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

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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 $J = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ .

1.1

$J$ 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 $J e_{1} = 0$ .

A3A2algebra

1.2

$J$ 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 $[0, + \infty)$ .

A1A2algebra

1.3

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

A2algebra

2.1

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.

step 1.1step 1.2step 1.3

3.1

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.

step 2.1A1A4∎

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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)