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A compact, bounded, closed-range operator on a Hilbert space has finite rank

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]H[/itex] be an [itex]\infty[/itex]-dimensional Hilbert space and [itex]T:H\to{H}[/itex] be an operator.

    Show that if [itex]T[/itex] is compact, bounded and has closed range, then [itex]T[/itex] has finite rank. Do not use the open-mapping theorem.

    Let [itex]B(H)[/itex] denote the space of all bounded operators mapping [itex]H\to{H}[/itex], [itex]K(H)[/itex] denote the space of all compact operators mapping [itex]H\to{H}[/itex], [itex]R(H)[/itex] denote the space of all finite rank operators mapping [itex]H\to{H}[/itex].

    2. Relevant defintions

    *[itex]T\in{B(H)}[/itex] is compact if the closure of [itex]T(B(0,1))[/itex] is a compact set.
    *[itex]T\in{B(H)}[/itex] has finite rank if [itex]Range(T)=T(H)[/itex] is finite-dimensional.

    3. The attempt at a solution

    I'm not sure how to do the proof, but I think the following propositions in my lecture notes could be useful:

    *[itex]T\in{R(H)}[/itex] iff [itex]T\in{B(H)}[/itex] is the norm limit of a sequence of finite rank operators, i.e. [itex]K(H)[/itex] is the closure of [itex]R(H)[/itex].
    *Let [itex]T\in{R(H)}[/itex]. Then there is an orthonormal set [itex]{{e_1,...,e_L}}\in{H}[/itex] s.t. [tex]Tu=\sum\limits_{i,j=1}^{L}{c_{ij}(u,e_j)e_i}[/tex] where [itex]c_{ij}[/itex] are complex numbers.

    Any help with the proof would be greatly appreciated.

    Thank you in advance.
     
    Last edited: May 26, 2012
  2. jcsd
  3. May 26, 2012 #2

    micromass

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    Do you know that compact operators can never be invertible on Hilbert spaces??

    Given your compact operator T with closed range, can you modify the domain and codomain a bit such that T becomes a bijection between Hilbert spaces??
     
  4. May 26, 2012 #3
    Thank you for your reply.

    The next part of the question is to show that T is not invertible, so I can't use that fact.

    T would be surjective if its image equalled its range, but the question defines these to be equal (under heading 2 in the original post), so T is already surjective. I'm not sure how to make T injective.
     
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