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Thread: Compact?

  1. #1
    Member Mauritzvdworm's Avatar
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    Compact?

    Let $\displaystyle T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ and consider the space $\displaystyle B(L^{2}(T^{2}))$, using the quotient topology, can we show that $\displaystyle B(L^{2}(T^{2}))$ is compact? Where B is the set of all bounded linear operators mapping elements of $\displaystyle L^{2}(T^{2})$ to itself.
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  2. #2
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    Quote Originally Posted by Mauritzvdworm View Post
    Let $\displaystyle T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ and consider the space $\displaystyle B(L^{2}(T^{2}))$, using the quotient topology, can we show that $\displaystyle B(L^{2}(T^{2}))$ is compact? Where B is the set of all bounded linear operators mapping elements of $\displaystyle L^{2}(T^{2})$ to itself.
    The space $\displaystyle B(L^{2}(T^{2}))$ 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 $\displaystyle B(L^{2}(T^{2}))$ is compact in the weak operator topology.
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