Double Orthocomplement is Closed Linear Span/Corollary

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Corollary to Double Orthocomplement is Closed Linear Span

Let $H$ be a Hilbert space.

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

Then:

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

Proof

Since $A$ is a subspace of $H$, it is closed under linear combination, so:

$\map \span A = A$

We therefore have:

\(\ds \vee A\)

\(=\)

\(\ds \paren {\map \span A}^-\)

Definition of Closed Linear Span

\(\ds \)

\(=\)

\(\ds A^-\)

\(\ds \)

\(=\)

\(\ds A\)

since $A$ is closed, we have $A^- = A$ from Set is Closed iff Equals Topological Closure

while from Double Orthocomplement is Closed Linear Span, we have:

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

So, we obtain:

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

$\blacksquare$

Sources

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