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Math Help - Compact?

  1. #1
    Member Mauritzvdworm's Avatar
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    Compact?

    Let T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2} and consider the space B(L^{2}(T^{2})), using the quotient topology, can we show that B(L^{2}(T^{2})) is compact? Where B is the set of all bounded linear operators mapping elements of L^{2}(T^{2}) to itself.
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  2. #2
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    Quote Originally Posted by Mauritzvdworm View Post
    Let T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2} and consider the space B(L^{2}(T^{2})), using the quotient topology, can we show that B(L^{2}(T^{2})) is compact? Where B is the set of all bounded linear operators mapping elements of L^{2}(T^{2}) to itself.
    The space B(L^{2}(T^{2})) can never be compact in any topology, because it is unbounded. For example, the sequence (nI), where I is the identity operator, cannot have a convergent subsequence.

    The unit ball of B(L^{2}(T^{2})) is compact in the weak operator topology.
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