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Let $(\mathcal{H},\langle\cdot,\cdot\rangle)$ be a Hilbert space and let $U$ denote a closed subspace of $\mathcal{H}$. Show that: $$U=(U^{\bot})^{\bot}$$

Afterwards, then show for a subspace $V$ of $\mathcal{H}$ that:

$$\overline{V}=(V^{\bot})^{\bot}$$

where $\overline{V}$ denotes the closure of $V$.

My thoughts: So I don't know quite how to approach this. I've thought about show both inclusions but I don't see a good way to do so for the first equality. Also for the second part I know the closure of a set is closed however V itself is not closed so the second doesn't just follow immediately from the first. Any help would be appreciated!

Ben Grossmann
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CruZ
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  • See this post for the more difficult direction – Ben Grossmann Nov 12 '19 at 08:39
  • I would encourage you to try to prove the inclusion $U \subset U^{\perp \perp}$ on your own. Ultimately the proof comes down to characterizing $U^{\perp \perp}$, then showing that every $u \in U$ satisfies this characterization. – Ben Grossmann Nov 12 '19 at 08:42
  • Thank you, this helped to show that $U=(U^{\bot})^{\bot}$. However I don't see how the same argument works for $\overline{V}$ though. I can only use orthogonal decomposition if $V$ itself is closed, right? The only things that comes to mind is using sequences but maybe I am overlooking something easier :) – CruZ Nov 12 '19 at 09:19
  • It is easy to show that $V \subset V^{\perp \perp}$. For the other direction, note that $V^\perp = (\bar V)^\perp$ – Ben Grossmann Nov 12 '19 at 09:20
  • Wouldn't I have to show that $\overline{V} \subseteq V^{\bot \bot}$? And why is that note true? – CruZ Nov 12 '19 at 09:26
  • Once you have established that $(\bar V)^{\perp} = V^{\perp}$, it follows that $(\bar V)^{\perp \perp} = V^{\perp \perp}$, which is to say that $\bar V = V^{\perp \perp}$. Try to figure out the note for yourself; it comes down to the definition of the closure. – Ben Grossmann Nov 12 '19 at 09:30

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