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Weak and strong additivity of orthogonal projections

Statement

Assume Countable Choice. Let $(X, Σ)$ be a measurable space, let $H$ be a complex Hilbert space, and let $E : Σ \to B (H)$ take values in orthogonal projections and satisfy $E (\emptyset) = 0$ and $E (B \cap C) = E (B) E (C)$ for all $B, C \in Σ$ . Then the following two properties are equivalent:

(weak countable additivity) for every pairwise disjoint sequence $(B_{n})_{n \in N}$ in $Σ$ with union $B$ and all $x, y \in H$ , $⟨ E (B) x, y ⟩ = \sum_{n = 0}^{\infty} ⟨ E (B_{n}) x, y ⟩;$
(strong countable additivity) for every such sequence and every $x \in H$ , $E (B) x = \sum_{n = 0}^{\infty} E (B_{n}) x$ with the series converging in norm.

Facts & Assumptions

[A1]

An orthogonal projection value $P = E (B)$ satisfies $P^{2} = P = P^{*}$ and $⟨ P x, x ⟩ = ∥ P x ∥^{2} \geq 0$ , and it is contractive, $∥ P u ∥ \leq ∥ u ∥$ for all $u$ (Projection valued measure, Hilbert projections are linear, self-adjoint and contractive).

[A2]

The Hilbert adjoint satisfies $⟨ S u, v ⟩ = ⟨ u, S^{*} v ⟩$ for all $u, v$ , and conjugate symmetry gives $⟨ S^{*} u, v ⟩ = \overline{⟨ v, S^{*} u ⟩} = \overline{⟨ S v, u ⟩} = ⟨ u, S v ⟩$ ; for a self-adjoint $S$ the two pairings with $S$ coincide (The Hilbert-space adjoint of a bounded operator, Real and complex inner-product spaces and their induced length).

[A3]

Multiplicativity on intersections and the empty-set value hold: $E (B \cap C) = E (B) E (C)$ and $E (\emptyset) = 0$ ; if $C \subseteq B$ then $E (C) = E (B \cap C) = E (B) E (C) = E (C) E (B)$ (Projection valued measure).

[A4]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A5]

The pairing is linear in the first argument and conjugate-linear in the second, and $∥ u ∥^{2} = ⟨ u, u ⟩$ (Real and complex inner-product spaces and their induced length).

[A6]

Countable Choice is the declared standing hypothesis of this block of the page (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A measurable space $(X, Σ)$ , a complex Hilbert space $H$ , a map $E$ with orthogonal projection values, $E (\emptyset) = 0$ , $E (B \cap C) = E (B) E (C)$ , a pairwise disjoint sequence $(B_{n})$ in $Σ$ with union $B$ , vectors $x, y \in H$ , and partial sums $Q_{N} : = \sum_{n \leq N} E (B_{n})$ .

1.1A1A4A5

Strong implies weak: for every $N$ the difference of the two sides of the weak identity is $⟨ E (B) x - Q_{N} x, y ⟩$ , so by Cauchy–Schwarz $∣ ⟨ E (B) x, y ⟩ - \sum_{n \leq N} ⟨ E (B_{n}) x, y ⟩ ∣ \leq ∥ E (B) x - Q_{N} x ∥ ∥ y ∥$ , which tends to $0$ because $E (B) x - Q_{N} x \to 0$ in norm by hypothesis.

1.2A1A2A3A5

Weak implies strong: expanding $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x - Q_{N} x, E (B) x - Q_{N} x ⟩$ and using $P^{2} = P = P^{*}$ for each projection value gives $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x, x ⟩ - 2 \sum_{n \leq N} ⟨ E (B_{n}) x, x ⟩ + \sum_{n, m \leq N} ⟨ E (B_{m} \cap B_{n}) x, x ⟩$ , because $⟨ E (B) x, E (B) x ⟩ = ⟨ E (B)^{2} x, x ⟩$ , because $⟨ E (B_{n}) x, E (B) x ⟩ = ⟨ E (B) E (B_{n}) x, x ⟩$ with $B_{n} \subseteq B$ , and because $⟨ E (B_{n}) x, E (B_{m}) x ⟩ = ⟨ E (B_{m}) E (B_{n}) x, x ⟩ = ⟨ E (B_{m} \cap B_{n}) x, x ⟩$ .

2.1step 1.2A3A5

In the last sum the off-diagonal terms are $⟨ E (\emptyset) x, x ⟩ = 0$ and the diagonal terms are $⟨ E (B_{n}) x, x ⟩$ , so $∥ E (B) x - Q_{N} x ∥^{2} = ⟨ E (B) x, x ⟩ - \sum_{n \leq N} ⟨ E (B_{n}) x, x ⟩$ , which tends to $0$ by weak countable additivity applied with $y = x$ ; hence $Q_{N} x \to E (B) x$ in norm.

3.1step 1.1step 2.1A6∎

Both implications hold for an arbitrary pairwise disjoint sequence, so weak and strong countable additivity of $E$ are equivalent.

Weak and strong additivity of orthogonal projections

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources