Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(\mathcal H, \langle\cdot,\cdot\rangle)$ be a Hilbertspace, $U,V \subset \mathcal H$ are closed subspaces. I want to show $$U \subset V \Leftrightarrow V^\bot \subset U^\bot$$ $\Rightarrow$ is easy to show, no problems with that. But I am stuck at $\Leftarrow $. Since $\mathcal H$ can be any Hilbertspace, it doesn't have to be of finite dimension, so usually $(U^\bot)^\bot \neq U$. I am pretty sure that I have to use the fact that $U$ and $V$ are closed subspaces, but I am not sure how.

I tried $x \in U \Rightarrow \langle x,u \rangle = 0 \forall u \in U^\bot \Rightarrow \langle x,v \rangle = 0 \forall v \in V^\bot$. But as $(V^\bot)^\bot \neq V$, I can't conclude $x \in V$.

I would appreciate hints more than answers, as I want to solve this myself.

share|improve this question
    
Do you know that $\mathcal{H}=V\oplus V^\perp$ if $V$ is a closed subspace of $\mathcal{H}$? –  wj32 Oct 30 '12 at 11:18
    
No, we don't know that. I would have to prove that first to use it. –  Stefan Oct 30 '12 at 11:29
    
That statement follows easily from the existence of orthogonal projections. Are you allowed to use orthogonal projections? –  wj32 Oct 30 '12 at 11:33
    
A following question say: "We say a projection $P$ is an orthogonal projection if and only if $ker(P) = ran(P)^\bot$". But this is just a classification, nothing about existance. –  Stefan Oct 30 '12 at 11:36

1 Answer 1

up vote 1 down vote accepted

Try to prove this one:

$U$ is closed subspace of a Hilbert space if and only if $(U^\perp)^\perp = U$.

Or, rather

For any subspace $U$, its closure is just $(U^\perp)^\perp$.

share|improve this answer
    
Thanks, that seems like a good point to start. Should I try to take $x \in (U^\bot)^\bot \backslash U$ and try to find a sequence(in $U$) that converges to $x$? –  Stefan Oct 30 '12 at 11:31
    
Yes, something like that should work. –  Berci Oct 30 '12 at 11:42
    
Or, rather: take an $x\in (U^\perp)^\perp$, and show that $d(x,U)=0$. –  Berci Oct 30 '12 at 11:47
    
How did you finish this? –  Berci Oct 30 '12 at 17:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.