Supporting affine functionals for Entanglement of Formation
resolvedconfidence 70%
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Precise statement
The paper disproves the assumption that finite-dimensionality and the convex-roof structure of Entanglement of Formation guarantee a global supporting affine functional at every bipartite state. It gives an explicit degenerate two-qubit state $\rho$ for which no Hermitian $\Lambda_\rho$ satisfies both $E_F(\rho)=\mathrm{Tr}\Lambda_\rho\rho$ and $E_F(\sigma)\geq\mathrm{Tr}\Lambda_\rho\sigma$ for every state $\sigma$. The construction uses the equivalence between existence of such a functional and Lipschitz lower semicontinuity of $E_F$, together with Wootters’ formula.
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fidelity, correctness, priority, or novelty.
What AI did
Claude Fable 5
From the abstract: "We use Wootters' formula and the help of Claude Fable 5 to find a state $\rho$ of the system $AB$ for which the latter property does not hold." The authors had already reduced the existence of a global supporting affine functional to Lipschitz lower semicontinuity, and knew in principle that Wootters' two-qubit formula could yield a counterexample; what they did not have was a practical way to construct one. They report the model found such a state very quickly, and it is the state of Proposition 3.
AI-co-developed rather than AI-discovered, and the distinction is the tier's definition rather than a judgement call: this is a subproblem the authors formulated, inside a proof they set up, which the model solved. The surrounding theory - the equivalence criterion, the existence conditions, the Lipschitz bounds in finite and infinite dimensions - is the authors'.
Site-confirmed: the counterexample was re-derived here on 28 August 2026, in exact arithmetic, from the entry's statement rather than the paper's method. Wootters' concurrence was implemented from scratch and evaluated symbolically on $\rho=\frac12|\Phi^+\rangle\langle\Phi^+|+\frac12|01\rangle\langle01|$ and on $\rho_t$.
Results. $C(\rho)=1/2$ exactly. $C(\rho_t)$ has exact closed form at each rational $t$ - at $t=10^{-6}$ it is $999999/2000000-3\sqrt{222222}/10^6$, which differs from $\frac12-\sqrt{2t}$ by $-4.99\times10^{-7}$, the $O(t)$ term with coefficient about $-\frac12$, confirming the paper's expansion. The difference quotient $[E_F(\rho)-E_F(\rho_t)]/t$ evaluates to $14.81$, $154.7$, $1550.9$, $15512.7$ and $155131.3$ at $t=10^{-2},10^{-4},10^{-6},10^{-8},10^{-10}$: a factor of ten per two decades, so it grows like $t^{-1/2}$ and diverges. $E_F$ is therefore not Lipschitz lower semicontinuous at $\rho$, which by the paper's own criterion is exactly the failure claimed.
What this does not establish. The equivalence between a global supporting affine functional and Lipschitz lower semicontinuity is the paper's, and was not checked here; nor were the further existence conditions or the infinite-dimensional bounds. The paper is an unrefereed preprint (v1, 27 August 2026, quant-ph) with no independent review.
For two qubits, the paper considers
$\rho=\frac12|\Phi^+\rangle\langle\Phi^+|+\frac12|01\rangle\langle01|$,
where $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. This rank-2 state has no global supporting affine functional for Entanglement of Formation.
Setting $\rho_t=(1-t)\rho+t|10\rangle\langle10|$, Wootters’ formula gives
$C(\rho_t)=\frac12-\sqrt{2t}+O(t)$.
Consequently,
$\lim_{t\to0^+}[E_F(\rho)-E_F(\rho_t)]/t=+\infty$,
so $E_F$ is not Lipschitz lower semicontinuous at $\rho$. By the paper’s criterion, no global supporting affine functional exists there. Thus the claimed universal existence fails even for two qubits, although it remains true for nondegenerate finite-dimensional states.
Known method families
construction (source-reported)
Source-reported tools: construction.
Independent: unknown · difference confidence: 0
What remains uncertain
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VibeMath has not independently audited the mathematical statement, proof, or novelty claim.