# Show that $U \subset V \Leftrightarrow V^\bot \subset U^\bot$ for $U,V$ subspaces in a Hilbertspace

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.

• 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. 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. Oct 30 '12 at 11:36

$U$ is closed subspace of a Hilbert space if and only if $(U^\perp)^\perp = U$.
For any subspace $U$, its closure is just $(U^\perp)^\perp$.
• 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 $x$? Oct 30 '12 at 11:31
• Or, rather: take an $x\in (U^\perp)^\perp$, and show that $d(x,U)=0$. Oct 30 '12 at 11:47