Operators on Hilbert Spaces - user web page - AIMS

Section 2.3. Orthogonal complement and direct sum Page 9
Proof.
Let A be a closed subspace of a Hilbert space H. Then by Lemma 2.3.4 (i), A ⊂ A ⊥⊥ .
Let x ∈ A ⊥⊥ . Since A is a closed subspace of H, then by Theorem 2.3.8, H = A ⊕ A ⊥ and for
some y ∈ A and z ∈ A ⊥ we have that x = y + z. Now since A ⊂ A ⊥⊥ , we have that y ∈ A ⊥⊥
and z = x − y ∈ A ⊥⊥ . But z ∈ A ⊥ , so that z = 0. Hence x = y ∈ A and A ⊥⊥ = A.
Conversely let A = A ⊥⊥ . Since A ⊥ ⊂ H, by Lemma 2.3.6 we have that (A ⊥ ) ⊥ = A is closed.