Orthogonal of an Hilbert subspace and density

Question

If $V$ is a subspace of an Hilbert space $H$, I know that the orthogonal of $V$, $V$$^o$, is ($V$closed)$^o$, even if $V$ is not closed.
Does this mean that $V$ is always dense in $V$$^o$?

Thanks!

Answer 1

I'm not following you here. So let $H$ be a Hilbert space and $V$ a (not necessarily closed) subspace, then $V^{\perp}=\left\{w\in H\mid \left\langle v,w\right\rangle=0 \mbox{ for all }v\in V\right\}$ is the orthogonal complement. Using the continuity of the inner-product you can show that $V^{\perp}$ is a closed subspace of $H$. It follows that $V^{\perp\perp}=\overline{V}$ is the closure of $V$. However $V\cap V^{\perp}=\left\{0\right\}$ for any subspace $V$ of $H$, it follows that $V$ is never dense in $V^{\perp}$.

Answer 2

For any subset: $$A\subseteq\mathcal{H}:\quad A\cap A^\perp=\{0\}$$ As well as it is: $$A\subseteq\mathcal{H}:\quad\overline{\langle A\rangle}^\perp=A^\perp=\overline{\langle A^\perp\rangle}$$ Moreover it holds: $$A\subseteq\mathcal{H}:\quad A^{\perp\perp}=\overline{\langle A\rangle}$$