LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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The norm of a positive element is the supremum of its state values

Statement

Assume the Axiom of Choice. Let $A$ be a C*-algebra and let $a \in A$ with $a \geq 0$ (Self-adjoint positive unitary and normal elements, Positive calculus and order estimates in a C star algebra). Then $∥ a ∥ = \sup {ω (a) : ω a state of A}$ (States and positive functionals on a C star algebra). For the zero algebra the supremum of the empty subset of $[0, \infty)$ is understood as $0$ . No state-value formula for arbitrary non-self-adjoint elements is asserted.

Facts & Assumptions

Given: AC; a C*-algebra $A$ with ambient unital C*-algebra $B$ ( $B = A$ if $A$ is unital, $B = A^{+}$ otherwise); an element $a \in A$ with $a \geq 0$ .

[F1]

Positivity and order toolkit: $a \geq 0$ has $σ (a) \subseteq [0, \infty)$ and $∥ a ∥ = \max σ (a)$ ; for self-adjoint $h$ and continuous $f$ , the calculus element $f (h) \in C^{*} (1, h) \subseteq B$ satisfies $∥ f (h) ∥ = \sup_{σ (h)} ∣ f ∣$ and $σ (f (h)) = f (σ (h))$ ; conjugation preserves positivity; the unitization is a unital C*-algebra containing $A$ (Positive calculus and order estimates in a C star algebra, Minimal C star unitization).

[F2]

A functional $ω$ on $A$ is positive when $ω (x^{*} x) \geq 0$ for all $x$ , and a state when moreover $∥ ω ∥ = 1$ ; every state satisfies $∣ ω (x) ∣ \leq ∥ x ∥$ (States and positive functionals on a C star algebra).

[F3]

Under AC every bounded linear functional on a subspace of a normed space has a norm-preserving extension (A bounded complex linear functional on a subspace of a complex normed space extends with the same norm).

Proof

technique · direct

Given: AC, a C*-algebra $A$ with ambient unital C*-algebra $B$ , and $a \in A$ with $a \geq 0$ ; for the main argument assume $a \neq 0$ .

1.1F1F2F3

Put $λ_{0} : = ∥ a ∥ = \max σ (a)$ , so $λ_{0} \in σ (a)$ by [F1], and consider the closed unital $*$ -subalgebra $C^{*} (1, a) \subseteq B$ . Evaluation at $λ_{0}$ , $e v (g) : = g (λ_{0})$ for $g \in C^{*} (1, a)$ identified via the calculus with continuous functions on $σ (a)$ , is a linear functional with $e v (1) = 1$ , $e v (a) = λ_{0} = ∥ a ∥$ and $∣ e v (g) ∣ \leq ∥ g ∥$ , because $∥ g ∥ = \sup_{σ (a)} ∣ g ∣$ by [F1]; hence $∥ e v ∥ = 1 = e v (1)$ . By [F3] it extends to a bounded linear functional $f$ on $B$ with $∥ f ∥ = 1$ . Also, for every state $ω$ and every $x$ , $∣ ω (x) ∣ \leq ∥ x ∥$ by [F2], so $ω (a) \leq ∥ a ∥$ for the positive element $a$ ; this will give the upper bound.

2.1F1step 1.1

The functional $f$ is positive on $B$ . Let $h \in B$ be self-adjoint. For real $t$ the element $e^{i t h}$ has modulus one in the calculus, $∣ e^{i t h} ∣ = 1$ on $σ (h)$ , so $σ (e^{i t h}) \subseteq T$ and $∥ e^{i t h} ∥ = 1$ by [F1]; hence $∣ f (e^{i t h}) ∣ \leq 1$ . Writing $f (h) = x + i y$ with $x, y \in R$ , linearity and the norm convergence of the exponential series give $f (e^{i t h}) = 1 + i t f (h) + O (t^{2})$ as $t \to 0$ ; the real part is $1 - t y + O (t^{2})$ , and $1 - t y + O (t^{2}) \leq ∣ f (e^{i t h}) ∣ \leq 1$ yields $y = 0$ after letting $t \to 0$ through positive and negative values. Thus $f$ is real on self-adjoint elements. If now $0 \leq b \leq 1$ in $B$ , then $∥ 1 - b ∥ \leq 1$ by [F1], so $∣ 1 - f (b) ∣ = ∣ f (1 - b) ∣ \leq 1$ , and since $f (b)$ is real this gives $f (b) \geq 0$ ; rescaling any positive $b \neq 0$ to $b / ∥ b ∥$ shows $f (b) \geq 0$ , and $f (0) = 0$ .

3.1F2step 1.1step 2.1

Restrict $f$ to $A$ : the restriction $ω : = f ∣_{A}$ is positive because $x^{*} x \geq 0$ in $A$ and $f$ is positive on $B$ by step 2.1; and $∥ ω ∥ \leq ∥ f ∥ = 1$ while $∥ ω ∥ \geq ∣ f (a) ∣ / ∥ a ∥ = 1$ because $a \neq 0$ and $f (a) = e v (a) = ∥ a ∥$ . Hence $ω$ is a state of $A$ with $ω (a) = ∥ a ∥$ .

4.1step 1.1step 3.1

Therefore $\sup {ω (a) : ω$ a state $} \geq ω (a) = ∥ a ∥$ , while step 1.1 gives the reverse inequality for every state, so the supremum equals $∥ a ∥$ . If $a = 0$ and $A \neq {0}$ , choose $x \neq 0$ and apply step 3.1 to $x^{*} x \neq 0$ (using $∥ x^{*} x ∥ = ∥ x ∥^{2} \neq 0$ ) to obtain a state, and every state vanishes at $0$ , so the supremum is $0 = ∥ 0 ∥$ . If $A = {0}$ the set of state values of $0$ is empty and the stated empty-supremum convention gives $0 = ∥ 0 ∥$ .

5.1givenF3∎

The Axiom of Choice is used for the norm-preserving Hahn–Banach extension of step 1.1 and is inherited from the calculus and unitization suppliers; the positivity and supremum arguments use no further choice (The Axiom of Choice).

Depends on

Used by

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Dependency tree · two levels

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Sources

Bachir Bekka and Pierre de la Harpe, Unitary Representations of Groups, Duals, and Characters (arXiv:1912.07262v1, 16 December 2019) (standard reference, not scraped)
Bachir Bekka, Pierre de la Harpe and Alain Valette, Kazhdan's Property (T) (Cambridge University Press 2008; author-hosted complete text) (standard reference, not scraped)