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Math Help - Compact operators on Hilbert Spaces

  1. #1
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    Feb 2009
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    Compact operators on Hilbert Spaces

    Hi, here is my problem :

    Let H be a Hilbert space with basis {u_k}_k>0.
    Verify that the operator T defined by :
    T(u_k) = 1/k . u_k+1
    is compact but has no eigenvectors.

    To show it is compact, I tried to use the definition : T is compact if any bounded sequence {f_k} in H has a subsequence {f_nk}_k>0 such that {T(f_nk)} converges.
    I took a bounded sequence {f_k} in H and approximated the f_k's by linear combinations of the basis elements {u_k}. But I don't see where this is supposed to lead me.

    For the eigenvectors, I am lost.

    Any ideas ?
    Thanks.
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  2. #2
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    Feb 2009
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    For the eigenvectors, assume that Tx = \alpha x for some \alpha \in \mathbb{C}, and write x = \sum^{\infty}_{i=1} \lambda_i \varphi_i . Then the equation reads T(\sum \lambda_i \varphi_i) = \sum \alpha \lambda_i \varphi_i, which, carrying out the operation, brings us to  \sum_{i=1}^{\infty} \frac{\lambda_i}{i} \varphi_{i+1} = \sum^{\infty}_{i=1} \alpha \lambda_i \varphi_i . Now solve for \lambda_i. What do you get?
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