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The Gopalan-Servedio conjecture: is $\sum_i\widehat f(\{i\})=O(\sqrt{\deg f})$ for Boolean functions?

Theorem 1: for every $C>0$ there is a nonconstant $f:\{-1,1\}^n\to\{-1,1\}$ with $\sum_i\widehat f(\{i\})>C\sqrt{\deg f}$ ; equivalently $\sum_i|\widehat f(\{i\})|/\sqrt{\deg f}$ is unbounded over Boolean functions of positive degree, so no constant multiple of the conjectured bound holds. The dimension and copy counts in the construction are finite but not usefully bounded, so no growth rate of the violation in terms of degree is given. The true order between $\sqrt{\deg f}$ and the linear bound $\deg f$ is left open.

Exact FrontierDelta

Prior state unknown→disproved

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Occurred: Sep 26, 2026

Delta type: SOURCE CLAIM

Assumptions: VibeMathed verification: lean-checked. Publication: announcement. AI contribution: ai-discovered. VibeMathed editorial classifications, scores, notes, relations, and dataset structure are CC BY 4.0. Source statements and linked content retain their own rights.

Canonical aliases: The Gopalan-Servedio conjecture: is $\sum_i\widehat f(\{i\})=O(\sqrt{\deg f})$ for Boolean functions? · Gopalan-Servedio square-root degree conjecture

Confidence: Not scored

Registry verification: lean checked · announcement · candidate

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VibeMathed record: The Gopalan-Servedio conjecture: is $\sum_i\widehat f(\{i\})=O(\sqrt{\deg f})$ for Boolean functions?

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No independent mathematician has checked this yet. Checked here: Theorem 1 and the introduction were read against the conjecture as O'Donnell's list states it. Lean: lean/docs/192.md links ComparatorChallenges/SquareRootDegree (theorem OAI.SquareRootDegree.main, solution module OAI.Combinatorics.BooleanFunctions.Main, file exists at the pinned commit), not in formalization.yaml. The statement was read: for every $C>0$ there is a nonconstant $\pm1$ -valued function on a finite sign cube with $C\sqrt{\deg f}$ below the sum of its singleton Fourier coefficients, where degree is the real Fourier degree, and the ratio of the absolute singleton sum to $\sqrt{\deg f}$ is unbounded. That is the headline. Not rebuilt here.