Dense linear subspace of a Hilbert space

**RaisinBread** · November 20th 2011, 02:52 PM

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.

**Jose27** · November 20th 2011, 03:11 PM

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 $\overline{E}= E^{**}=H$ .

**RaisinBread** · November 20th 2011, 04:17 PM

Quick question how can you tell $E^{**}=\overline{E}$

**Jose27** · November 20th 2011, 06:46 PM

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 $E^{**} \subset \overline{E}^{**}=\overline{E}$ .

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