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Section 2.3. Orthogonal complement and direct sum Page 7
Theorem 2.3.2 ([Kre78], 3·3-2 Lemma (Orthogonality)) In Theorem 2.3.1, let M be a complete
subspace Y and x ∈ X fixed. Then z = x − y is orthogonal to Y.
Proof.
Suppose by contradiction that z is not orthogonal to Y. Then there would be a y1 ∈ Y such that
y1 = 0 since otherwise 〈z, y1〉 = 0.
Now for any scalar α,
z − αy1 2 = 〈z − αy1, z − αy1〉
〈z, y1〉 = β = 0. (2.12)
= 〈z, z〉 − ¯α〈z, y1〉 − α[〈y1, z〉 − ¯α〈y1, y1〉]
= 〈z, z〉 − ¯αβ − α[ ¯ β − ¯α〈y1, y1〉]. (2.13)
Let us choose α = β
〈y1,y1〉 , so that ¯α = ¯ β
〈y1,y1〉 . Then the expression ¯ β − ¯α〈y1, y1〉 becomes zero.
From Theorem 2.3.1, we have that z = x − y = δ for some y ∈ Y, so that equation (2.13)
becomes
z − αy1 2 = z 2 − |β|2
〈y1, y1〉 < δ2 .
But this is impossible since z − αy1 = x − y2, where y2 = y + αy1 ∈ Y so that z − αy1 ≥ δ,
by definition of δ. Hence equation (2.12) cannot hold, and the lemma is proved.
Definition 2.3.3 ([Erd80], Orthogonal complement)
Given a subspace M of H, the orthogonal complement M ⊥ is defined by
M ⊥ = {x ∈ X : 〈x, m〉 = 0 for all m ∈ M } . (2.14)
Lemma 2.3.4 ([Erd80], 1·9 Lemma) For subsets A and B of a Hilbert space H, if B ⊃ A, then
(i) A ⊂ A ⊥⊥ ,
(ii) B ⊥ ⊂ A ⊥ ,
(iii) A ⊥⊥⊥ = A ⊥ .
Proof.
(i) Let a ∈ A. Then a ⊥ A ⊥ and by Definition 2.3.3 a ∈ A ⊥⊥ . Hence A ⊂ A ⊥⊥ .
(ii) Let y ∈ B ⊥ and a ∈ A. Then A ⊂ B implies that a ∈ B. Thus y ⊥ a. Hence y ∈ A ⊥ .
(iii) Applying (i) to A ⊥ gives A ⊥ ⊂ A ⊥⊥⊥ . Also from (i), A ⊂ A ⊥⊥ and applying (ii) gives
A ⊥ ⊃ A ⊥⊥⊥ . Hence A ⊥⊥⊥ = A ⊥ .