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

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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:

$\map \ker {I - P} = \paren {A^\perp}^\perp$

$\Box$

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,:

$\map \ker {I - P} = P \sqbrk H$

$\Box$

Finally, from Range of Orthogonal Projection:

$P \sqbrk H = A$

$\Box$

To conclude:

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

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Corollary $2.9$