Show that $S^{\perp \perp} \equiv (S^\perp)^\perp$ is the closure of $S$.

Asked Sep 11 '14 at 04:19

Active Sep 11 '14 at 04:19

Viewed 1,600 times

Suppose $S$ is a (not neccessarily closed) subspace of a Hilbert space $H$. Show that $S^{\perp \perp} \equiv (S^\perp)^\perp$ is the closure of $S$.

I know that if $X\in H$, that $X^\perp$ is a closed subspace, but not really sure where to go from there?

asked Sep 11 '14 at 04:19

user3784030

So how is $S^{\perp}$ related to $S$? – almagest Sep 11 '14 at 04:39
1

$S^\perp$ is the set of all vectors $y$ such that $x\perp y$ (or $<x,y>=0$) for all $x\in S$. – user3784030 Sep 11 '14 at 05:04
Just so. So what can you say about $(S^{\perp})^{\perp}$? – almagest Sep 11 '14 at 05:06
@user3784030 This post can be helpful. – Pedro Sep 11 '14 at 05:06
1

So then $(S^\perp )^\perp$ would be the set of all $z\in S^\perp$ such that $z\perp y \implies <z,y>=0$ and as $<z,y>=<x,y>$ then $z=x$? – user3784030 Sep 11 '14 at 05:08
You need to know some facts about Hilbert Space; otherwise I'm afraid the answer to your question is going to involve too much. Do you know that $L\oplus L^{\perp}=X$ if $L$ is a closed linear subspace of the Hilbert space $X$? Or, equivalently, do you know that every orthonormal subset of $X$ can be completed to an orthonormal basis? – Disintegrating By Parts Sep 11 '14 at 12:34
1

See also: How can I show $U^{\bot \bot}\subseteq \overline{U}$? – Martin Sleziak Jan 18 '15 at 14:30

0 Answers0