Closures and complements in Hilbert spaces

Question

Let $H$ be a Hilbert space, and $V \subseteq H$ a linear subspace. Denoting $$ V^{\perp} := \{w \in H; \text{such that } \langle w,v\rangle = 0, \text{ for all } v \in V\}. $$ Is it clear that $V^{\perp}$ is a closed set? If it is, then is it true that $$ (V^{\perp})^{\perp} = \overline{V}, $$ where $\overline{V}$ denotes the closure of $V$?

Answer 1

The orthogonal complement is closed, and as far as if this is clear, it depends on your definition on clear- it is not hard to prove.

Here's a simple proof: let $(x_j)$ be a sequence in $V^\perp$ converging to $x\in H$. By definition, $\langle x_j, y\rangle=0$ for all $y\in V,$ for each $j$. Using the continuity of the inner product, we conclude by taking the limit that $\langle x,y\rangle =0$, and so $x\in V^\perp.$

Also, as you conjectured in your post, it is true that $(V^\perp)^\perp=\overline{V}.$ Indeed, $V^\perp=\overline{V}^\perp,$ and so $\left(V^\perp\right)^\perp=\left(\overline{V}^\perp\right)^\perp=\overline{V},$ where we have used that $\left(C^\perp\right)^\perp=C$ when $C$ is a closed set.