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

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

    From here, I guess that I have to sow that $\displaystyle E^*=\{0\}$ iff the closure of $\displaystyle E$ is $\displaystyle H$, 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 $\displaystyle E$ is dense then $\displaystyle E^{**}=\overline{E}= H$ so $\displaystyle E^{***}=H^* = \{ 0\}$, but $\displaystyle E^*$ is always closed so $\displaystyle E^{*}=E^{***}$. For the other direction is the same, if $\displaystyle E^{*}=\{ 0 \}$ then $\displaystyle \overline{E}= E^{**}=H$.
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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 $\displaystyle E^{**}=\overline{E}$
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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 $\displaystyle E^{**}=\overline{E}$
    If you already know that $\displaystyle F^{**}=F$ when $\displaystyle F$ is closed then, just note that $\displaystyle \overline{E}\subset E^{**}$ because the latter is closed and contains $\displaystyle E$, for the other inclusion $\displaystyle E\subset \overline{E}$ implies $\displaystyle \overline{E}^*\subset E^*$ which implies $\displaystyle E^{**} \subset \overline{E}^{**}=\overline{E}$.
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