# Dense linear subspace of a Hilbert space

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• Nov 20th 2011, 03:52 PM
RaisinBread
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.
• Nov 20th 2011, 04:11 PM
Jose27
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 $\overline{E}= E^{**}=H$.
• Nov 20th 2011, 05:17 PM
RaisinBread
Re: Dense linear subspace of a Hilbert space
Quick question how can you tell $E^{**}=\overline{E}$
• Nov 20th 2011, 07:46 PM
Jose27
Re: Dense linear subspace of a Hilbert space
Quote:

Originally Posted by RaisinBread
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 $E^{**} \subset \overline{E}^{**}=\overline{E}$.