
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    Show that SS_n converges to Se^S in \mathbb{B}(E) and similarly that S_nS converges to e^SS in \mathbb{B}(E)
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  7. #7
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    Quote Originally Posted by girdav View Post
    Show that SS_n converges to Se^S in \mathbb{B}(E) and similarly that S_nS converges to e^SS in \mathbb{B}(E)
    Done! Thank you again.

    And, could you explain a bit how to do the other part?

    Thanks!
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  8. #8
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    Thank you..

    Sorry double post.
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  9. #9
    Super Member girdav's Avatar
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    Since S and T commute, you can try a proof which is similar to the case S,T\in\mathbb{C}. The main step consists of showing that
    e^Se^T = \sum_{n=0}^{+\infty}\sum_{k=0}^n\dfrac{S^k}{k!} \cdot\dfrac{T^{n-k}}{(n-k)!}.
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