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Math Help - Compact, bounded, closed-range operators on Hilbert spaces have finite rank

  1. #1
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    Compact, bounded, closed-range operators on Hilbert spaces have finite rank

    Hi,

    I am trying to prove that if T:H\to{H} is a compact, bounded operator with closed range, where H is an \infty-dimensional Hilbert space, then T has finite rank. I want to prove this without using the open-mapping theorem. T is not necessarily linear.

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

    The definitions I have in my lecture notes are:

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

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

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

    Any help with the proof would be greatly appreciated.

    Thank you in advance.
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  2. #2
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    Re: Compact, bounded, closed-range operators on Hilbert spaces have finite rank

    Suppose Range(T) is infinite dimensional. Since Range(T) is closed, the map

    T': (\ker T)^\perp \rightarrow Range(T)
    T'(x) = T(x)

    is a bijection between Hilbert spaces. Take a sequence  \{ T' x_i \} \in Range(T) of pairwise orthogonal elements such that ||T' x_i || =1 . By the bounded inverse theorem, ||x_i|| \leq || T'^{-1}|| \ ||T'x||= || T'^{-1}||, so since T' is compact  \{ T' x_i \} should have a convergent subsequence. This is a contradiction since  \{ T' x_i \} has no subsequence that is Cauchy.
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