CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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.

Self adjointness cannot be dropped from the order calculus

Statement refuted

Assume Countable Choice. For every bounded operator whose spectrum is a subset of $[0, + \infty)$ , the quadratic form is nonnegative; equivalently, spectral nonnegativity alone characterises positivity and self-adjointness may be dropped from the order calculus.

Facts & Assumptions

[A1]

Positivity of an operator is the quadratic-form condition that $⟨ T x, x ⟩$ is a real number in $[0, + \infty)$ for every $x$ , and the order relation is defined only for self-adjoint pairs (Self-adjoint, positive, unitary and normal operators, Order on bounded self adjoint operators).

[A2]

The adjoint is characterised by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ (The Hilbert-space adjoint of a bounded operator).

[A3]

$λ \in ρ (T)$ exactly when $λ I - T$ is bijective with bounded inverse (Spectrum and resolvent of a bounded operator).

[A4]

Countable Choice is the declared choice hypothesis of this pair's order calculus (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Counterexample

technique · direct

Given: The two-dimensional complex inner-product space with orthonormal basis $(e_{1}, e_{2})$ and the Jordan nilpotent $J = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ .

1.1A3A2algebra

$J$ has real spectrum ${0} \subseteq [0, + \infty)$ : $J^{2} = 0$ gives the inverse $z^{- 1} (I + z^{- 1} J)$ of $z I - J$ for every $z \neq 0$ , while $z = 0$ is a spectral value because $J e_{1} = 0$ .

1.2A1A2algebra

$J$ is not positive: for $x : = \frac{1}{\sqrt{2}} (e_{1} + i e_{2})$ one has $J x = \frac{i}{\sqrt{2}} e_{1}$ and hence $⟨ J x, x ⟩ = \frac{i}{\sqrt{2}} \cdot \overline{\frac{1}{\sqrt{2}}} = \frac{i}{2}$ , which is not a real number, while positivity of a bounded operator requires the value of the quadratic form at every vector to be a real number in $[0, + \infty)$ .

1.3A2algebra

$J$ is not self-adjoint: the defining pairing on the standard orthonormal basis gives $J^{*} e_{1} = e_{2}$ and $J^{*} e_{2} = 0$ , so $J^{*} = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}) \neq J$ .

2.1step 1.1step 1.2step 1.3

The witness $J$ therefore has nonnegative real spectrum but is neither positive nor self-adjoint, so nonnegativity of the spectrum alone does not give the quadratic-form inequalities of the order calculus.

3.1step 2.1A1A4∎

The statement is refuted: $J$ satisfies its spectral antecedent but fails its quadratic-form conclusion, so spectral nonnegativity alone cannot extend the self-adjoint order definition to all bounded operators.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3–5.4, printed pp.235–255 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–13 (standard reference, not scraped)