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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.

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  • $\begingroup$ Do you know that $\mathcal{H}=V\oplus V^\perp$ if $V$ is a closed subspace of $\mathcal{H}$? $\endgroup$
    – wj32
    Oct 30 '12 at 11:18
  • $\begingroup$ No, we don't know that. I would have to prove that first to use it. $\endgroup$
    – Stefan
    Oct 30 '12 at 11:29
  • $\begingroup$ That statement follows easily from the existence of orthogonal projections. Are you allowed to use orthogonal projections? $\endgroup$
    – wj32
    Oct 30 '12 at 11:33
  • $\begingroup$ 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. $\endgroup$
    – Stefan
    Oct 30 '12 at 11:36
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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 $(U^\perp)^\perp$.

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  • $\begingroup$ 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 $x$? $\endgroup$
    – Stefan
    Oct 30 '12 at 11:31
  • $\begingroup$ Yes, something like that should work. $\endgroup$
    – Berci
    Oct 30 '12 at 11:42
  • $\begingroup$ Or, rather: take an $x\in (U^\perp)^\perp$, and show that $d(x,U)=0$. $\endgroup$
    – Berci
    Oct 30 '12 at 11:47
  • $\begingroup$ How did you finish this? $\endgroup$
    – Berci
    Oct 30 '12 at 17:38

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