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I'm trying to prove that $S^{\perp\perp} = \overline{S}$, where S is a subspace of a Hilbert Space, where $S^{\perp\perp} = \{f: \langle f,g\rangle = 0, \forall g\in S^\perp\}$ and $\overline{S}$ is the closure under limits of sequences. I can easily show that $f\in \overline{S} \implies f \in S^{\perp\perp}$ but I don't know how to prove the converse. Can someone help point me in the right direction?

Martin Sleziak
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Keith Z
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  • you probably meant a inner product space, note that a Hilbert space is complete, and hence if $S$ is a subspace then $S = \overline{S}$. – reuns Apr 27 '16 at 20:12
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    @user1952009 No, subspaces of Hilbert spaces do not need to be complete. For example $C[0,1]$ is a non-complete subspace of $L^2[0,1]$. This question is posed correctly. – Ian Apr 27 '16 at 20:13
  • @Ian : ok but you didn't convince me, under what definition is $C[0,1]$ a subspace of $L^2[0,1]$ ? whatever it is not so important – reuns Apr 27 '16 at 20:15
  • @user1952009 Every element of $C[0,1]$ is an element of $L^2[0,1]$; sums of elements of $C[0,1]$ are elements of $C[0,1]$; scalar multiples of elements of $C[0,1]$ are elements of $C[0,1]$. One might pedantically disagree with my first statement through something about equivalence classes but it is not really important. – Ian Apr 27 '16 at 20:15
  • @user1952009 $C[0,1]$ is a subspace means that $C[0,1]$ is a vector subspace; we need not allow infinite sums in the definition of a subspace and generally do not. – Ben Grossmann Apr 27 '16 at 20:16
  • yes sorry it's the different between a subspace and a closed subspace, whatever, $S^{\perp}$ is complete ? we agree on that ? – reuns Apr 27 '16 at 20:17
  • @user1952009 yes. – Ben Grossmann Apr 27 '16 at 20:18
  • http://math.stackexchange.com/questions/1043940/double-orthogonal-complement-is-equal-to-topological-closure – Ian Apr 27 '16 at 20:24
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    You know that $S\subset \overline{S}$ and hence $S^{\perp\perp}\subset \overline{S}^{\perp\perp}$; if you know that $S^{\perp\perp} = S$ for $S$ closed then you have the reverse inclusion that you need... Not sure if that helps though. – John Martin Apr 27 '16 at 20:28

1 Answers1

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Ok, so here is the summary.

One should first note the following easy properties:

  1. $S \subseteq S^{\perp \perp}$

  2. $S_1 \subseteq S_2 \implies S_2^{\perp} \subseteq S_1^{\perp}$

  3. For a closed subspace $M$ the orthogonal projection theorem gives that $M^{\perp \perp} = M$.

  4. $S^{\perp}$ is a closed subset for every $S$.

Then one proceeds as follows.

$\Rightarrow)$ Always $S \subseteq \overline{S}$ and by 2. $S^{\perp \perp} \subseteq \overline{S}^{\perp \perp}$ but since $\overline{S}$ is a closed subset 3. gives that $\overline{S}^{\perp \perp} = \overline{S}$. This gives the desired inclusion.

$\Leftarrow)$ Since by 4. $S^{\perp \perp}$ is closed and by 1. $S \subseteq S^{\perp \perp}$ by the monotonicity of topological closure we have that $\overline{S} \subseteq \overline{S^{\perp \perp}} = S^{\perp \perp}$. This gives the opposite inclusion.

Selrach Dunbar
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user1868607
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  • You should have $$\overline {S} \subseteq \overline {\left (S^{\perp}\right )^{\perp}} = \left (S^{\perp}\right )^{\perp}$$ at the very last line. Right? – Anacardium Sep 10 '20 at 19:54
  • Nice! (To remove distractions for anyone referencing this answer in the future, I corrected the typo in the original answer that prompted the two comments before this one.) – Selrach Dunbar Jun 24 '23 at 21:46