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Math Help - Dense linear subspace of a Hilbert space

  1. #1
    Junior Member RaisinBread's Avatar
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    Dense linear subspace of a Hilbert space

    Hey, I have difficulties with the following question;

    Let H be a Hilbert space with associated inner product \langle x,y\rangle, and let E be a linear subspace of H. Define the orthogonal set of E as E^*=\{y\in H| \langle y,x\rangle=0 ~ ~\forall x\in E\}. Show that E is dense in H if and only if E^*=\{0\}.

    I've already shown that E^* is always a closed linear subspace of H, and I am told I can use the fact that, if a given linear subspace M is closed, then (M^*)^*=M.

    From here, I guess that I have to sow that E^*=\{0\} iff the closure of E is H, but I can't manage to get started on either side of the iif.
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  2. #2
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    Re: Dense linear subspace of a Hilbert space

    If E is dense then E^{**}=\overline{E}= H so E^{***}=H^* = \{ 0\}, but E^* is always closed so E^{*}=E^{***}. For the other direction is the same, if E^{*}=\{ 0 \} then \overline{E}= E^{**}=H.
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  3. #3
    Junior Member RaisinBread's Avatar
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    Re: Dense linear subspace of a Hilbert space

    Quick question how can you tell E^{**}=\overline{E}
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  4. #4
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    Re: Dense linear subspace of a Hilbert space

    Quote Originally Posted by RaisinBread View Post
    Quick question how can you tell E^{**}=\overline{E}
    If you already know that F^{**}=F when F is closed then, just note that \overline{E}\subset E^{**} because the latter is closed and contains E, for the other inclusion E\subset \overline{E} implies \overline{E}^*\subset E^* which implies E^{**} \subset \overline{E}^{**}=\overline{E}.
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