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Let $H$ be a Hilbert space and $U$ a subspace. Let $U^{\bot}$ denote its orthogonal complement.

I had no trouble showing $\overline{U}\subseteq U^{\bot\bot}$. But now I'm stuck for $\supseteq$.

Please could someone help me finish this argument?

This is what I am trying to do:

Let $x \in U^{\bot \bot}$. The goal is to construct a sequence $u_n\in U$ such that $u_n \to x$.

Since $\overline{U} \subset U^{\bot \bot}$ either $x\in \overline{U}$ or $x \notin \overline{U}$. If $x \in \overline{U}$ then we're done. If $x \notin \overline{U}$ I don't know what I can do.

2 Answers2

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Hint: If $x\not\in \overline{U}$, then by Hilbert projection theorem we have $x=y+z$ with $y\in \overline{U}$ and $0\ne z\perp \overline{U}$.

Since $z\ne 0$ one has $$ \langle x,z\rangle=\langle y+z,z\rangle=\| z\|^2\ne 0. $$ Hence $z\in U^\perp$ does not annihilate $x$ and therefore $x\not\in U^{\perp\perp}$.

Janko Bracic
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  • Hilbert projection theorem seems to be shooting the pigeon with a canon. Is there a way to prove it from first principles? Otherwise I will instead have to prove Hilbert projection theorem first... and that seems more difficult than the problem at hand. –  Jan 18 '15 at 12:15
  • @student Somehow you have to show that in the case $x\not\in \overline{U}$ there exists $0\ne z\in U^\perp$ such that $\langle x,z\rangle \ne 0$ and therefore $x\not\in U^{\perp\perp}$. If you will be able to see $\overline{U}=U^{\perp\perp}$ in some different way, then from this you can deduce Hilbert projection theorem, at least partially, I guess. – Janko Bracic Jan 18 '15 at 12:23
  • How do you get the result you state in the answer from the Hilbert projection theorem? –  Jan 18 '15 at 23:32
  • From the theorem we get a $y \in \overline{U}$ such that $|x-y|$ is minimal. It's not clear why $z = x-y$ should be orthogonal to $\overline{U}$. –  Jan 18 '15 at 23:37
  • I see. If it minimizes it can't be non-orthogonal? –  Jan 18 '15 at 23:39
  • How to finish the argument? We have to show that $\langle x, u\rangle = 0$ for $u \in U^\bot$. But $$ \langle y + z, u\rangle = \langle z,u \rangle$$ and I don't see why that should be zero... –  Jan 18 '15 at 23:44
  • @student I have extended the answer. However I guess that you have already solved the problem. – Janko Bracic Jan 19 '15 at 05:42
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For closed subspaces: $$Z=\overline{Z}\implies\mathcal{H}=Z\oplus_\perp Z^\perp$$

For orthogonal decompositions: $$\mathcal{H}=Z\oplus_\perp Z'\implies Z'=Z^\perp$$

Taking both together one gets: $$\overline{U}=\left(\overline{U}^\perp\right)^\perp=U^{\perp\perp}$$ (This works even for the closure on the span of plain sets.)

freishahiri
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    This requires the Hilbert projection theorem mentioned in the previous answer. – Olivier Bégassat Jan 19 '15 at 00:07
  • @OlivierBégassat: Yep, I know; but I think it is important to mention the order of reasoning here. (Especially since the projection theorem does only half of the story.) – freishahiri Jan 19 '15 at 00:11