bibkey: verjovsky2026mobiussmoothing authors: Alberto Verjovsky year: 2026 title: “How Random Is the Möbius Function? Smoothing, Probability, and the Riemann Hypothesis” doi: null url: https://arxiv.org/abs/2607.25002v2 claim: “Theorem 3.1 states the Möbius Laplace Lp criterion for every 1≤p<2; Proposition 2.2 gives the zero-free half-plane from one exponent. The actual FIB multiplier is an invertible dilation filter, not an unconditional membership proof.” strata_touched: [] license: citation-only triage: anchor
The existing smoothed Möbius criterion and its FIB filter
The primary version is arXiv:2607.25002v2, submitted 14 August 2026, 28 pages. The inspected original scope is Sections 2–3: Propositions 2.1–2.2, Theorem 3.1 and Remarks 3.2–3.3. The selected proofs were read; the rest of the manuscript and its probabilistic and dynamical interpretations are not independently audited. No Lean verification or original FIB criterion is asserted.
For
Proposition 2.1 gives the classical Mellin transform for . Proposition 2.2 states that one bound, , implies on . Theorem 3.1 states
Remark 3.2 explicitly excludes an endpoint assertion at . A fixed exponent below two does not give the full criterion, and a random coefficient statement is not a statement about the deterministic Möbius sequence.
The actual FIB coefficients satisfy , with and the already established absolutely summable inverse . Define . Absolute convergence at every and the existing convolution identities give the two dilation formulas
These are applications of the established filter, not new analytic criteria. The standard dilation norm and the triangle inequality give, whenever the right side is finite,
Both filter constants are finite by the existing FIB inverse budget. Thus the source supplies a reusable alternative test for the same actual arithmetic input. It does not supply the still missing membership for all exponents below two, a signed Robin bound, or a geometric rule that forces either. This adaptation has no new retained Lean wrapper.