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., $I h = h$).

Also, let $P : H \to A$ be the orthogonal projection.

Then $I - P : H \to A^\perp$ is the Orthogonal Projection onto Orthocomplement.

By Kernel of Orthogonal Projection:
 * $\map \ker {I - P} = \paren {A^\perp}^\perp$

From Orthogonal Projection is Projection:
 * $h \in P \sqbrk H \implies h = P h$

where $P \sqbrk H$ denotes the image of $H$ under $P$.

Also:
 * $0 = \paren {I - P} h \iff h = P h$

Therefore,:
 * $\map \ker {I - P} = P \sqbrk H$

Finally, from Range of Orthogonal Projection:
 * $P \sqbrk H = A$

To conclude:
 * $\paren {A^\perp}^\perp = \map \ker {I - P} = P \sqbrk H = A$