Double Orthocomplement is Closed Linear Span

Theorem

Let $H$ be a Hilbert space.

Let $A \subseteq H$ be a subset of $H$.

Then the following identity holds:

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

Here $A^\perp$ denotes orthocomplementation, and $\vee A$ denotes the closed linear span.

Corollary

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

Then:

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

Proof

From Orthocomplement is Closed Linear Subspace:

$\paren {A^\perp}^\perp$ is a closed linear subspace of $H$.

Also:

$\paren {A^\perp}^\perp \supseteq A$

Indeed, for each $a \in A$, we have:

$\forall a' \in A^\perp : \innerprod a {a'} = 0$

by definition of $A^\perp$.

This also means that:

$a \in \paren{A^\perp}^\perp$

by definition of $\paren{A^\perp}^\perp$.

Hence:

$\vee A \subseteq \paren{A^\perp}^\perp$

$\Box$

For the converse direction, note:

$A \subseteq \vee A$

Now apply Orthocomplement Reverses Subset twice:

$\paren {A^\perp}^\perp \subseteq \paren {\paren {\vee A}^\perp}^\perp$

By Double Orthocomplement of Closed Linear Subspace:

$\paren{A^\perp}^\perp \subseteq \vee A$

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Corollary $2.10$