Double Orthocomplement of Closed Linear Subspace

Theorem
Let $H$ be a Hilbert space. Let $A \subseteq H$ be a closed linear subspace of $H$.

Then:


 * $\paren {A^\perp}^\perp = A$

Proof
Let $I : H \to H$ be the identity operator (viz., $Ih=h$).

Let $P : H \to A$ be the orthogonal projection (see Definition:Orthogonal Projection).

Then $I-P : H \to A^\perp$ is the Orthogonal_Projection_onto_Orthocomplement.

By Kernel of Orthogonal Projection, then, $\ker(I-P)=\paren{A^\perp}^\perp$.

Also, $\ker(I-P)=P(H)$, since $0=(I-P)h$ iff $h=Ph$ and $h=Pf$ (some $f \in H$) implies $h=Ph$, by Orthogonal Projection is Projection.

Conclude that $\paren{A^\perp}^\perp = \ker(I-P) = P(H) = A$.

Here, the last equality uses Range of Orthogonal Projection.