LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Positive-state Cauchy–Schwarz and left radical

Form conventions

A sesquilinear form $B : A \times A \to C$ is taken to be linear in its first variable and conjugate-linear in its second. It is Hermitian when $B (a, b) = \overline{B (b, a)}$ for all $a, b \in A$ , positive semidefinite when $B (a, a) \in [0, \infty)$ for every $a \in A$ , and has radical

$rad B : = {a \in A : B (a, b) = 0 for every b \in A} .$

For a Hermitian form this left radical also equals ${a \in A : B (b, a) = 0 for every b \in A}$ .

Statement

Let $A$ be a unital complex star algebra and let $ω$ be a positive normalized state on $A$ (Unital star algebra and positive normalized state). Define $q (a) : = ω (a^{*} a)$ and $N_{ω} : = {a \in A : q (a) = 0}$ . Then

$∣ ω (b^{*} a) ∣^{2} \leq q (a) q (b) (a, b \in A) .$

The form $(a, b)_{ω} : = ω (b^{*} a)$ is Hermitian and positive semidefinite, $N_{ω}$ is its radical and a complex linear subspace, and $N_{ω}$ is a left ideal: $c a \in N_{ω}$ whenever $a \in N_{ω}$ and $c \in A$ .

Facts & Assumptions

Given: A unital complex star algebra $A$ and a positive normalized state $ω : A \to C$ .

[F1]

Unital star algebra and positive normalized state: multiplication is complex-bilinear and associative, $(a b)^{*} = b^{*} a^{*}$ , and $ω$ is complex-linear with $ω (1_{A}) = 1$ and real nonnegative $ω (a^{*} a)$ for every $a \in A$ .

Proof

technique · direct

1.1givenF1algebra

Put $X : = ω (a^{*} b)$ and $Y : = ω (b^{*} a)$ . Positivity makes both $q (a + b) = q (a) + X + Y + q (b)$ and $q (a + i b) = q (a) + i X - i Y + q (b)$ real; their imaginary parts give $Im (X + Y) = 0$ and $Re (X - Y) = 0$ , hence $Y = \overline{X}$ . Thus $(a, b)_{ω} = ω (b^{*} a)$ is Hermitian.

2.1givenF1step 1.1algebra

Suppose $q (b) > 0$ . For every $z \in C$ , positivity and the star-algebra laws give $0 \leq q (a + z b) = q (a) + z X + \overline{z} \overline{X} + ∣ z ∣^{2} q (b)$ . Taking $z = - \overline{X} / q (b)$ yields $0 \leq q (a) - ∣ X ∣^{2} / q (b)$ , so $∣ ω (b^{*} a) ∣^{2} = ∣ Y ∣^{2} = ∣ X ∣^{2} \leq q (a) q (b)$ .

3.1givenF1step 1.1step 2.1algebra

Suppose $q (b) = 0$ . If $X \neq 0$ , taking $z = - t \overline{X}$ for real $t > q (a) / (2 ∣ X ∣^{2})$ makes $q (a + z b) = q (a) - 2 t ∣ X ∣^{2} < 0$ , contradicting positivity. Hence $X = Y = 0$ , so the same Cauchy–Schwarz inequality holds when $q (b) = 0$ ; if instead $q (a) = 0$ and $q (b) > 0$ , step 2.1 forces $Y = 0$ as well.

4.1F1step 2.1step 3.1algebra∎

If $a \in N_{ω}$ , steps 2.1 and 3.1 give $(a, b)_{ω} = 0$ for every $b \in A$ ; conversely $(a, a)_{ω} = 0$ means $a \in N_{ω}$ . Thus $N_{ω}$ is exactly the radical and is a linear subspace. For $c \in A$ , set $d : = c^{*} c a$ . Applying Cauchy–Schwarz to $a, d$ gives $∣ ω (d^{*} a) ∣^{2} \leq q (a) q (d) = 0$ , while $d^{*} a = a^{*} c^{*} c a = (c a)^{*} (c a)$ ; therefore $c a \in N_{ω}$ .

Depends on

Unital star algebra and positive normalized state