0
$\begingroup$

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.

  • 0
    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.2012-10-30

1 Answers 1

1

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 $(U^\perp)^\perp$.

  • 0
    How did you finish this?2012-10-30