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Bounded Hilbert operators form a C star algebra

Statement

Assume Countable Choice. For every nonzero complex Hilbert space $H$ , $B (H)$ , with the operator norm, composition, identity, and Hilbert adjoint, is a unital C*-algebra and $∥ T^{*} T ∥ = ∥ T ∥^{2}$ .

Facts & Assumptions

[A1]

A Hilbert space is an inner-product space complete for its induced norm, that is, a Banach space for that norm (Hilbert space, Banach space).

[A2]

$B (X, Y)$ is the vector space of bounded linear operators with pointwise operations and the operator norm $∥ T ∥ = \sup {∥ T x ∥ : ∥ x ∥ \leq 1}$ , which is the least bound of $T$ , so that $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A3]

If $Y$ is a Banach space then $B (X, Y)$ is complete for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach).

[A4]

The Hilbert adjoint $T^{*}$ is the unique operator with $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ ; the assignment $T \mapsto T^{*}$ is conjugate-linear, involutive and isometric, satisfies $(S T)^{*} = T^{*} S^{*}$ , and obeys $∥ T^{*} T ∥ = ∥ T ∥^{2}$ (Hilbert-adjoint identities).

[A5]

A unital complex Banach algebra is a nonzero complex Banach algebra with submultiplicative norm and a unit of norm one; a complex C*-algebra is a complex Banach algebra carrying a conjugate-linear involution with $(a^{*})^{*} = a$ , $(a b)^{*} = b^{*} a^{*}$ and $∥ a^{*} a ∥ = ∥ a ∥^{2}$ (Unital Banach algebra, C star algebra).

[A6]

Countable Choice is the hypothesis under which the Hilbert-adjoint and completeness suppliers below are stated (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: A nonzero complex Hilbert space $H$ and operators $S, T \in B (H)$ .

1.1A2algebra

Composition in $B (H)$ is bilinear and associative, and the operator norm is submultiplicative: $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ .

1.2A2algebra

The identity operator $I$ lies in $B (H)$ and satisfies $I T = T I = T$ , and $∥ I ∥ = 1$ : since $H \neq {0}$ every nonzero $x$ has $∥ x / ∥ x ∥ ∥ = 1$ , so the unit-ball supremum defining $∥ I ∥$ equals $1$ . Thus $B (H)$ is a nonzero algebra whose unit $I$ has norm one.

1.3A4A6

The Hilbert adjoint is a map $B (H) \to B (H)$ which is conjugate-linear, involutive and isometric, satisfies $(S T)^{*} = T^{*} S^{*}$ , and satisfies $∥ T^{*} T ∥ = ∥ T ∥^{2}$ for every $T$ .

2.1step 1.1step 1.2A1A3A5

The space $H$ is Banach for its norm by [A1], so $B (H)$ is complete for the operator norm by [A3]; together with the submultiplicativity, the identity of norm one and the nonvanishing just recorded, this makes $B (H)$ a unital complex Banach algebra in the sense of [A5].

3.1step 2.1step 1.3A5∎

The space $B (H)$ , with the operator norm, composition, the identity and the Hilbert adjoint, fulfils every axiom of a unital complex C*-algebra listed in [A5], and the identity $∥ T^{*} T ∥ = ∥ T ∥^{2}$ holds.

Bounded Hilbert operators form a C star algebra

Statement

Facts & Assumptions

Proof

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