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Thread: 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 $\displaystyle $\in$$ B(E).

    If the sum $\displaystyle $\sum\limits_{n = 1}^\infty {\frac{{{S^k}}}{{k!}}}$$ convergs to the operator $\displaystyle $e^S$$, show that $\displaystyle $Se^S$$ = $\displaystyle $e^S S$$. And if T,S $\displaystyle \in $ B(E) conmute then $\displaystyle $e^{S+T}$$= $\displaystyle $e^S$$$\displaystyle $e^T$$. In particular prove that $\displaystyle $e^S$$ 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 $\displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}$. Show that for all $\displaystyle n\in\mathbb N$ we have $\displaystyle SS_n=S_nS$. Then use the fact that the maps $\displaystyle T\mapsto ST$ and $\displaystyle T\mapsto TS$ are continuous for the usual norm on $\displaystyle \mathcal B(E)$.
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  3. #3
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    Quote Originally Posted by girdav View Post
    For the first question, put $\displaystyle S_n :=\sum_{k=0}^n\frac{S^k}{k!}$. Show that for all $\displaystyle n\in\mathbb N$ we have $\displaystyle SS_n=S_nS$. Then use the fact that the maps $\displaystyle T\mapsto ST$ and $\displaystyle T\mapsto TS$ are continuous for the usual norm on $\displaystyle \mathcal B(E)$.
    Sorry to bother again, but I canīt show $\displaystyle $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 $\displaystyle SS_n$ converges to $\displaystyle Se^S$ in $\displaystyle \mathbb{B}(E)$ and similarly that $\displaystyle S_nS$ converges to $\displaystyle e^SS$ in $\displaystyle \mathbb{B}(E)$
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  7. #7
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    Quote Originally Posted by girdav View Post
    Show that $\displaystyle SS_n$ converges to $\displaystyle Se^S$ in $\displaystyle \mathbb{B}(E)$ and similarly that $\displaystyle S_nS$ converges to $\displaystyle e^SS$ in $\displaystyle \mathbb{B}(E)$
    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 $\displaystyle S$ and $\displaystyle T$ commute, you can try a proof which is similar to the case $\displaystyle S,T\in\mathbb{C}$. The main step consists of showing that
    $\displaystyle e^Se^T = \sum_{n=0}^{+\infty}\sum_{k=0}^n\dfrac{S^k}{k!} \cdot\dfrac{T^{n-k}}{(n-k)!}$.
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