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Math Help - finding a specific operator

  1. #1
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    finding a specific operator

    Hi! I am kinda going mad trying to find the following operator. Here goes the problem: Let (z_n) be a complex seq. such that \lim z_n=0 and let (x_n) be a seq. of pos. numbers such that \lim x_n=\infty. Show that there exists (i.e. find) a compact operator P on l^2 so that the following two are satisfied:
    (a) \lim_n \frac{\|(z_n I-P)^{-1}\|}{ x_n}=\infty
    and
    (b) All the z_n lie in the resolvent of P.

    Thank You!
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  2. #2
    Super Member girdav's Avatar
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    Re: finding a specific operator

    I've not done the computations, but did you try with P diagonal?
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  3. #3
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    Re: finding a specific operator

    yes I did try... I'm not sure how that helps.
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  4. #4
    Super Member girdav's Avatar
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    Re: finding a specific operator

    It would reduce the problem to find a sequence instead of an operator.
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  5. #5
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    Re: finding a specific operator

    I'm sorry but I just can't get it to work. How am I supposed to find it?
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  6. #6
    Super Member girdav's Avatar
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    Re: finding a specific operator

    There is something I don't understand: if z_n=0 for all n, then  P has to be both compact and invertible in order to make the expression have a sense. It's impossible since \ell^2 is infinite dimensional.
    Last edited by girdav; January 15th 2012 at 07:08 AM.
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  7. #7
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    Re: finding a specific operator

    The assumptio is not z_n=0  \forall n, it is that \lim_{n\rightarrow \infty} z_n=0
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  8. #8
    Super Member girdav's Avatar
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    Re: finding a specific operator

    Yes, but z_n=0 for all n is a particular case of \lim_{n\to \infty}z_n=0. So we have to assume that z_n\neq 0 for all n otherwise P would be invertible.
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  9. #9
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    Re: finding a specific operator

    I see what you mean. Ok, how would I proceed after removing that special case.
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  10. #10
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    Re: finding a specific operator

    Quote Originally Posted by Zeke View Post
    Hi! I am kinda going mad trying to find the following operator. Here goes the problem: Let (z_n) be a complex seq. such that \lim z_n=0 and let (x_n) be a seq. of pos. numbers such that \lim x_n=\infty. Show that there exists (i.e. find) a compact operator P on l^2 so that the following two are satisfied:
    (a) \lim_n \frac{\|(z_n I-P)^{-1}\|}{ x_n}=\infty
    and
    (b) All the z_n lie in the resolvent of P.

    Thank You!
    As girdav pointed out, none of the z_n can be 0. So make that assumption.

    Following girdav again, let P be the operator on \l^2(\mathbb{N}) whose matrix (with respect to the standard basis) is diagonal, with diagonal elements \alpha_n. Let Z = \{z_n\}_{n\in\mathbb{N}}\cup\{0\}. For each n, choose \alpha_n so that \alpha_n\notin Z and |\alpha_n-z_n| < (nx_n)^{-1}.

    Then P is compact (because \alpha_n\to0). The spectrum of P is A = \{\alpha_n\}_{n\in\mathbb{N}}\cup\{0\}. Since, for each n, z_n\notin A, it follows that (b) holds. Finally, \|(z_nI-P)^{-1}\| = \max_{k\in\mathbb{N}}|z_n-\alpha_k|^{-1}\geqslant|z_n-\alpha_n|^{-1}>nx_n.
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